1)What is the output of following program?
|
1 2 3 4 5 |
main() { int x = -10; printf("%dn", ~x+1); } |
a)-11
b)-9
c)10
d)Size of the integer is required to
Ans:c
2)What is the output of following program?
|
1 2 3 4 |
main() { printf("%cn", '1' + 1); } |
a)ASCII value of ‘1’ is required to
b)2
c)50
d)Syntax Error
Ans: b
3)What is the output of following program?
|
1 2 3 4 5 6 |
main() { char x = 0x80; int y = x; printf("%dn", y); } |
a)0
b)128
c)-128
d)Unpredictable
Ans: c
4)What is the output of following program in a little endian machine?
|
1 2 3 4 5 6 |
main() { int x = 0xdeadbeef; char y = x; printf("%xn", y); } |
a)de
b)ef
c)0
d)Unpredictable
Ans: b
5)What is the output of following program?
|
1 2 3 4 5 6 7 8 |
main() { char y[10] = "abcdefghi"; char *p = y; p = p + 9; printf("%cn",*p); } |
a)i
b)Program will have runtime error
c)Unpredictable
d)No visible output
Ans: d
6)What is the output of following program?
|
1 2 3 4 5 6 7 8 |
main() { int y[2][2] = { {1,2}, {3,4} }; int *p = &y[1]; p = p + 1; printf("%dn",*p); } |
a)4
b)3
c)The program doesn’t compile
d)Output is unpredicatable
Ans: a
7)
|
1 2 3 4 5 6 |
typedef struct date { int day; int month; int year; } MYDATE; |
Given the above the type definition, which of the following statements are correct declarations?
1. date x;
2. MYDATE x;
3. struct date x;
4. struct MYDATE x;
a)2 & 3
b)1 & 2
c)3 & 4
d)1 & 4
Ans: a
8)
|
1 2 3 4 |
main() { printf("%dn",f1(10,10)); } |
Assuming f1 is defined in another file, which of the following statements are acceptable by compiler as function prototype of f1.
1. extern int f1();
2. extern f1(int, int);
3. extern f1();
4. extern int f1(int, int);
a)All of them
b)4
c)3
d)1 & 4
Ans:a
9)What is output of following program?
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 |
int y = 10; main() { int x = 10; int y = 20; x = x + y; if (x >= 30) { int y = 30; x = x + y; } else { int y = 40; x = x + y; } printf("%dn", x); } |
a)40
b)50
c)60
d)70
Ans: c
10)What is the output of following program?
|
1 2 3 4 5 6 7 8 9 |
main() { int x = 10; int y = 20; if (x & y) if(x | y) x = 20; else x = 30; printf("%dn", x); } |
a)10
b)20
c)30
d)Unpredicatable
Ans: a
11)What is the output of following program?
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 |
main() { int x = 1; int y = 2; switch(x | y) { case 1: x = 2; case 2: x = 3; case 3: x = 4; case 4: x = 5; default:x = 6; } printf("%dn", x); } |
a)6
b)4
c)5
d)3
Ans: a
12)
|
1 2 3 |
#define MAKEDOUBLE(x) x+x a = MAKEDOUBLE(4) * 3; VALUE OF a |
a)24
b)16
c)8
d)None
Ans: b
13)What happens when the following code segment is executed?
|
1 2 3 4 5 |
main() { int i = 8; i = i++; } |
a)i takes value 8
b)i takes value 9
c)response is compiler dependent and hence
unpredictable
unpredictable
d)syntax is invalid
Ans:c
14)Consider the following C code segment:
|
1 2 3 4 5 6 7 8 9 10 |
void function(int *ip) { static int myValue = 100; ip = &myValue; } main() { int *myPointer; function(myPointer); } |
What happens when the following call is made:
a)myPointer value remains unchanged
b)myPointer gets modified to point to address of myValue
c)Response is compiler dependent
d)Code does not compile
Ans: a
15)What is the output of the following program?
|
1 2 3 4 5 |
int x[100]; main() { printf("%dn",x[99]); } |
a)Unpredicatable
b)Runtime error
c)0
d)99
Ans: c
16)What is the output of the following program?
|
1 2 3 4 |
main() { printf("%dn", 100 / 10 / 10); } |
a)1
b)100
c)10
d)0
Ans: a
17)
|
1 2 3 4 5 6 7 8 |
main() { int a=5; if (a=1) { printf("%d", a); } } |
In the above code
a)The printf statement will never get executed
b)The printf statement will always get executed and the value of a will be printed as 5
c)The program will encounter syntax error
d)The printf statement will always get executed and the value of a will be printed as 1
Ans: d
18)
|
1 2 3 4 5 6 7 8 |
int main() { int a[10]; int i; for (i=0; i<10; i++) a[i] = i; printf("%d", a[-1]); } |
The above program
a)Will encounter a compilation error
b)Will encounter segmentation violation when run
c)Will have unpredictable behavior
d)Will get into infinite loop
Ans: c
19)What is the output of the following program?
|
1 2 3 4 5 6 7 |
main() { int a=3, b=5, c=1; int x=8; x = a < (b < c); printf("%d", x); } |
a)1
b)8
c)Will encounter run time error
d)0
Ans: d
20)What is the octal equivalent of the binary number 10111000
a)b8
b)560
c)270
d)None of the above
Ans:c
21)How many times the “Hello world” is printed in following program.?
|
1 2 3 4 5 6 7 8 |
main() { unsigned int i = 5; while (--i >= 0) { printf("Hello Worldn"); } } |
a)5
b)6
c)Infinite
d)Program will not compile
Ans: c
22)What would be the Output of the following program?
|
1 2 3 4 5 6 7 8 9 10 11 12 |
main() { extern int change(float); int x; x = change(5.42); printf("%dn", x); } int change(float x) { return ((int)x); } <span style="line-height: 1.5;"> </span> |
a)0
b)5
c)5.42
d)None of the above
Ans: b
23)Binary equivalent of 5.375 is
a)101.101110111
b)101.011
c)101011
d)None of the above
Ans: b
24)The preprocessor can trap simple errors like missing declarations, nested comments or mismatch of braces
a)True
b)False
c)Depends on the compiler
d)None of the above
Ans: b
25)What is the output of the following program.
|
1 2 3 4 5 |
main() { char str[7] = "Strings"; printf("%s", str); } |
a)Strings
b)Not predicatable
c)Compilation error
d)None of the above
Ans: b
26)What is the output of following program?
|
1 2 3 4 5 6 |
main() { int y = 100; const int x = y; printf("%dn", x); } |
a)100
b)Garbage value
c)Error
d)0
Ans: a
27)How many times “Hello World” will be printed in the following program?
|
1 2 3 4 5 6 7 8 |
main() { int i, j, k; for(i = 1; i <= 2; ++i) for(j = 1; j <= i; ++j) for(k = i; k <= j; ++k) printf("Hello Worldn"); } |
a)2
b)8
c)16
d)4
Ans: a
28)The keyword ‘static’ when used with a variable defined outside of any function in C means:
a)The variable’s value cannot be changed by any function in the program
b)The variable is stored in non-volatile memory that retains value across power cycles
c)The variable is accessible only to functions defined in the same file
d)Can be used only by functions that are statically linked
Ans: c
29)The keyword ‘volatile’ when used with a variable in C means:
a)The variable is stored in volatile memory (RAM) that does not retain value across a power cycle
b)The value of the variable can change asynchronous to program execution
c)The variable’s value is not initialized by the start-up code
d)Suggests that the compiler stores the value in registers
Ans: b
30)What does the value stored in a function pointer variable contain?
a)Address of the Variable that contains a function address
b)Starting address of the function
c)Address of the link table
d)Stack Frame of the function
Ans: b
31)In a C file, the following line is present:
#define FAST_COMPUTE
When trying to compile the program,
a)The C preprocessor gives an error
b)The C compiler gives an error
c)The program compiles normally
d)The assembler gives an error
Ans: c
32)In a C Program, stack is used for
a)Storing local variables only
b)Storing local variables and passing parameters
c)Passing parameters only
d)Storing local static variables
Ans: b
33)Memory leak in a program is mostly due to
a)RAM becoming bad during program execution
b)linker defects
c)Improper use of dynamic memory in the
code or the libraries used
code or the libraries used
d)Compiler Defects
Ans: c
34)
|
1 2 3 4 5 6 7 8 |
int main() { int i = 0; int sizeint; sizeint = sizeof i++; printf("%dn", i); return 0; } |
Considering that integer requires 4 bytes of memory, the value of i printed by the above program is:
a)1
b)4
c)0
d)The above program encounters a compilation error
Ans: c
35)
|
1 2 3 4 5 6 7 8 |
int main() { int i = 0; int sizeint; sizeint = sizeof(i++); printf("%dn", sizeint); retrun 0; } |
a)1
b)2
c)4
d)The output will be machine dependent
Ans: d
36)
|
1 2 3 4 5 6 7 8 |
int main() { int j = 6; int i; i = 5, j++; printf("%dn", i); return 0; } |
Which of the following statements are true about the code snippet given above
a)The code does not compile.
b)5
c)6
d)The result in machine dependent.
Ans: b
37)
|
1 2 3 4 5 6 7 8 |
int main() { int j=6; int i; i = (5, j); printf("%d %dn", i, j); return 0; } |
The output from the above program is:
a)5 6
b)6 6
c)6 7
d)The output is machine dependent
Ans: b
38)
|
1 2 3 4 5 6 7 8 |
int main() { int j=6; int i; i = (5, j++); printf("%d %dn", i, j); return 0; } |
The output from the above program is:
a)5 6
b)6 6
c)6 7
d)The output is machine dependent
Ans: c
39)What is the output of following program?
|
1 2 3 4 5 6 |
main() { int x=10; x= x++; printf("%dn", x); } |
a)10
b)11
c)12
d)Depends on the compiler.
Ans: d
40)What is the output of following program?
|
1 2 3 4 5 6 7 |
main() { int x = 0; int y; y = (x++, x++); printf("%d", y); } |
a)0
b)1
c)2
d)Compiler Dependent
Ans:b
41)What is the output of following program?
|
1 2 3 4 5 6 7 8 9 10 |
main() { int x = 0; int y = 2; if (1 < ++x | 2 < y++) { ++y; } printf("%d %dn", x, y); } |
a)1 2
b)1 3
c)1 4
d)Compiler Error
Ans: b
42)What is the output of following program?
|
1 2 3 4 5 6 7 8 9 10 |
main() { int x = 0; int y = 0; if (x++ && ++y) { ++x; } printf("%d %dn",x, y); } |
a)0 0
b)1 0
c)2 0
d)2 1
Ans: b
43)What is the output of following program?
|
1 2 3 4 5 6 7 8 9 10 |
main() { int x = 0; int y = 0; if (++x || ++y) { x++; } printf("%d %dn",x, y); } |
a)1 1
b)1 0
c)2 1
d)2 0
Ans: d
44)What is the output of following program?
|
1 2 3 4 5 6 7 8 9 10 |
main() { int x = 0; int y = 0; if (!(++x && ++y)) { x++; } printf("%d %dn",x, y); } |
a)1 1
b)1 0
c)2 1
d)2 0
Ans: a
45)What is the output of following program?
|
1 2 3 4 5 6 7 8 9 10 |
main() { int x = 0; int y = 0; if (!(++x || ++y)) { x++; } printf("%d %dn",x, y); } |
a)1 0
b)1 1
c)2 1
d)2 0
Ans: a
46)What is the output of following program?
|
1 2 3 4 5 6 |
main() { int x = 0x01C0FFEE; int y = x >> 15 + 1 & 0xFF; printf("%xn", y); } |
Priority is +, >>, &
a)C0
b)82
c)1C0
d)382
Ans: a
47)What is the output of following program?
|
1 2 3 4 5 6 |
main() { int x = 0xdeadbeef; int y = ~0xdeadbabe ^ x & 0xFFFF; printf("%xn", y); } |
a)2152FBAE
b)FBAE
c)Compiler Error
d)0
Priority is ~, & , ^.
Ans: a
48)What is the output of following program?
|
1 2 3 4 5 6 7 8 9 10 |
main() { int x = 10; int y = 20; if (x <= y == 1) { ++x; } printf("%xn", x); } |
a)11
b)10
c)Compiler Dependent
d)Syntax Error
priority is <=, ==
Ans: a
49)What is the output of following program?
|
1 2 3 4 5 6 7 8 9 10 |
main() { int x = 1; int y = 0; if (x | y++) { ++y; } printf("%d %dn", x); } |
a)Compiler Dependent
b)1 1
c)1 0
d)1 2
Ans: d
50)What is the output of following program?
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 |
int x; int f() { x += 10; return x; } int g() { x /= 10 ; return x; } int h() { x *=2; return x; } int main() { x = 10; printf("%dn", f() + g() * h()); } |
a)14
b)28
c)52
d)88
Priorty *, +
Ans: b
51)What is the output of the following program?
|
1 2 3 4 5 6 7 |
int main() { int x=10; int *y=&x; x = x+ *y*x/ *y ; printf("%dn",x); } |
a)Syntax Error
b)20
c)Runtime error
d)Unpredictable Output
Ans: b
52)What is the output of the following program?
|
1 2 3 4 5 6 7 |
int main() { int a = 0; int b; b = (a++)? a++: a++; printf("%d %dn", a,b); } |
a)2 1
b)2 2
c)Syntax Error
d)Unpredictable
Ans: a
53)What is the output of the following program?
|
1 2 3 4 5 6 7 8 |
int main() { int a = 0; int b = 1; int c; c = a?1:b?2:3; printf("%dn", c); } |
a)1
b)2
c)3
d)Syntax Error
tenary operator from right to left
Ans: a
54)What is the output of following program?
|
1 2 3 4 5 |
#define DOUBLE(x) x + x main() { printf("%dn", DOUBLE(10)*2); } |
a)20
b)30
c)40
d)Syntax error
Ans:b
55)What is the output of following program?
|
1 2 3 4 5 6 7 |
#define SQUARE(x) (x * x) main() { int x = 10; int y = 5; printf("%dn", SQUARE(x +y)); } |
a)225
b)45
c)65
d)Syntax error
Ans: c
56)What is the output of following program?
|
1 2 3 4 5 6 7 8 |
#define ABC(x) DEF(x) #define DEF(x) GHI(x) #define GHI(x) printf x main() { int x = 100; ABC(("value of x is %dn", x)); } |
a)value of x is 100
b)Syntax error
c)Linker error
d)Runtime error
Ans: a
57)What is the output of following program?
|
1 2 3 4 5 6 7 8 9 |
#define ABC(x) DEF(x) #define DEF(x) GHI(x) #define GHI(x) printf(x) main() { int x = 100; int y = 200; ABC(("Sum of x + y is %d", x + y)); } |
a)Sum of x + y is 300
b)Syntax error
c)Linker error
d)Runtime error or unpredictable behaviour
Ans: d
58)What is the output of following program?
|
1 2 3 4 5 6 7 |
#define ABC(x,y) printf(x":printed from ABCn", y) main() { int x = 100; int y = 200; ABC("Sum of x + y is %d", x + y); } |
a)Sum of x + y is 300:printed from ABC
b)Syntax error
c)Linker error
d)Behaviour is compiler dependent and hence unpredicatable.
Ans: a
59)What is the output of following program?
|
1 2 3 4 5 6 7 |
#define KING_KONG 0xDEADCA1F #define KING_ARTHUR 0xDEADBEAD #define ABC(x) (KING_##x) main() { printf("0x%Xn", ABC(ARTHUR)); } |
a)0xDEADBEAD
b)Syntax error
c)Linker error
d)Behaviour is compiler dependent and hence unpredicatable.
Ans: a
60)What is the output of following program?
|
1 2 3 4 5 |
#define MYPRINT(x) printf(#x) main() { MYPRINT(This appears like syntax errorn); } |
a)This appears like syntax error
b)Compilation error
c)Linker error
d)Behaviour is compiler dependent and hence
unpredicatable.
unpredicatable.
Ans: a
61)Which of the following statements are not true for a preprocessor.
1. Can detect syntax errors
2. Can direct compiler to do certain implementation specific actions
3. Can expand macros only if macro is declared in single line.
4. Supports nested macros
a)All of the above
b)1 & 3
c)1, 2 & 3
d)3 & 4
Ans: b
62)What is the output of following program?
|
1 2 3 4 5 6 7 8 9 10 11 |
#define MYPRINTF(x) printf("%s: printed at time:%dn", #x, __TIME__) main() { int i = 5; while(i--) { MYPRINTF(Hello World); sleep(1); } } |
a)The program prints the same output 5 times
b)The program prints Hello World along with time at which the printf is executed.
c)Program doesn’t link due to undefined symbol.
d)Syntax error
Ans: a
63)What is the output of following program?
Assume, sizeof(int) is 4 and sizeof(char) is 1.
Assume, sizeof(int) is 4 and sizeof(char) is 1.
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 |
#pragma pack(1) typedef struct { char day; char month; int year; }DATE; #pragma pack() typedef struct { int empno; DATE doj; /* date of joining */ int salary; } EmpRec; main() { printf("%d %dn", sizeof(DATE), sizeof(EmpRec)); } |
a)8 16
b)6 16
c)6 14
d)8 20
Ans: b
64)What is the output of the following program?
|
1 2 3 4 5 6 7 8 9 10 |
#include <stdio.h> #define a(x,y) x##y #define b(x) #x #define c(x) b(x) int main() { printf("%st",c(a(34,56))); printf("%sn",b(a(34,56))); return 0; } |
a)34 56 34
b)3456 3456
c)3456 a(34,56)
d)Compiler Error
Ans: c
65)Assuming the code below is stored in a file “a.c” and the command “gcc a.c” is used to compile the file, what is the output of this program:
|
1 2 3 4 5 6 7 8 9 10 11 12 13 |
#ifdef DEBUG #define DEBUG_PRINT(x) printf("debug :%sn", (x)); #else #define DEBUG_PRINT(x) #endif main() { int x; DEBUG_PRINT("In main"); x=10; printf("%dn", x); DEBUG_PRINT("Exiting main"); } |
a)Program encounters compilation error since DEBUG is not defined
b)Program does not print anything
c)Program prints 10
d)Program prints
debug :In main
10
debug :Exiting main
Ans: c
66)Assuming the code below is stored in a file “a.c” and the command “gcc a.c” is used to compile the file, what is the output of this program:
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 |
#ifdef DEBUG #define DEBUG_PRINT(x) printf("debug :%sn", (x)); #else #define DEBUG_PRINT(x) #endif main() { #define DEBUG DEBUG_PRINT("In main"); int x; x=10; printf("%dn", x); DEBUG_PRINT("Exiting main"); } |
a)Program encounters compilation error since DEBUG is not defined
b)Program does not print anything
c)Program prints 10
d)Program prints
debug :In main
10
debug :Exiting mainn”
Ans: c
67)What is the output of the following program:
|
1 2 3 4 5 6 7 8 9 |
main() { int x; #if 0 x=10; #endif printf("%dn", x); DEBUG_PRINT("Exiting main"); } |
a)Program encounters compilation error
b)The program prints some junk value
c)Program gives run-time error
d)Program prints 10
Ans: b
68)What is the output of the following program?
|
1 2 3 4 5 |
main() { int xyz = 0xA0A0A0A0; printf("0x%X", xyz >> 1); } |
a)0x50505050
b)0xD0505050
c)0x40404040
d)0xC0C0C0C0
Ans: b
69)What is the output of following program?
|
1 2 3 4 5 |
main() { unsigned int xyz = 0; printf("0x%Xn",--xyz); } |
a)-0
b)0xFFFFFFFF
c)Runtime Error or Unpredicatable output
d)Syntax Error
Ans: b
70)What is the output of following program?
|
1 2 3 4 5 6 |
main() { int number = 0xdeadbeef; int mask = 0x7ff0; printf("0x%Xn", (number & mask) >> 16); } |
a)0
b)0x3EEF000
c)0xBEEF
d)0x7EEF000
Ans: a
71)For what kind of numbers, the following program will print “Yes”
|
1 2 3 4 5 6 7 8 9 10 11 12 13 |
main() { int x; scanf("%d", &x); if (x & 3) { printf("Non"); } else { printf("Yesn"); } } |
a)Multiples of 2
b)Multiples of 4
c)Multiples of 8
d)Multiples of 3
Ans: b
72)What is the output of following program?
|
1 2 3 4 5 6 |
main() { int number = 0xdeadbeef; int mask = 0x7ff0; printf("0x%Xn", (number & mask) << 16); } |
a)0
b)0x3EE00000
c)0xBEEF
d)0x7EEF000
Ans: b
73)What is the size of the following structure?
|
1 2 3 4 5 6 7 8 |
typedef struct { int ctrl1; char flag; int ctrl2; char flag2; int ctrl3; } MYREG; |
a)14
b)20
c)24
d)16
Ans: b
74)What is the size of the following structure?
|
1 2 3 4 5 6 7 8 |
typedef struct { int ctrl1; char flag; char flag2; int ctrl2; int ctrl3; } MYREG; |
a)14
b)20
c)24
d)16
Ans: d
75)What would this function return:
|
1 2 3 4 |
int foo(int a) { return !(a & (a - 1)) && a; } |
a)Returns non zero when a is odd
b)Returns non zero when a is even
c)Returns non zero when a is power of 2
d)Always return zero
Ans: c
76)What would this function return:
|
1 2 3 4 5 6 7 8 9 |
int foo(int a) { int c; for (c = 0; a; c++) { a &= a - 1; } return c; } |
a)Number of bits set in variable a
b)location of most significant set bit
c)always return sizeof(v)*8
d)if a is odd it will retunr 1 else sizeof(v)*8
Ans: a
77)What would be the output of the following piece of code:
|
1 2 3 4 5 6 7 8 9 |
void main() { int x, y,z; z = (x=0xaaaa) && (y=0x5555); printf("logical AND result: %x;", z); x =0, y =0; z = (x=0xaaaa) & (y=0x5555); printf(" AND operator result:%x n", z); } |
a)logical AND result: 1; AND operator result: ffff
b)logical AND result: 0; AND operator result: ffff
c)logical AND result: 1; AND operator result: 0
d)logical AND result: 0; AND operator result: 0
Ans: c
78)What would be the output of the following piece of code:
|
1 2 3 4 5 6 7 8 9 |
void main() { int x, y,z; z = (x=0xaaaa) || (y=0x5555); printf("logical OR result:%x;", z); x =0, y =0; z = (x=0xaaaa) | (y=0x5555); printf(" bitwise OR result:%x n", z); } |
a)a. logical OR result: 1; bitwise OR result: ffff
b)b. logical OR result: 0; bitwise OR result: ffff
c)c. logical OR result: 1; bitwise OR result: 0
d)d. logical OR result: 0; bitwise OR result: 0
Ans: a
79)If you do not understand and address alignment issues in your software, which of the following scenarios are possible?
a)Your software may run slower, lock up
b)Your software may crash
c)Your software may silently fail, yielding
incorrect results
incorrect results
d)All of the above
Ans: d
80)Consider the following piece of code:
|
1 2 3 4 5 6 |
#define MASK 0xFFFFFFFF void bit_op(unsigned int p) { unsigned int i; i = MASK ^ ((p ^ MASK) + p); } |
Which of the following statements are true with respect to the above code?
a)At the end of the function, all the bits in ‘i’ will be set to 0
b)At the end of the function, all the bits in ‘i’ will be set to 1
c)At the end of the function, the value of ‘i’ will be equal to that of ‘p’
d)At the end of the function, the value of ‘i’ will be equal to that of ‘p’ with all the bits toggled
Ans: a
81)What does the following code output (on a 32-bit Intel/Linux system)?
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 |
struct x { unsigned int a:2; unsigned int b:2; unsigned int c:1; unsigned int d:1; int e; }; int main() { struct x foo; memset(&foo, 0, sizeof(foo)); foo.a = 2; foo.b = 7; foo.c = 3; printf("a=%d, b=%d, c=%d, d=%dn", foo.a, foo.b, foo.c,foo.d); return 0; } |
a)a=2, b=7, c=3, d=0
b)a=2, b=3, c=1, d=0
c)a=3, b=3, c=1, d=0
d)a=2, b=7, c=1, d=0
Ans: b
the correct answer is b. The bit fields will be filled only as much as the size specified in the definition. So only 2
bits are used for ‘b’ and one bit for ‘c’. Make sure that you are careful about using such code while writing portable applications.
bits are used for ‘b’ and one bit for ‘c’. Make sure that you are careful about using such code while writing portable applications.
82)What does the following code output (on a 32-bit Intel/Linux system)?
|
1 2 3 4 5 6 7 8 9 10 11 12 13 |
struct x { int a:2; int b:2; int c; int d:1; int e:1; }; int main() { printf("%dn", sizeof(struct x)); return 0; } |
a)5
b)8
c)12
d)16
Ans: c
83)What does the following code output (on a 32-bit Intel/Linux system)?
|
1 2 3 4 5 6 7 8 9 10 11 12 |
struct record { char *name; int refcount:4; unsigned int active:1; }; int main() { printf("%dn", sizeof(struct record)); return 0; } |
a)5
b)6
c)8
d)12
Ans: c
the correct answer is c. If enough space remains, a bit-field that immediately follows another bit-field in a structure
will be packed into adjacent bits of the same unit. Here, since refcount uses only 4 bits of the int, active will be packed into the same int. So name uses 4 bytes (pointer to a char), and refcount and active together use up another 4
bytes.
will be packed into adjacent bits of the same unit. Here, since refcount uses only 4 bits of the int, active will be packed into the same int. So name uses 4 bytes (pointer to a char), and refcount and active together use up another 4
bytes.
84) Which of the following statements is most accurate about the belowcode?
|
1 2 3 4 |
int fn(int x) { return (x && ((x ^ (x - 1)) == ((2 * x) - 1))); } |
a)It checks if x is an even number or odd number
b)It checks if at least one bit is set to 1 in x
c)It checks if all the bits are set to 1 in x
d)It checks if x is a power of 2
Ans: d
85)What does the following code output (on a 32-bit Intel/Linux system)?
|
1 2 3 4 5 6 7 8 9 10 11 12 13 |
struct st { char a[3]; short int b; long int c; char d[1]; }; int main() { printf("%dn", sizeof(struct st)); return 0; } |
a)10
b)16
c)18
d)20
Ans: b
86)Which of the following represents the logic for a function getbits(x, p, n)that returns the right adjusted n-bit field of x that begins at position p.
Assume that bit position 0 is at the right end and that p and n are sensiblevalues. For example, if x = 174, p = 4, n =
3, then getbits() should return 3bits in positions 4, 3 and 2 (101″011″10), i. e., getbits(174, 4, 3) = 3.
3, then getbits() should return 3bits in positions 4, 3 and 2 (101″011″10), i. e., getbits(174, 4, 3) = 3.
a)return (x << (p + 1 – n)) & ~(~0 >> n);
b)return (x >> (p + 1 – n)) & ~(~0 << n);
c)return (x << (p – n)) & ~(~0>> n);
d)return (x >> (p – n)) & ~(~0<< n);
Ans: b
87)Consider the following piece of code (on a 32-bit Intel/Linux system):
|
1 2 3 4 5 |
#define BIT(n) (1 << (n)) void bit_op(unsigned int a, unsigned int n) { a |= BIT(n); } |
Which of the following statements is most accurate about the above code?
a)The function sets the n-th bit of a, assuming 0 to be the least significant bit position
b)The function sets the n-th bit of a, assuming 0 to be the most significant bit position
c)The function clears the n-th bit of a, assuming 0 to be the least significant bit position
d)The function clears the n-th bit of a, assuming 0 to be the most significant bit position
ans : a
88)Consider the following piece of code (on a32-bit Intel/Linux system):
|
1 2 3 4 5 6 7 8 9 10 11 12 13 |
#define BITS_IN_CHAR 8 int fn1(int a) { int i; i = -(a < 0); return i; } int fn2(int a) { int i; i = a >> (sizeof(int) * BITS_IN_CHAR - 1); return i; } |
Which of the following statements is most accurate about the above code?
a)Both the functions compute the sign of a given integer & return -1 for negative values and 0 for positive values
b)Both the functions compute the sign of a given integer & return -1 for negative values and 1 for positive values
c)Both the functions compute the sign of a given integer & return -1 for negative values and 0 for positive values.
fn1() will avoid branching on CPUs with flag registers.
fn1() will avoid branching on CPUs with flag registers.
d)Both the functions compute the sign of a given integer & return -1 for negative values and 0 for positive values.
fn2() will avoid branching on CPUs with flag registers.
fn2() will avoid branching on CPUs with flag registers.
Ans: d
89)Consider the following piece of code (on a 32-bit Intel/Linux system):
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 |
struct mystruct { char a; int b; short c; char d; }; int main() { struct mystruct foo[3]; char ch; int i; printf("%p %p %p %p %pn", &foo[0], &foo[1], &foo[2], &ch, &i); return 0; } |
Which of the following would be the most likely candidate for its output?
a)0xfeeece90 0xfeeece9c 0xfeeecea8 0xfeeece8f 0xfeeece88
b)0xfeeece90 0xfeeece9c 0xfeeecea8 0xfeeeceb4 0xfeeeceb8
c)0xfeeece90 0xfeeece9c 0xfeeecea8 0xfeeeceb4 0xfeeeceb5
d)0xfeeece90 0xfeeece9c 0xfeeecea8 0xfeeece8c 0xfeeece88
Ans: a
the correct answer is a. Remember that the stack grows towards lower memory addresses here. So ch would be at a location above foo, and i above ch. ch would be at a location aligned to its natural (1-)byte boundary. Similarly, i would be a location aligned to the 4-byte boundary.
90)Consider the following piece of code (on a 32-bit Intel/Linux system):
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 |
/* Assume a > 0 */ unsigned long bit_op(unsigned long a) { int n = 0; if ((a & 0xffff) == 0) { n += 16; a >>= 16; } if ((a & 0xff) == 0) { n += 8; a >>= 8; } if ((a & 0xf) == 0) { n += 4; a >>= 4; } if ((a & 0x3) == 0) { n += 2; a >>= 2; } if ((a & 0x1) == 0) n += 1; return n; } |
Which of the following statements is mostaccurate about the above code?
a)The function returns the position of the first rightmost bit(LSB) set to 1
b)The function returns the position of the first rightmost(LSB) bit set to 0
c)The function returns the position of the first leftmost(MSB) bit set to 1
d)The function returns the position of the first leftmost(MSB) bit set to 0
Ans: a
91)Consider the following piece of code (on a 32-bit Intel/Linux system):
|
1 2 3 4 5 6 7 8 9 10 11 12 13 |
# pragma pack (1) struct x { char a; int b; short c; }; # pragma pack () int main() { printf("%dn", sizeof(struct x)); return 0; } |
What is the output of the above code?
a)8
b)7
c)12
d)10
Ans: b
the correct answer is b. The #pragma pack(n) directive indicates to the compiler the alignment needs. When #pragma pack(1) is specified, it tells the compiler to align on a 1-byte boundary.
#pragma pack() is used to get back to default alignment. Since struct x is defined within #pragma pack(1) directive, it will align to 1 byte boundary. No padding will be done between the members a, b or c of the structure. Hence a
will take up 1 byte, b will take up 4 bytes and c will take up 2 bytes.
#pragma pack() is used to get back to default alignment. Since struct x is defined within #pragma pack(1) directive, it will align to 1 byte boundary. No padding will be done between the members a, b or c of the structure. Hence a
will take up 1 byte, b will take up 4 bytes and c will take up 2 bytes.
92)Consider an unsigned int v. For values of v=255, what would the values of ~v, ~v+1 and ~(v+1) be?
a)ffffff00, ffffff01, fffffeff
b)fffffffe, fffffffd, ffffffed
c)0, 1, 255
d)None of the above
Ans: a
93)What is the output of following program?
|
1 2 3 4 5 |
main() { char **p = 0; printf("%dn", ++p); } |
a)4
b)Syntax Error
c)Runtime Error
d)Unpredictable output
Ans: a
the correct answer is a. p is double pointer (pointer to pointer). Sizeof any pointer in 32 bit machines is 4 bytes. So, by incrementing p, it moves 4 bytes. Since, p is initialised to zero, value of p after increment is 4.
94)What is the output of following program?
|
1 2 3 4 5 6 7 |
main() { int x, y; int *p1 = &x; int *p2 = &y; printf("%dn", p1-p2); } |
a)4
b)1
c)-4
d)Syntax or Runtime error
Ans: b
95)What is the output of following program?
|
1 2 3 4 5 6 7 |
main() { int x, y ; int *p1 = &x; int *p2 = &y; printf("%dn", (int)p1-(int)p2); } |
a)4
b)1
c)-4
d)Syntax or Runtime error
Ans: a
the correct answer is a. All the local variables are stored in stack. So, x, y, p1 and p2, all these variables will be part of the stack. Since, stack follows “last in first out” principle, the first local variable is pushed into the stack. i.e x is pushed first, followed by y, followed by p1 and p2. So, &p2 and &p1 would differ by 4 bytes. Since
p1 is pushed earlier than p2, it will have higher memory address than p2. Since the pointers are cast to integer, by subtracting them, they just behave like normal numbers and hence output would be 4.
p1 is pushed earlier than p2, it will have higher memory address than p2. Since the pointers are cast to integer, by subtracting them, they just behave like normal numbers and hence output would be 4.
96)What is the output of following program?
|
1 2 3 4 5 6 7 |
main() { int x, y ; char *p1 = &x; char *p2 = &y; printf("%dn", p1-p2); } |
a)4
b)1
c)-4
d)Syntax or Runtime error
Ans: a
the correct answer is a. All the local variables are stored in stack. So, x, y, p1 and p2, all these variables will be part of the stack. Since, stack follows “last in first out” principle, the first local variable is pushed into the stack. i.e x is pushed first, followed by y, followed by p1 and p2. So, &p2 and &p1 would differ by 4 bytes. Since
p1 is pushed earlier than p2, it will have higher memory address than p2. However, subtracting two pointers of the same type, will give the “number of elements” between them. Since sizeof char is 1 byte, output would be 4.
p1 is pushed earlier than p2, it will have higher memory address than p2. However, subtracting two pointers of the same type, will give the “number of elements” between them. Since sizeof char is 1 byte, output would be 4.
97)What is the output of the following program?
|
1 2 3 4 5 6 7 8 9 10 11 12 |
typedef struct { int day; int month; int year; } MYDATE; main() { MYDATE *dp = 0xB8000000; printf("0x%xn", &dp->year); } |
a)Unpredictable output
b)Runtime error
c)0xB8000004
d)0xB8000008
Ans: d
the correct answer is d. The first field starts at 0xB8000000, second field starts at 0xB8000004 and hence third field would start 0xB8000008.
98)What is the output of following program?
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 |
typedef struct { int day; int month; int year; } MYDATE; int x = 0x10; int y = 0x20; int z = 0x30; int a = 0x40; int b = 0x50; int c = 0x60; int d = 0x70; MYDATE abc = { 1, 1, 2006 }; main() { MYDATE *dp = &x; ++dp; ++dp ; printf("%xn",dp->day); } |
a)0x30
b)0x60
c)0x70
d)Syntax or Runtime error
Ans: c
the correct answer is c. Size of MYDATE is 12 bytes. So, if incremented twice, dp moves 24 bytes or 6 integers. So,
dp->day will print 0x70.
dp->day will print 0x70.
99)What is the output of the following program in little endian architecture based processor ?
|
1 2 3 4 5 6 7 |
main() { int xyz = 0xC0FFEE; unsigned char *p = &xyz; ++p; printf("0x%Xn",*p); } |
a)0xC0
b)0x00
c)0xEE
d)0xFF
Ans: d
100)What is the output of the following program?
|
1 2 3 4 5 6 |
main() { int x = 10; int *p = &x; printf("%d %d %dn", x, *(&x), **(&p)); } |
a)Syntax Error
b)Runtime Error
c)10 10 10
d)Unpredictable output as addresss of x and p are not known.
Ans: c
101)What is the output of the following program?
|
1 2 3 4 5 6 7 8 |
main() { int x = 10; int y = 20 ; int *p = &y ; &x = p ; printf("%d %d %dn", x, y, *p); } |
a)Syntax Error
b)Runtime Error
c)20 20 20
d)10 20 20
Ans: a
the correct answer is a. & operator doesn’t have lvalue. i.e & of variable cannot be changed.
102)What is the output of the following program?
|
1 2 3 4 5 6 7 8 9 |
main() { int x = 10 ; int y = 20 ; int *p = &x ; int **pp=&p ; *pp = &y ; printf("%d %dn", *p, **pp); } |
a)Syntax Error or Runtime Error
b)10 20
c)10 10
d)20 20
Ans: d
the correct answer is d. First, p is initialised to &x and pp is initialised to &p. i.e *pp is &p. After the assignment *pp=&y, value of p changes to &y. So, both *p and **pp is 20.
103)What is the output of the following program?
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 |
typedef struct { int day; int month; int year; } MYDATE; MYDATE abc = {11, 9, 2006}; main() { MYDATE *dp = &abc; dp = (MYDATE *)((char *)dp + 8); printf("%dn", dp->day); } |
a)Syntax Error or Runtime Error
b)11
c)2006
d)Unpredictable output
Ans: c
104)What is the output of the following program?
|
1 2 3 4 5 6 7 8 |
char *def[5] = {"pqrs","rstu", "tuvw", "vwxyz", "xyzab" }; char abc[5][5] = {"abc","def", "ghi", "jkl", "mno"}; main() { char *p =(char *)def; p = p + 40; printf("%sn", p); } |
a)Syntax Error or Runtime Error
b)mno
c)jkl
d)ghi
Ans: b
the correct answer is b. def is array of character pointer. Its size is 20. (sizeof(def[0] is 4, as each element is a pointer). abc is two dimensional array and its size is 25. The size of each element in abc is 5. (i.e sizeof(abc[0]) is 5. p is initialised to def and moved by 40 bytes. This offset of 40 bytes makes it cross 20 bytes of def and 20 bytes in
abc. i.e it just crosses a[3] and points to a[4]. Hence the above program would print “mno” which is the last element of abc.
abc. i.e it just crosses a[3] and points to a[4]. Hence the above program would print “mno” which is the last element of abc.
105)What is the output of the following program?
|
1 2 3 4 5 6 7 8 9 10 11 |
f1(char *p) { char abc[6]="Hello"; p = abc; } main() { char *p="World"; f1(p); printf("%sn", p); } |
a)World
b)Hello
c)Runtime error
d)Hello World
Ans: a
106)What is the output of the following program?
|
1 2 3 4 5 6 7 8 9 10 11 12 |
char *f1() { char abc[6]="Hello"; char *p = abc; return abc; } main() { char *p; p = f1(); printf("%sn", p); } |
a)H
b)Hello
c)Runtime error Or Unpredictable output
d)Syntax error
Ans: c
the correct answer is c. Note that local variables go into the stack and is valid only in the scope of that function. Once the function returns, contents of the called function “stack frame” is not guaranteed. Since content pointed by p is not guaranteed to be “Hello” after the f1 returns, output is unpredictable or can have runtime error. In general, whenever function returns an address of local variable, it is definitely an error.
107)What is the output of the following program?
|
1 2 3 4 5 6 7 8 9 10 11 |
char *f1() { char *abc ="Hello"; return abc; } main() { char *p; p = f1(); printf("%sn", p); } |
a)H
b)Hello
c)Runtime error Or Unpredictable output
d)Syntax error
Ans: b
the correct answer is b. Note that string constants go to the constant data segment. Therefore the string hello will be
kept in the constant data segment and the address of it would get assigned to abc. The same is being returned by the function.
kept in the constant data segment and the address of it would get assigned to abc. The same is being returned by the function.
108)The output of the following program is:
|
1 2 3 4 5 6 7 |
main() { char *p; char buff[10] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h'}; p = (buff + 2)[5]; printf("%c" , p); } |
a)f
b)g
c)h
d)None of the others
Ans: c
109)The output of the following program is:
|
1 2 3 4 5 6 7 |
main() { int x[][3] = {1, 2, 3, 4, 5, 6}; int (*ptr)[3] = x; printf("%d ", sizeof(ptr)); printf("%dn", sizeof(x)); } |
Considering size of integer is 4 bytes,
a)12 24
b)12 12
c)4 12
d)4 24
Ans: d
the correct answer is d. ptr is a pointer to an array of 3 integers. Hence its size is 4 bytes (size of a pointer). x is
a 2-d array containing 6 integers. Hence its size is 6*4 = 24.
a 2-d array containing 6 integers. Hence its size is 6*4 = 24.
110)The output of the following program is:
|
1 2 3 4 5 6 7 8 |
main() { int x[][4] = {1, 2, 3, 4, 5, 6, 7, 8}; int (*ptr)[4] = x; printf("%d %d ", (*ptr)[1], (*ptr)[2]); ++ptr; printf("%d %dn", (*ptr)[1], (*ptr)[2]); } |
Considering size of integer as 4 bytes,
a)5 6 5 6
b)2 3 6 7
c)2 3 4 5
d)2 3 5 6
Ans: b
the correct answer is b. ptr is a pointer to an array of 4 integers. It gets assigned the address of array x. (*ptr)[1]
gets expanded to (*((*ptr)+1)). So ((*ptr)+1) would point to address of 2. Since ptr is a pointer to an array of 4 integers, ptr++ would make it move by 4 integers. This is nothing but address of 6.
gets expanded to (*((*ptr)+1)). So ((*ptr)+1) would point to address of 2. Since ptr is a pointer to an array of 4 integers, ptr++ would make it move by 4 integers. This is nothing but address of 6.
111)What is the output of the followoing program:
|
1 2 3 4 5 6 7 8 |
main() { int i; char str[80] = "I am feeling good today"; for(i=0; i <strlen(str); i++) printf("%c", i[str]); printf("n"); } |
a)I am feeling good today
b)Compilation error
c)Run time error generating core
d)Unexpected results
Ans: a
112)Considering that malloc is successful, What would be the output of the following program (consider sizeof(int) is 4):
|
1 2 3 4 5 6 7 8 |
main() { int StatArr[2]; int *DynArr; DynArr = (int*)malloc(2*sizeof(int)); printf("sizeof(StatArr) : %d, sizeof(DynArr): %dn", sizeof(StatArr), sizeof(DynArr)); } |
a)sizeof(StatArr) : 8, sizeof(DynArr): 8
b)sizeof(StatArr) : 8, sizeof(DynArr): 4
c)sizeof(StatArr) : 4, sizeof(DynArr): 8
d)sizeof(StatArr) : 8, sizeof(DynArr): 8
Ans: b
113)What is the output of following program?
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 |
int i = 10; int f1(int(*p)()) { --i; if (i > 0) { int j; j = p(p); return(j); } return 1; } main() { printf("%dn", f1(f1)); } |
a)1
b)Syntax Error
c)10
d)Runtime Error
Ans: a
the correct answer is a. The function takes a pointer to function as an argument. f1(f1) will eventually call f1 and p(p) will call f1 untill value of i becomes 0. When value of i becomes 0, f1 returns 1. The same is returned throughout the previous invocation of f1 and hence output is 1.
114)The output of the following program is:
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 |
int f1(int); int f2(int); int (*fp[2])(int) = { f1, f2 }; int f1(int x) { return(x+10); } int f2(int x) { return(x+20); } main() { int sum = 0; int (*fp1)(int); fp1 = fp[0]; ++fp1; sum = fp1(sum); printf("%dn",sum); } |
a)Runtime error or unpredictable output
b)20
c)10
d)Syntax Error
Ans: a
the correct answer is a. If we increment a function pointer, it moves only by one byte. Because, a function pointer can point to any function (as long as return type and argument types are matched) and hence sizeof a function will be
different. Since fp1 is moved only by one byte, the control would go to a wrong entry point and the program willl
have run time error.
different. Since fp1 is moved only by one byte, the control would go to a wrong entry point and the program willl
have run time error.
115)The output of the following program is:
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 |
int f1(int); int f2(int); int (*fp[2])(int) = { f1, f2 }; int f1(int x) { return(x+10); } int f2(int x) { return(x+20); } main() { int sum = 0; int (**fp1)(int); fp1 = fp; ++fp1; sum = (**fp1)(sum); printf("%dn",sum); } |
a)Runtime error or unpredictable output
b)20
c)10
d)Syntax Error
Ans: b
the correct answer is b. If we increment a pointer to a function pointer, it moves only by 4 bytes. Hence, *fp1 would point to f2. By calling (**fp1)(sum) or (*fp1)(sum), both will result in calling f2 and hence output would be 20.
116)What is the output of the following program:
|
1 2 3 4 5 6 7 |
main() { int i=15; char *str; strcpy(str, "Hello, World!"); printf("i=%dn", i--); } |
a)15
b)14
c)The program does not compile
d)Program behaviour is unpredictable
Ans: d
the correct answer is d. str being a pointer has only 4-bytes allocated for it to store an address. Copying a string
into it will corrupt the stack
into it will corrupt the stack
117)What is the output of the following program:
|
1 2 3 4 5 6 7 |
#include <stdio.h> int i=0; main() { int i=5; printf("%dn", i++); } |
a)0
b)1
c)5
d)6
Ans: c
118)What is the output of the following program:
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 |
#include <stdio.h> int i=0; main() { printf("%d ", i); { int i=2; printf("%d ", i); { i++; printf("%d ", i); } printf("%d ", i); } printf("%dn", i); } |
a)0 2 3 3 0
b)0 2 3 2 0
c)0 2 3 2 2
d)None of the above
Ans: a
119)What is the output of the following program:
|
1 2 3 4 5 6 7 8 9 10 11 |
int i=5; int f1(int i) { i=i/2; return i; } main() { int i=10; printf("%d, %dn", f1(i), i); } |
a)5, 5
b)5, 10
c)10, 10
d)None of the above
Ans: b
120)What is the output of the following program:
|
1 2 3 4 5 6 7 8 9 10 11 12 13 |
int i=10; int f1() { static int i = 15; printf("f1:%d ", i); return i--; } main() { int i, j; i = 5; printf("%d %d %d", f1(), f1(), i); } |
a)f1:15 f1:14 14 15 5
b)f1:15 14 f1:15 14 5
c)f1:15 f1:14 14 14 5
d)None of the above
Ans: a
the correct answer is a. Here the catch is that the parameters are evaluated right to left. So, when f1 is first called it
is the call made from the 3rd parameter of printf, During this call, 15 is printed, and the static variable i gets decremented to 14. Again, f1 is called and this time, 14 gets printed. 14 is also returned. In the sequence in which
values are returned, the printf statement prints them. Hence the output.
is the call made from the 3rd parameter of printf, During this call, 15 is printed, and the static variable i gets decremented to 14. Again, f1 is called and this time, 14 gets printed. 14 is also returned. In the sequence in which
values are returned, the printf statement prints them. Hence the output.
121) what is the output of the program
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 |
File a.c contains: static int i=12; forward(void) { return (i+=1); } backward(void) { return (i-=1); } f1(int i) { static int j=2; return(i=j+=i); } File b.c contains: extern int i; int restore() { return i; } Another file c.c contains: #include <stdio.h> int i=5; main() { int i, j; i=restore(); printf("%d ", forward()); printf("%d ", backward()); printf("%d ", f1(i)); } |
When all the above three files are linked together and executable is created what is the output of the program?
a)13 13 7
b)13 12 7
c)12 12 7
d)13 7 7
Ans: b
the correct answer is b. When restore() is first called, it returns the global value of i and assigns it to the local
value of i in main(). The scope of i in a.c is that of the static variable with visibility in the file a.c. In file b.c the global variable i defined in a.c holds. Within the function main() the scope of i is that of the on-stack local variable. With this in mind, when forward() is called, the static value of i(initialized to 12) in a.c gets incremented and printed. Then, when backward() is called the same gets decremented. When f1() is called, 2 gets added to the value of i which was assigned 5 during call to restore.
value of i in main(). The scope of i in a.c is that of the static variable with visibility in the file a.c. In file b.c the global variable i defined in a.c holds. Within the function main() the scope of i is that of the on-stack local variable. With this in mind, when forward() is called, the static value of i(initialized to 12) in a.c gets incremented and printed. Then, when backward() is called the same gets decremented. When f1() is called, 2 gets added to the value of i which was assigned 5 during call to restore.
122)what is the output of the program ?
|
1 2 3 4 5 6 7 8 9 10 11 12 |
File a.c contains int f1(int x, int y) { printf("%d %d", x, y); return x+y; } File b.c contains extern int f1(); main() { printf(" %dn", f1(10, 20, 3)); } |
What is the output of the above program when a.c and b.c are compiled and linked together
a)Linker Error
b)Compilation Error
c)10 20 23
d)10 20 30
Ans: d
123)what is the output of the program?
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 |
void Recurse(int x) { int y = 10; printf("%d %pn", y, &y); if (x == 0) { ++y; Recurse(1); } } main() { Recurse(0); } |
The above program prints two lines. Which of the following statements is true for the above program:
a)Value of y and &y will remain same across the two output lines.
b)Value of y will remain same across two output lines, but not &y
c)Value of &y will remain same across two output lines, but not y
d)Both value of y and &y will be different across both output lines.
Ans: b
the correct answer is b. Here, y is a local variable. Copies of it are created each time a function is called recursively.
Its value is being printed each time it is assigned the value 10. Therefore, the value of y printed will always be 10, but its address will vary for each call to the function.
Its value is being printed each time it is assigned the value 10. Therefore, the value of y printed will always be 10, but its address will vary for each call to the function.
124) what is the output of the program ?
|
1 2 3 4 5 6 7 8 9 10 |
int func(int x) { char a[10]; strcpy(a, "Hello, This is indeed a very big world"); return (x+1); } main() { printf("%dn", func(1)); } |
Which of the following statement is true with respect to the above program.
a)There is a stack corruption, but it does not corrupt x. So, output will be 2.
b)Value of x is indeterministic, but program control returns to main and some junk value is printed on the screen.
c)There is a stack corruption, it would corrupt return address as well. Hence there will be runtime error.
d)None of the above
Ans: c
the correct answer is c. The return address is stored at the location (ebp+1). Remember that stack grows towards lower memory addresses and local variables start from (ebp-1). Since a is just a 10 byte array, and the length of the string being copied is much beyond that the return address will get corrupted.
125) what is the output of the program ?
|
1 2 3 4 5 6 7 8 9 10 11 12 |
int * func(int *xp) { int y = 10 + *xp; return (&y); } main() { int x = 10; int *xp = func(&x); printf("%dn", x); printf("%dn", *xp); } |
a)20
b)10
c)Runtime Error
d)Unpredictable
Ans: d
the correct answer is d. Please note that a normal gcc compiler will always give a warning whenever the address of a local variable is returned. When func() is called and then printf() is called right after, the stack would have got over-written during the execution of printf(). Hence when contents of that location where y was present during call to func() are read, either some junk value will be read or an unpredictable behavior may be seen.
126) what is the output of the program ?
|
1 2 3 4 5 6 7 8 9 10 |
void func(int a[10]) { a[0] = 1; } main() { int a[10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }; func(a); printf("%dn", a[0]); } |
a)Syntax error
b)Runtime error
c)0
d)1
Ans: d
the correct answer is d. When call to func(a), a is nothing but &a[0]. So when func modifies a[0], the same is reflected in the array that is referred to in main.
127)What is the output of the following program:
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 |
#include <stdarg.h> int func(int x, ...) { char y; int z; va_list ap; va_start(ap, x); y = (char)va_arg(ap, int); z = va_arg(ap, int); printf("%c 0x%xn", y, z); va_end(ap); } main() { func(0x10, 0x32, 0x31); } |
a)0x32 0x31
b)Not predictable
c)2 0x31
d)Syntax Error
Ans: c
the correct answer is c. When va_arg is made to interpret the location pointed to by 0x32, it will interpret it as a
character and print 2. The next argument is interpreted as an integer and hence its hex value is printed. The cast to char is necessary since va_arg is capable of interpreting only fully promoted types.
character and print 2. The next argument is interpreted as an integer and hence its hex value is printed. The cast to char is necessary since va_arg is capable of interpreting only fully promoted types.
128) what is the output of the program
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 |
#include <stdarg.h> int foo(char *fmt, ...) { va_list ap; int d; int count=0; char c, *p, *s; va_start(ap, fmt); while (*fmt) switch(*fmt++) { case '%': if((*fmt) && (*(fmt) != '%')) count++; else if (*fmt) fmt++; break; } va_end(ap); return count; } |
a)Counts the number of arguments passed from the format string
b)Counts the number of %s in the given format string
c)Encounters a linker error
d)Encounters compilation error
Ans:a
129)What is wrong with the following program:
|
1 2 3 4 5 |
main() { char *s = "Hello Worldn"; printf("%s", s); } |
a)s has memory for only 4-bytes so this will not compile.
b)Its behaviour is unpredictable since s has memory for only 4 bytes
c)Program will encounter segmentation violation
d)It will work without any problem
Ans: d
130)What is wrong with the following program:
|
1 2 3 4 5 6 |
main() { char *s = (char *) malloc(sizeof("Hello Worldn")); strcpy(s, "Hello Worldn"); printf("%s", s); } |
a)It will work fine since memory sufficient to hold “Hello Worldn” is requested from heap
b)Its behaviour is unpredictable since s has memory for only 4 bytes
c)Program will encounter segmentation violation
d)The malloc statement is fine, but subsequently the program could have problem
Ans: d
131)What is wrong with the following program:
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 |
main() { char *p = (char *)malloc(sizeof ("Hello Worldn")); if (p) { strcpy(p,"Hello World"); printf("%s", p); } p= (char *)malloc(sizeof("Good Morningn")); if(p) { strcpy("p, "Good Morning"); printf("%s", p); } if (p) free(p); } |
a)There is nothing wrong in the program since all memory allocated is being freed
b)The memory allocated for “Hello Worldn” is lost forever
c)p should be an array of characters instead of a pointer to char
d)None of the above
Ans: b
132)What is the problem with the following piece of code?
|
1 2 3 4 5 6 7 8 9 |
main() { char *p = (char *)malloc(strlen("Hello Worldn")); if (p) { strcpy(p, "Hello Worldn"); free(p); } } |
a)strlen returns 12, but the entire string needs 13 bytes, so strcpy will corrupt memory.
b)strlen returns 12, the string needs 12 bytes, so there is no problem in the program.
c)strlen returns 13, the string needs 13 bytes, so there is no problem in the program.
d)None of the above
ans: a
133)Which of the following statements is true:
a)Memory corruption will immediately result in a crash – hence debugging it is very easy
b)Memory corruption can never result in a crash
c)Memory corruption will not always result in a crash, problem occurs only when corrupted memory gets accessed
d)It is OK to have memory corruption since the problems caused by it are very rare
Ans: c
the correct answer is c. Since it is not always necessary to have misbehaviour with memory corruption, it is very
necessary to be very discplined during coding.
necessary to be very discplined during coding.
134)Having access to the memory allocated through the pointer returned is important because:
a)The address returned by malloc can be accessed only through it
b)To check the return value of malloc
c)Freeing that memory is completely dependent on this address
d)All of the above
Ans: d
135)A loader is
a)Independent of the OS and hence loaders can work across OS platforms
b)It is dependent on the OS and understandsthe OS specific process loading requirements
c)Is part of the compiler and takes care of the last phase of compilation
d)None of the above
Ans: b
the correct answer is b. Loader is responsible for loading the executable file generated into the memory to
prepare it for running. Hence it needs to understand the OS specifics to enable running of the program.
prepare it for running. Hence it needs to understand the OS specifics to enable running of the program.
136)What is the likely problem with this code:
|
1 2 3 4 5 6 7 8 9 10 11 12 13 |
main() { char *p = (char *)malloc(1024); if (p) { strcpy(p,"Hellon"); p += 10; strcpy(p,"Worldn"); p += 10; strcpy(p,"That is alln"); free (p); } } |
a)Incrementing p will result in runtime error.
b)There is no problem with the code.
c)The pointer being given to free does not contain the address returned by malloc.
d)None of the above.
Ans: c
137)What happens when the following program is compiled as a.out and run as ./a.out
|
1 2 3 4 5 6 7 |
main(int argc, char *argv[]) { int flag; char filename[10]; strcpy(filename, argv[1]); printf("%s", filename); } |
a)If the file is stored in a file by name “a.c”, that name will get copied into parameter filename
b)Since the file executable filename is “a.out”, that will get copied to filename in the strcpy command.
c)Any junk value on the stack will be copied on to filename and subsequent behavior of the program is unpredictable
d)None of the above
Ans: c
the correct answer is c. Any junk value supposedly present in the location of argv[1] will get copied on to filename
and the subsequent behaviour of the program would be unpredictable.
and the subsequent behaviour of the program would be unpredictable.
138)What happens when the following program is compiled as a.out and run as./a.out /home/user/harry/next.c
|
1 2 3 4 5 6 7 |
main(int argc, char *argv[]) { int flag; char filename[10]; strcpy(filename, argv[1]); printf("%s", filename); } |
a)If the file is stored in a file by name “a.c”, that name will get copied into parameter filename
b)The entire string “/home/user/harry/next.c” will be copied into filename corrupting flag and other contents of the stack
c)Just the path “next.c” will be copied into filename and rest of the path is omitted.
d)Since the file executable filename is “a.out”, that will get copied to filename in the strcpy command.
Ans: b
139)Consider the code snippet:
|
1 2 3 4 5 6 7 8 9 10 11 |
main() { struct { int arr[4]; int key; } name_key; int i; name_key.key = 0; for (i=0; i<=4; i++) name_key.arr[i] = i+5; } |
What is the value of “name_key.key” after the above chunk of code is executed?
a)0
b)9
c)The compiler gives a warning since array bounds are crossed in the loop and does not generate code.
d)The compiler gives an error since array bounds are crossed in the loop and does not generate code.
Ans: b
140)Which of the following is meant by DoS?
a)Dual Office Service
b)Denial of Service
c)Denial Office Symdrome
d)Denial of Security
Ans: b
141)Which of the following statements is true ?
a)The C-compiler checks for array bounds and prevents overwriting of memory that is not part of the array if the array is on the stack
b)The C-compiler checks for sizes of various variables, and prevents a particular variable’s memory being illegally
overwritten when another variable is being written to.
overwritten when another variable is being written to.
c)The C-compiler checks for array bounds and prevents overwriting of memory that is not part of the array if the array is on the heap
d)None of the above
Ans: d
Permalink
Thanks for posting all the concepts at one place …
Good Job !!!
Permalink
Thank you Rajesh!!
Please suggest if anything needs to be added
Permalink
Please give the explanation of the ans