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What will be the output of the program?
#include<stdio.h> int i; int fun1(int); int fun2(int); int main() { extern int j; int i=3; fun1(i); printf("%d,", i); fun2(i); printf("%d", i); return 0; } int fun1(int j) { printf("%d,", ++j); return 0; } int fun2(int i) { printf("%d,", ++i); return 0; } int j=1;
Options- A. 3, 4, 4, 3
- B. 4, 3, 4, 3
- C. 3, 3, 4, 4
- D. 3, 4, 3, 4
- Correct Answer
- 4, 3, 4, 3
ExplanationStep 1: int i; The variable i is declared as an global and integer type.Step 2: int fun1(int); This prototype tells the compiler that the fun1() accepts the one integer parameter and returns the integer value.
Step 3: int fun2(int); This prototype tells the compiler that the fun2() accepts the one integer parameter and returns the integer value.
Step 4: extern int j; Inside the main function, the extern variable j is declared and defined in another source file.
Step 5: int i=3; The local variable i is defines as an integer type and initialized to 3.
Step 6: fun1(i); The fun1(i) increements the given value of variable i prints it. Here fun1(i) becomes fun1(3) hence it prints '4' then the control is given back to the main function.
Step 7: printf("%d,", i); It prints the value of local variable i. So, it prints '3'.
Step 8: fun2(i); The fun2(i) increements the given value of variable i prints it. Here fun2(i) becomes fun2(3) hence it prints '4' then the control is given back to the main function.
Step 9: printf("%d,", i); It prints the value of local variable i. So, it prints '3'.
Hence the output is "4 3 4 3".
Functions problems
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- 1. What will be the output of the program?
#include<stdio.h> int addmult(int ii, int jj) { int kk, ll; kk = ii + jj; ll = ii * jj; return (kk, ll); } int main() { int i=3, j=4, k, l; k = addmult(i, j); l = addmult(i, j); printf("%d, %d\n", k, l); return 0; }
Options- A. 12, 12
- B. 7, 7
- C. 7, 12
- D. 12, 7 Discuss
Correct Answer: 12, 12
Explanation:
Step 1: int i=3, j=4, k, l; The variables i, j, k, l are declared as an integer type and variable i, j are initialized to 3, 4 respectively.The function addmult(i, j); accept 2 integer parameters.
Step 2: k = addmult(i, j); becomes k = addmult(3, 4)
In the function addmult(). The variable kk, ll are declared as an integer type int kk, ll;
kk = ii + jj; becomes kk = 3 + 4 Now the kk value is '7'.
ll = ii * jj; becomes ll = 3 * 4 Now the ll value is '12'.
return (kk, ll); It returns the value of variable ll only.
The value 12 is stored in variable 'k'.
Step 3: l = addmult(i, j); becomes l = addmult(3, 4)
kk = ii + jj; becomes kk = 3 + 4 Now the kk value is '7'.
ll = ii * jj; becomes ll = 3 * 4 Now the ll value is '12'.
return (kk, ll); It returns the value of variable ll only.
The value 12 is stored in variable 'l'.
Step 4: printf("%d, %d\n", k, l); It prints the value of k and l
Hence the output is "12, 12".
- 2. If int is 2 bytes wide.What will be the output of the program?
#include <stdio.h> void fun(char**); int main() { char *argv[] = {"ab", "cd", "ef", "gh"}; fun(argv); return 0; } void fun(char **p) { char *t; t = (p+= sizeof(int))[-1]; printf("%s\n", t); }
Options- A. ab
- B. cd
- C. ef
- D. gh Discuss
Correct Answer: cd
Explanation:
Since C is a machine dependent language sizeof(int) may return different values.The output for the above program will be cd in Windows (Turbo C) and gh in Linux (GCC).
To understand it better, compile and execute the above program in Windows (with Turbo C compiler) and in Linux (GCC compiler).
- 3. What will be the output of the program?
#include<stdio.h> int main() { int i=2; printf("%d, %d\n", ++i, ++i); return 0; }
Options- A. 3, 4
- B. 4, 3
- C. 4, 4
- D. Output may vary from compiler to compiler Discuss
Correct Answer: Output may vary from compiler to compiler
Explanation:
The order of evaluation of arguments passed to a function call is unspecified.Anyhow, we consider ++i, ++i are Right-to-Left associativity. The output of the program is 4, 3.
In TurboC, the output will be 4, 3.
In GCC, the output will be 4, 4.
- 4. What will be the output of the program?
#include<stdio.h> int main() { int i=-3, j=2, k=0, m; m = ++i && ++j && ++k; printf("%d, %d, %d, %d\n", i, j, k, m); return 0; }
Options- A. -2, 3, 1, 1
- B. 2, 3, 1, 2
- C. 1, 2, 3, 1
- D. 3, 3, 1, 2 Discuss
Correct Answer: -2, 3, 1, 1
Explanation:
Step 1: int i=-3, j=2, k=0, m; here variable i, j, k, m are declared as an integer type and variable i, j, k are initialized to -3, 2, 0 respectively.Step 2: m = ++i && ++j && ++k;
becomes m = -2 && 3 && 1;
becomes m = TRUE && TRUE; Hence this statement becomes TRUE. So it returns '1'(one). Hence m=1.
Step 3: printf("%d, %d, %d, %d\n", i, j, k, m); In the previous step the value of i,j,k are increemented by '1'(one).
Hence the output is "-2, 3, 1, 1".
- 5. What will be the output of the program?
#include<stdio.h> int main() { int a=100, b=200, c; c = (a == 100 || b > 200); printf("c=%d\n", c); return 0; }
Options- A. c=100
- B. c=200
- C. c=1
- D. c=300 Discuss
Correct Answer: c=1
Explanation:
Step 1: int a=100, b=200, c;
Step 2: c = (a == 100 || b > 200);
becomes c = (100 == 100 || 200 > 200);
becomes c = (TRUE || FALSE);
becomes c = (TRUE);(ie. c = 1)
Step 3: printf("c=%d\n", c); It prints the value of variable i=1
Hence the output of the program is '1'(one).- 6. What will be the output of the program?
#include<stdio.h> int sumdig(int); int main() { int a, b; a = sumdig(123); b = sumdig(123); printf("%d, %d\n", a, b); return 0; } int sumdig(int n) { int s, d; if(n!=0) { d = n%10; n = n/10; s = d+sumdig(n); } else return 0; return s; }
Options- A. 4, 4
- B. 3, 3
- C. 6, 6
- D. 12, 12 Discuss
Correct Answer: 6, 6
- 7. What will be the output of the program?
#include<stdio.h> int addmult(int ii, int jj) { int kk, ll; kk = ii + jj; ll = ii * jj; return (kk, ll); } int main() { int i=3, j=4, k, l; k = addmult(i, j); l = addmult(i, j); printf("%d %d\n", k, l); return 0; }
Options- A. 12 12
- B. No error, No output
- C. Error: Compile error
- D. None of above Discuss
Correct Answer: 12 12
- 8. What will be the output of the program?
#include<stdio.h> int main() { void fun(char*); char a[100]; a[0] = 'A'; a[1] = 'B'; a[2] = 'C'; a[3] = 'D'; fun(&a[0]); return 0; } void fun(char *a) { a++; printf("%c", *a); a++; printf("%c", *a); }
Options- A. AB
- B. BC
- C. CD
- D. No output Discuss
Correct Answer: BC
- 9. What will be the output of the program?
#include<stdio.h> int fun(int, int); typedef int (*pf) (int, int); int proc(pf, int, int); int main() { printf("%d\n", proc(fun, 6, 6)); return 0; } int fun(int a, int b) { return (a==b); } int proc(pf p, int a, int b) { return ((*p)(a, b)); }
Options- A. 6
- B. 1
- C. 0
- D. -1 Discuss
Correct Answer: 1
- 10. What will be the output of the program?
#include<stdio.h> int i; int fun(); int main() { while(i) { fun(); main(); } printf("Hello\n"); return 0; } int fun() { printf("Hi"); }
Options- A. Hello
- B. Hi Hello
- C. No output
- D. Infinite loop Discuss
Correct Answer: Hello
Explanation:
Step 1: int i; The variable i is declared as an integer type.Step 1: int fun(); This prototype tells the compiler that the function fun() does not accept any arguments and it returns an integer value.
Step 1: while(i) The value of i is not initialized so this while condition is failed. So, it does not execute the while block.
Step 1: printf("Hello\n"); It prints "Hello".
Hence the output of the program is "Hello".
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- 1. What will be the output of the program?