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- Question
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
- Correct Answer
- 6, 6
- 1. 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 Discuss
- 2. 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
- 3. 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
- 4. 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
- 5. 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
- 6. 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
- 7. 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
- 8. 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
- 9. 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
- 10. What will be the output of the program?
#include<stdio.h> int main() { int i=1; if(!i) printf("CuriousTab,"); else { i=0; printf("C-Program"); main(); } return 0; }
Options- A. prints "CuriousTab, C-Program" infinitely
- B. prints "C-Program" infinetly
- C. prints "C-Program, CuriousTab" infinitely
- D. Error: main() should not inside else statement Discuss
Functions problems
Search Results
Correct Answer: 4, 3, 4, 3
Explanation:
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".
Correct Answer: 12, 12
Explanation:
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".
Correct Answer: cd
Explanation:
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).
Correct Answer: Output may vary from compiler to compiler
Explanation:
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.
Correct Answer: -2, 3, 1, 1
Explanation:
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".
Correct Answer: 12 12
Correct Answer: BC
Correct Answer: 1
Correct Answer: Hello
Explanation:
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".
Correct Answer: prints "C-Program" infinetly
Explanation:
Step 2: if(!i) Here the !(NOT) operator reverts the i value 1 to 0. Hence the if(0) condition fails. So it goes to else part.
Step 3: else { i=0; In the else part variable i is assigned to value 0(zero).
Step 4: printf("C-Program"); It prints the "C-program".
Step 5: main(); Here we are calling the main() function.
After calling the function, the program repeats from step 1 to step 5 infinitely.
Hence it prints "C-Program" infinitely.
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