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- Question
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
- Correct Answer
- Hello
ExplanationStep 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".
Functions problems
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- 1. 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
- 2. 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
- 3. 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
- 4. 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
- 5. 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
Correct Answer: 4, 3, 4, 3
Explanation:
Step 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".
- 6. 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
Correct Answer: prints "C-Program" infinetly
Explanation:
Step 1: int i=1; The variable i is declared as an integer type and initialized to 1(one).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.
- 7. What will be the output of the program?
#include<stdio.h> int fun(int(*)()); int main() { fun(main); printf("Hi\n"); return 0; } int fun(int (*p)()) { printf("Hello "); return 0; }
Options- A. Infinite loop
- B. Hi
- C. Hello Hi
- D. Error Discuss
Correct Answer: Hello Hi
- 8. What will be the output of the program?
#include<stdio.h> int fun(int i) { i++; return i; } int main() { int fun(int); int i=3; fun(i=fun(fun(i))); printf("%d\n", i); return 0; }
Options- A. 5
- B. 4
- C. Error
- D. Garbage value Discuss
Correct Answer: 5
Explanation:
Step 1: int fun(int); This is prototype of function fun(). It tells the compiler that the function fun() accept one integer parameter and returns an integer value.Step 2: int i=3; The variable i is declared as an integer type and initialized to value 3.
Step 3: fun(i=fun(fun(i)));. The function fun(i) increements the value of i by 1(one) and return it.
Lets go step by step,
=> fun(i) becomes fun(3) is called and it returns 4.
=> i = fun(fun(i)) becomes i = fun(4) is called and it returns 5 and stored in variable i.(i=5)
=> fun(i=fun(fun(i))); becomes fun(5); is called and it return 6 and nowhere the return value is stored.
Step 4: printf("%d\n", i); It prints the value of variable i.(5)
Hence the output is '5'.
- 9. What will be the output of the program?
#include<stdio.h> int func1(int); int main() { int k=35; k = func1(k=func1(k=func1(k))); printf("k=%d\n", k); return 0; } int func1(int k) { k++; return k; }
Options- A. k=35
- B. k=36
- C. k=37
- D. k=38 Discuss
Correct Answer: k=38
Explanation:
Step 1: int k=35; The variable k is declared as an integer type and initialized to 35.Step 2: k = func1(k=func1(k=func1(k))); The func1(k) increement the value of k by 1 and return it. Here the func1(k) is called 3 times. Hence it increements value of k = 35 to 38. The result is stored in the variable k = 38.
Step 3: printf("k=%d\n", k); It prints the value of variable k "38".
- 10. What will be the output of the program in 16 bit platform (Turbo C under DOS)?
#include<stdio.h> int main() { int fun(); int i; i = fun(); printf("%d\n", i); return 0; } int fun() { _AX = 1990; }
Options- A. Garbage value
- B. 0 (Zero)
- C. 1990
- D. No output Discuss
Correct Answer: 1990
Explanation:
Turbo C (Windows): The return value of the function is taken from the Accumulator _AX=1990.But it may not work as expected in GCC compiler (Linux).
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- 1. What will be the output of the program?