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- Question
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
- Correct Answer
- 5
ExplanationStep 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'.
Functions problems
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- 1. 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
- 2. 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.
- 3. 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".
- 4. 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
- 5. 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
- 6. 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".
- 7. 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).
- 8. What will be the output of the program?
#include<stdio.h> void fun(int*, int*); int main() { int i=5, j=2; fun(&i, &j); printf("%d, %d", i, j); return 0; } void fun(int *i, int *j) { *i = *i**i; *j = *j**j; }
Options- A. 5, 2
- B. 10, 4
- C. 2, 5
- D. 25, 4 Discuss
Correct Answer: 25, 4
Explanation:
Step 1: int i=5, j=2; Here variable i and j are declared as an integer type and initialized to 5 and 2 respectively.Step 2: fun(&i, &j); Here the function fun() is called with two parameters &i and &j (The & denotes call by reference. So the address of the variable i and j are passed. )
Step 3: void fun(int *i, int *j) This function is called by reference, so we have to use * before the parameters.
Step 4: *i = *i**i; Here *i denotes the value of the variable i. We are multiplying 5*5 and storing the result 25 in same variable i.
Step 5: *j = *j**j; Here *j denotes the value of the variable j. We are multiplying 2*2 and storing the result 4 in same variable j.
Step 6: Then the function void fun(int *i, int *j) return back the control back to main() function.
Step 7: printf("%d, %d", i, j); It prints the value of variable i and j.
Hence the output is 25, 4.
- 9. What will be the output of the program?
#include<stdio.h> int main() { int fun(int); int i = fun(10); printf("%d\n", --i); return 0; } int fun(int i) { return (i++); }
Options- A. 9
- B. 10
- C. 11
- D. 8 Discuss
Correct Answer: 9
Explanation:
Step 1: int fun(int); Here we declare the prototype of the function fun().Step 2: int i = fun(10); The variable i is declared as an integer type and the result of the fun(10) will be stored in the variable i.
Step 3: int fun(int i){ return (i++); } Inside the fun() we are returning a value return(i++). It returns 10. because i++ is the post-increement operator.
Step 4: Then the control back to the main function and the value 10 is assigned to variable i.
Step 5: printf("%d\n", --i); Here --i denoted pre-increement. Hence it prints the value 9.
- 10. What will be the output of the program?
#include<stdio.h> int reverse(int); int main() { int no=5; reverse(no); return 0; } int reverse(int no) { if(no == 0) return 0; else printf("%d,", no); reverse (no--); }
Options- A. Print 5, 4, 3, 2, 1
- B. Print 1, 2, 3, 4, 5
- C. Print 5, 4, 3, 2, 1, 0
- D. Infinite loop Discuss
Correct Answer: Infinite loop
Explanation:
Step 1: int no=5; The variable no is declared as integer type and initialized to 5.Step 2: reverse(no); becomes reverse(5); It calls the function reverse() with '5' as parameter.
The function reverse accept an integer number 5 and it returns '0'(zero) if(5 == 0) if the given number is '0'(zero) or else printf("%d,", no); it prints that number 5 and calls the function reverse(5);.
The function runs infinetely because the there is a post-decrement operator is used. It will not decrease the value of 'n' before calling the reverse() function. So, it calls reverse(5) infinitely.
Note: If we use pre-decrement operator like reverse(--n), then the output will be 5, 4, 3, 2, 1. Because before calling the function, it decrements the value of 'n'.
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- 1. What will be the output of the program?