Output of following program?#includevoid print(int n){ if (n 4000)return;printf("%d ", n);print(2*n);printf("%d ", n);}int main(){print(1000);getchar();return 0;}a)1000 2000 4000b)1000 2000 4000 4000 2000 1000c)1000 2000 4000 2000 1000d)1000 2000 2000 1000Correct answer is option 'B'. Can you explain this answer?

Question

Output of following program?#include<stdio.h>void print(int n){  if (n > 4000)return;printf("%d ", n);print(2*n);printf("%d ", n);}int main(){print(1000);getchar();return 0;}a)1000 2000 4000b)1000 2000 4000 4000 2000 1000c)1000 2000 4000 2000 1000d)1000 2000 2000 1000Correct answer is option 'B'. Can you explain this answer?

Answer

First time n=1000 Then 1000 is printed by first printf function then call print(2*1000) then again print 2000 by printf function then call print(2*2000) and it prints 4000 next time print(4000*2) is called. Here 8000 is greater than 4000 condition becomes true and it return at function(2*4000). Here n=4000 then 4000 will again print through second printf. Similarly print(2*2000) after that n=2000 then 2000 will print and come back at print(2*1000) here n=1000, so print 1000 through second printf. Option (B) is correct.

Answer

Explanation:

print() function:
- The print() function takes an integer n as input.
- It first checks if n is greater than 4000, if yes, it returns from the function.
- Then, it prints the value of n.
- It recursively calls the print() function with the value 2*n.
- Finally, it prints the value of n again.

Main function:
- In the main function, print(1000) is called.
- The print() function is called with n=1000.
- It prints 1000 and then calls print(2000).
- Inside the print(2000) function, it prints 2000 and then calls print(4000).
- Inside the print(4000) function, it prints 4000 and since 4000 is greater than 4000, it returns without printing.
- Now, as we unwind the recursion, it prints 4000 (the second time), then 2000, and finally 1000.

Therefore, the output of the program is: 1000 2000 4000 4000 2000 1000

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