Output of following program?
C
#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 4000
(B)
1000 2000 4000 4000 2000 1000
(C)
1000 2000 4000 2000 1000
(D)
1000 2000 2000 1000
Answer: (B)
Explanation:
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 print.
Hence Option (B) is correct.
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