Answer / susie

Answer :

1

Explanation:

before entering into the for loop the checking condition is
"evaluated". Here it evaluates to 0 (false) and comes out of
the loop, and i is incremented (note the semicolon after the
for loop).

#include<stdio.h> void fun(int); int main() { int a; a=3; fun(a); printf("\n"); return 0; } void fun(int i) { if(n>0) { fun(--n); printf("%d",n); fun(--n); } } the answer is 0 1 2 0..someone explain how the code is executed..?

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