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#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..?
- 1 Answers
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- Wipro, I also Faced
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Answer / siva kumar
for example :
Here a is three(3).
calling fun(3) in main function.
FUN(3) {
calling fun(2) {
calling fun(1) {
calling fun(0) {
N>0 found.
}
printed 0
calling second fun(-1) {
N>0 found.
}
}
printed 1
calling second fun(0) {
N>0 found.
}
}
printed 2
calling second fun(1) {
calling fun(0) {
N>0 found.
}
printed 0
calling second fun(-1) {
N>0 found.
}
}
}
| Is This Answer Correct ? | 27 Yes | 12 No |
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