#include<stdio.h>
int f(int,int);
int main()
{
printf("%d",f(20,1));
return 0;
}
int f(int n,int k)
{
if(n==0) return 0;
else if(n%2)return f(n/2,2*k)+k;
else return f(n/2,2*k)-k;
}
how this program is working and generating output as 9....?
Re: #include<stdio.h>
int f(int,int);
int main()
{
printf("%d",f(20,1));
return 0;
}
int f(int n,int k)
{
if(n==0) return 0;
else if(n%2)return f(n/2,2*k)+k;
else return f(n/2,2*k)-k;
}
how this program is working and generating output as 9....?
n=20, k =1
if, else if false. so it will call
f(n/2,2*k)-k ==> f(10,2)-1
if, else if false
f(n/2,2*k)-k ==> f(5, 4)-2
if is false. else if is true
f(n/2,2*k)+k ==> f(2, 8)+4
if, else if false
f(n/2,2*k)-k ==> f(1, 16)-8
if is false. else if is true
f(n/2,2*k)+k ==> f(0, 32)+16
now n is zero.
what would be the output of the follwing
struct st
{
char name[20];
int i;
float f;
};
main()
{
struct st emp = {"forum"};
printf("%d %f",emp.i,emp.f);
}
what is the difference between : func (int list[], ...) or
func (int *list , ....) - what is the difference if list is an array and if also if list is a pointer
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