Why does the below code not produce any output ?

[radha gogia]

#include<stdio.h>
int gcd(int m ,int n)
{
if(m%n==0)
return n;
else return(n,m%n);
}
 
int main()
{
int x,y;
printf("enter two numbers");
scanf("%d",&x);
scanf("%d",&y);
int a=gcd(x,y);
if(a==1)
{
printf("the numbers are co-prime");
}
return 0;
}

Here when I run the code just after taking input there is no output so what is the reason for it , I gave input as 7 and 9 and I didn't get any output ,plz clarify this .

[weaknessforcats]

What is gcd returning? If m is 7 and n is 9 then m%n is 7%9 which is 7. So then m%n ==0 is false. Therefore you return (n,m%n). This looks odd but it is a valid a valid expression using the comma operator. SO first there's n and then there's m%n and that's what is returned.

Two things, first write your code such that you can see the return value from your functions. Second, I have the feeling this gcd function is recursive but you have omitted the gcd in that return statement.

Post again and let me know what happened.

[radha gogia]

First it would be
gcd(7,9) --> gcd(9,7)--> gcd(7,2) -->gcd(2,1) now since m%n==0 hence it should return n which is equal to 1 , where's the problem then ?why is the code not producing the output ?

[weaknessforcats]

If you look inside the gcd function you see:

int gcd(int m ,int n)
 {
 if(m%n==0)
 return n;
 else return(n,m%n);   <----------------
 }
 

So it's not doing this: gcd(7,9) --> gcd(9,7)--> gcd(7,2) -->gcd(2,1)

I think the code should be:

int gcd(int m ,int n)
 {
 if(m%n==0)
 return n;
 else return gcd(n,m%n);   <----------------revised
 }
 

This is a case where stepping though the code with a debugger can save a lot of time.