Why there is no error when int used

Note also that if x >= 1 then after completing the loop you fall out of the function fun, but you do not have a return statement. So whatever value happens to be in a particular machine register at that point will be received by the caller, and printed as a character. You need to remove the else statement before return c; .

Have also a look at this page: Implicit conversions - cppreference.com[^].

Quote:

infact when i am using char in place of int it is showing error.

Because of your compiler oddities.
A C compiler should not even compile your code.

#include<stdio.h>
// Here is missing the func prototype,
int main()
{
    int n,i;
    printf("enter n\n");
    scanf("%d",&n);
    fun(n);
}
int fun(int x)
{int i;
char c='*';
    if(x>=1)
    {for(i=1;i<=x;i++)
    printf("%c",fun(x-1));
// here is missing a return
    }
    else 
    return c;
}

Quote:

why it is not showing error as return value is not of type int.

Because of compiler internals, it can handle some things and not some others.
No one can predict what will be ok or not without empirical experiment.