Question

i have this piece of code

int main()
{
    char a[100];
    a[0]='a';
    a[1]='b';
    fun(a[0]);
}

void fun(char *a)
{
    printf("%c",a);
}

but im passing a character to a pointer.will the pointer not be expecting an address???

Answer 1

a[0] holds the value 97 ('a' in ASCII). fun will receive the value 97 in a but interpret it as an address. However, since you’re only passing it to printf, and happen to incorrectly be using the %c formatter which will interpret a as a char, you’ll end up printing a anyway.

Of course, on most compilers you should receive warnings that:

1. You are converting an integer into a pointer (when you call fun) without casting it to a pointer.
2. The %c formatter in printf should take a char, not a char *.