memchr implementation

I have a small doubt in the implementation of memchr function implementation in the code shown below.

[insert]

Code:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

void * xmemchr(const void *buffer, int ch, size_t count);

int main(int argc, char *argv[])
{
	char *p;
	char *s = "This is a test";

	//p = memchr(s,'e',14);
	p = xmemchr(s,'e',14);
	printf("\n %p \t %s \t %c \n", p, p, *p);
	return 0;
}

void * xmemchr(const void *buffer, int ch, size_t count)
{
	int i;
	char c;
	char *s = buffer;
	
	for(i = 0; i< count ; i++)
	{
		if(*s == ch){
			
			return s;
		}
		else
			s++;
	}
}

In the line
if(*s == ch)

i was assuming how can i compare the char contained in s and the int value that ch has or is it that the compiler does this check on its own. I mean how is it possible to compare a char and an integer. Though i got it working but this point is yet unclear to me.

They're both numbers, aren't they? If they're the same number, then they're equal; if not, they're not.

Yes they are both numbers in the internal machine language but when i say *s is it not that i am getting the value contained at address in s which is a character for example 'e' in this case.

My point was that would the check condition like this be wrong

*s == 'e' for the above case.

*s is the value that s points to, yes. s is a char*, so *s must be a char, yes.

So, for instance, if s was "Hello world", then the first time *s would be 72, then 101, then 108, then 108 again, and so on. s is a movable pointer, so s++ makes it point to each character in turn.

Thanks tabstop i get the point and also i tried the following

*s == 'e' and it seems to work. So my thinking that this should work was correct as well.