C: Finding end of string: *s++ VS *s then s++

Question

I'm writing a simple string concatenation program.

The program works the way I have posted it. However, I first wrote it using the following code to find the end of the string:

while (*s++)
    ;

However, that method didn't work. The strings I passed to it weren't copied correctly. Specifically, I tried to copy "abc" to a char[] variable that held "\0".

From reading the C K&R book, it looks like it should work. That compact form should take the following steps.

1. *s is compared with '\0'
2. s points to the next address

So why doesn't it work? I am compiling with gcc on Debian.

I found that this version does work:

strncat(char *s, const char *t, int n)
{
    char *s_start = s;

    while (*s)
        s++;

    for ( ; n > 0 && *t; n--, s++, t++)
        *s = *t;

    *(s++) = '\0';

    return s_start;
}

Thanks in advance.

Answer 1

After the end of while (*s++);, s points to the character after the null terminator. Take that into account in the code that follows.

Answer 2

There is difference. In the first case, s will point to the position after '\0', while the second stops right at '\0'.

Answer 3

The problem is that

 while (*s++)
     ;

Always Increments s, even when s is zero (*s is false)

 while (*s)
    s++;

only increments s when *s is nonzero

so the first one will leave s pointing to first character after the first \0, while the second one will leave s pointing to the first \0.

Answer 4

In addition what has been said, note that in C it is technically illegal for a pointer to point to unallocated memory, even if you don't dereference it. So be sure to fix your program, even if it appears to work.

Answer 5

As John Knoeller said, at the end of the run it'll s will point to the location after the NULL. BUT There is no need to sacrifice performance for the correct solution.. Take a look for yourself:

while (*s++) --s;

Should do the trick.