[ILUG] Re: C string concat question ??

[kevin lyda]

On Fri, May 04, 2001 at 11:52:09AM +0000, Conor Daly wrote:
> only strncpy() needs to be followed by a null terminator and only if it
> reads less than the length of src.  so, assuming 
> 
> char *src;
> 
> strncpy(dest, src, strlen(src) * sizeof(char));
> 
> dest will be null terminated.  Or does strlrn() leave out the null
> terminator when returning a length?

either you don't understand or you're leaving out information.  so...

#define STRSIZE 80
char *
strdup2(char *s1, char *s2)
{
    char *s;

    if ((s=(char *)malloc(STRSIZE + 1)) == NULL) {
	return NULL;
    }
    /* the following initialises the string.  s[0]=0;s[STRSIZE]=0; *
     * would work too.                                             */
    memset(s, 0, STRSIZE + 1);
    /* from here on in we don't let str* routines near s[STRSIZE]  *
     * so we know it will stay 0.                                  */
    strncpy(s, s1, STRSIZE);
    strncpy(s, s2, STRSIZE - strlen(s));
    return s;
}

strlen works by counting chars until it reaches '\0'.  therefore don't use
    it if you're not sure the string is null terminated.
sizeof(char) is not required.

kevin

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