Why is passing a string literal into a char* argument only sometimes a compiler error?

Posted by Brian Postow on Stack Overflow See other posts from Stack Overflow or by Brian Postow
Published on 2010-05-03T19:04:21Z Indexed on 2010/05/03 20:28 UTC
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I'm working in a C, and C++ program. We used to be compiling without the make-strings-writable option. But that was getting a bunch of warnings, so I turned it off.

Then I got a whole bunch of errors of the form "Cannot convert const char* to char* in argmuent 3 of function foo". So, I went through and made a whole lot of changes to fix those.

However, today, the program CRASHED because the literal "" was getting passed into a function that was expecting a char*, and was setting the 0th character to 0. It wasn't doing anything bad, just trying to edit a constant, and crashing.

My question is, why wasn't that a compiler error?

In case it matters, this was on a mac compiled with gcc-4.0.

EDIT: added code:

char * host = FindArgDefault("EMailLinkHost", "");
stripCRLF(linkHost, '\n');

where:

char *FindArgDefault(char *argName, char *defVal) 
{// simplified
    char * val = defVal;
    return(val);
}

and

void stripCRLF(char *str, char delim)
{
    char *p, *q;

    for (p = q = str; *p; ++p) {
        if (*p == 0xd || *p == 0xa) {
            if (p[1] == (*p ^ 7)) ++p;
            if (delim == -1) *p = delim;
            }
        *q++ = *p;
        }
    *q = 0;  // DIES HERE
}

This compiled and ran until it tried to set *q to 0...

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