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