strncat Wformat-overflow warning when using gcc 8.2.1

Question

I'm using gcc 8.2.1 and trying to build this code:

std::string dir = "Documents";
char * _tempname = static_cast <char*> (malloc( dir.length() + 14));
strncpy (_tempname, dir.c_str(), dir.length()+1 );
strncat (_tempname, "/hellooXXXXXX", 13);

but it gives me this warning:

warning: 'char* strncat(char*, const char*, size_t)' specified bound 13 equals source length [-Wstringop-overflow=]

After searching I found that it's an overflow problem to have the size_t equals the source length according to the discussion in this link but I couldn't understand why this is considered a problem and why this overflows the destination. And how could I remove this warning without changing my code?

Emia

strncat expects a C-string as a first argument, but you're passing some uninitialized char array to it. Also why you're allocating memory manually when you could simply do std::string tempname = dir + "/hellooXXXXXX"? – Engage

I forgot a line of code and I edited it now sorry, regarding the code why I'm not using string, actually I'm maintaining this code not writing it from scratch. – Emia

Answer 1

Apparently GCC understands that strncat(_tempname, "/hellooXXXXXX", 13); is no different from strcat(_tempname, "/hellooXXXXXX");, and finds it suspicious that you're using former instead of the latter.

If you can change the code, use strcat instead (or even better, rewrite the thing to use std::string).

If you can't change the code, use -Wno-stringop-overflow flag to disable the warning.

Engage

Answer 2

my understanding is that gcc issues this warning only because it's a common mistake for users to set the bound equal to the src length, nothing more.

Underhung

Answer 3

The function expects the space left in the destination not the length of the source string, so use

// ...
strncat(_tempname, "/hellooXXXXXX", dir.length() + 14 - strlen(_tempname) - 1);` 

instead. No, forget that. Use std::string instead.

Putrescine