Can a C++ user-defined literal operator ever be passed a null pointer?
This is really happening with an experimental version of g++ (gcc version 4.7.0 20111114 (experimental) [trunk revision 181364] (Debian 20111114-1)) but I am unsure if this is a bug (90% sure) or some strange expected behavior.
Example program:
#include <iostream>
#include <stdexcept>
#include <string>
std::string operator "" _example (const char * text) {
using std::cerr;
using std::endl;
cerr << "text (pointer) = " << static_cast<const void *>(text) << endl;
if (!text) throw std::runtime_error("Is a null pointer really expected?");
cerr << "text (string) = \"" << text << '"' << endl;
return text;
}
int main() {
2_example;
1_example;
0_example;
}
Output (probably a bug in gcc ... but maybe not?!, hence the question):
text (pointer) = 0x8048d49
text (string) = "2"
text (pointer) = 0x8048d4b
text (string) = "1"
text (pointer) = 0
terminate called after throwing an instance of 'std::runtime_error'
what(): Is a null pointer really expected?
Aborted
It's not just "0_example"; it's whenever the literal value is zero. For example, it still happens even when the literal is "0x0000_example".
Is this a bug? Or some strange special-case when the literal's value is zero?
