In "User-defined literals" on cppreference.com, what does it mean by this?
b) otherwise, the overload set must include either, but not both, a raw literal operator or a numeric literal operator template. If the overload set includes a raw literal operator, the user-defined literal expression is treated as a function call
operator "" X("n")
Please, I need a simple example that illustrate this text.
CodePudding user response:
unsigned long long operator "" _w(unsigned long long);
unsigned operator "" _u(const char*);
int main() {
12_w; // calls operator "" _w(12ULL)
12_u; // calls operator "" _u("12")
}
A little bit changes based on the example in your link.
Here 12_w
calls operator "" _w(12ULL)
since there is a literal operator with the parameter type unsigned long long
, while 12_u
calls operator "" _u("12")
since there is only a raw literal operator.