User defined literals for variadic char template

Question

Recently, in the gcc-trunk sources was implemented the "user defined literals". Tell me please, do I understand correctly that I can`t define a "user defined literals" for variadic char template?

template<char... chars>
int operator"" _call() { return sizeof...(chars); }
...
std::cout << "method"_call;

Up.

I don`t understand why this expression is allowed:

template<char... chars>
int operator"" _call() { return sizeof...(chars); }
...
std::cout << 12345566_call;

and this one is disallowed:

template<char... chars>
int operator"" _call() { return sizeof...(chars); }
...
std::cout << method_call;

?

What's the point?

Up. this is because of the ambiguity?

Thanks.

Answer 1

method_call is a valid identifier As is for example some_call or my_call. Now imagine how much code would be broken if such identifiers were allowed to be redefined by operator"".

Answer 2

No, it doesn't really make sense. String literals are passed as two arguments to operator"", and one of them is size, so what you want is:

size_t operator"" _call(const char*, size_t len) {
    return len;
}

Standard quote time (2.14.8.5):

5 If L is a user-defined-string-literal, let str be the literal without its ud-suffix and let len be the number of code units in str (i.e., its length excluding the terminating null character). The literal L is treated as a call of the form

operator "" X (str, len)

The variadic template forms are considered only for user-defined-integer-literal (2.14.8.3) and user-defined-floating-literal (2.14.8.4).

As for method_call, method is not a literal.