什么是C ++ 20的字符串文字运算符模板？

Question

What is C++20's string literal operator template? Cppreference's example in this respect is quite concise and not very clear to me:

struct A { A(const char *); auto operator<=>(const A&) const = default; };

template<A a> A operator ""_a(); 

In trying to understand what this feature is I've just learned that you can have numeric literal operator templates in C++, which make each digit of numerical constant be passed as a non-type argument to a template (cf. a better explanation here ). Currently, literal operator templates do not work with character literals, though there are compilers extensions enabling that. I don't think C++20's string literal operator templates have anything to do with that as I've learned that proposals to extend literal operator templates to work with character literals were voted down in the commitee?

Answer 1

There were two separate proposals:

• Allowing string literals as non-type template parameters ( P0424 )
• Allowing class types as non-type template parameters ( P0732 )

The first proposal was partially merged into the second. String literals still are not valid arguments as non-type template parameters, but they are valid arguments into class types. The example from [temp.arg.nontype]/4 might help:

 template<class T, T p> class X { /* ... */ }; X<const char*, "Studebaker"> x; // error: string literal as template-argument const char p[] = "Vivisectionist"; X<const char*, p> y; // OK struct A { constexpr A(const char*) {} friend auto operator<=>(const A&, const A&) = default; }; X<A, "Pyrophoricity"> z; // OK, string literal is a constructor argument to A 

However, the part of the first proposal which extended the literal operators was what was merged into the second, [lex.ext]/5 :

If S contains a literal operator template with a non-type template parameter for which str is a well-formed template-argument, the literal L is treated as a call of the form operator "" X<str>()

So using this:

struct A { A(const char *); auto operator<=>(const A&) const = default; };     
template<A a> A operator ""_a() { return a; }

We can write "Hello"_a , which will be interpreted as calling operator "" _a<A("Hello")> .

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Note that these rules are slightly in flux, as the defaulted <=> requirement will be changing to a defaulted == requirement as per P1185 .