What is C++20's string literal operator template?
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?
c++ c++20 user-defined-literals
asked 2 days ago
lukeglukeg
1,7961826
add a comment |
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?
c++ c++20 user-defined-literals
asked 2 days ago
lukeglukeg
1,7961826
Does the proposal answer your question?
– Lightness Races in Orbit
2 days ago
@LightnessRacesinOrbit That's the version that got voted down.
– cpplearner
2 days ago
Okay, how about this one?
– Lightness Races in Orbit
2 days ago
add a comment |
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?
c++ c++20 user-defined-literals
asked 2 days ago
lukeglukeg
1,7961826
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?
c++ c++20 user-defined-literals
c++ c++20 user-defined-literals
asked 2 days ago
lukeglukeg
1,7961826
asked 2 days ago
lukeglukeg
1,7961826
asked 2 days ago
lukeglukeg
1,7961826
asked 2 days ago
lukeglukeg
1,7961826
asked 2 days ago
lukeglukeg
1,7961826
1,7961826
Does the proposal answer your question?
– Lightness Races in Orbit
2 days ago
@LightnessRacesinOrbit That's the version that got voted down.
– cpplearner
2 days ago
Okay, how about this one?
– Lightness Races in Orbit
2 days ago
add a comment |
Does the proposal answer your question?
– Lightness Races in Orbit
2 days ago
@LightnessRacesinOrbit That's the version that got voted down.
– cpplearner
2 days ago
Okay, how about this one?
– Lightness Races in Orbit
2 days ago
Does the proposal answer your question?
– Lightness Races in Orbit
2 days ago
Does the proposal answer your question?
– Lightness Races in Orbit
2 days ago
@LightnessRacesinOrbit That's the version that got voted down.
– cpplearner
2 days ago
@LightnessRacesinOrbit That's the version that got voted down.
– cpplearner
2 days ago
Okay, how about this one?
– Lightness Races in Orbit
2 days ago
Okay, how about this one?
– Lightness Races in Orbit
2 days ago
add a comment |
1 Answer
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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")>.
Note that these rules are slightly in flux, as the defaulted <=> requirement will be changing to a defaulted == requirement as per P1185.
answered 2 days ago
BarryBarry
180k19312572
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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")>.
Note that these rules are slightly in flux, as the defaulted <=> requirement will be changing to a defaulted == requirement as per P1185.
answered 2 days ago
BarryBarry
180k19312572
add a comment |
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")>.
Note that these rules are slightly in flux, as the defaulted <=> requirement will be changing to a defaulted == requirement as per P1185.
answered 2 days ago
BarryBarry
180k19312572
add a comment |
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")>.
Note that these rules are slightly in flux, as the defaulted <=> requirement will be changing to a defaulted == requirement as per P1185.
answered 2 days ago
BarryBarry
180k19312572
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")>.
Note that these rules are slightly in flux, as the defaulted <=> requirement will be changing to a defaulted == requirement as per P1185.
answered 2 days ago
BarryBarry
180k19312572
answered 2 days ago
BarryBarry
180k19312572
answered 2 days ago
BarryBarry
180k19312572
answered 2 days ago
BarryBarry
180k19312572
180k19312572
add a comment |
add a comment |
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Does the proposal answer your question?
– Lightness Races in Orbit
2 days ago
@LightnessRacesinOrbit That's the version that got voted down.
– cpplearner
2 days ago
Okay, how about this one?
– Lightness Races in Orbit
2 days ago