Why does type aliasing determine whether output is L-value or R-value?
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I created a struct with two static functions for testing purposes. The first instance of f is called when an l-value reference is passed. The second instance is called when an r-value is passed:
template <typename _Tp>
struct T {
static constexpr void f(_Tp&) { std::cout << "f(T&) is called!n"; }
static constexpr void f(_Tp&&) { std::cout << "f(T&&) is called!n"; }
};
When I was experimenting with strong types, I found out the first instance, T::f(_Tp&) was called when I attempted to create the strong type implicitly. Why is this? (See following)
using T_int = T<int>;
T_int::f(
typename strong_types::create_strong_type<int, struct tag>(5)()
); // calls f::(T&) (?)
using KG = typename strong_types::create_strong_type<double, struct KG_tag>;
T_int::f(KG(4.2)()); // calls f(T&&)
Note that operator() returns the value given through the constructor.
Feel free to ask if I need to elaborate.
EDIT: strong_types is a namespace. It exists among other things of the alias create_strong_type:
namespace strong_type {
template <typename T, typename tag>
using create_strong_type = Strong_Type<T, tag>;
...
}
...
template <typename T, typename tag>
struct Strong_Type {
constexpr explicit Strong_Type(const T& value) : _value(value) {}
constexpr explicit Strong_Type(T&& value) : _value(std::move(value)) {}
constexpr T& operator()() noexcept { return _value; }
private:
T _value;
};
c++ reference c++14 lvalue type-alias
asked Nov 19 at 16:57
Lourens Dijkstra
769
|
show 3 more comments
up vote
3
down vote
favorite
I created a struct with two static functions for testing purposes. The first instance of f is called when an l-value reference is passed. The second instance is called when an r-value is passed:
template <typename _Tp>
struct T {
static constexpr void f(_Tp&) { std::cout << "f(T&) is called!n"; }
static constexpr void f(_Tp&&) { std::cout << "f(T&&) is called!n"; }
};
When I was experimenting with strong types, I found out the first instance, T::f(_Tp&) was called when I attempted to create the strong type implicitly. Why is this? (See following)
using T_int = T<int>;
T_int::f(
typename strong_types::create_strong_type<int, struct tag>(5)()
); // calls f::(T&) (?)
using KG = typename strong_types::create_strong_type<double, struct KG_tag>;
T_int::f(KG(4.2)()); // calls f(T&&)
Note that operator() returns the value given through the constructor.
Feel free to ask if I need to elaborate.
EDIT: strong_types is a namespace. It exists among other things of the alias create_strong_type:
namespace strong_type {
template <typename T, typename tag>
using create_strong_type = Strong_Type<T, tag>;
...
}
...
template <typename T, typename tag>
struct Strong_Type {
constexpr explicit Strong_Type(const T& value) : _value(value) {}
constexpr explicit Strong_Type(T&& value) : _value(std::move(value)) {}
constexpr T& operator()() noexcept { return _value; }
private:
T _value;
};
c++ reference c++14 lvalue type-alias
asked Nov 19 at 16:57
Lourens Dijkstra
769
3
what isstrong_types?
– Piotr Skotnicki
Nov 19 at 17:32
Sure, but is it a third party library ?
– Piotr Skotnicki
Nov 19 at 17:40
Oo, I am sorry for my misunderstanding.strong_typesis a namespace I created myself.
– Lourens Dijkstra
Nov 19 at 17:41
1
Also unrelated: Identifiers beginning with underscore, followed by uppercase letter are reserved and you are not allowed to use them.
– user10605163
Nov 19 at 17:45
1
Please fix the typos in your edits, the identifiers do not match, e.g.strong_typevsstrong_typesandStrong_Typeand probablyStrong_Typeshould be part of the namespace and beforecreate_strong_type?. It doesn't compile the way it is even if generously moving around code.
– user10605163
Nov 19 at 17:51
|
show 3 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I created a struct with two static functions for testing purposes. The first instance of f is called when an l-value reference is passed. The second instance is called when an r-value is passed:
template <typename _Tp>
struct T {
static constexpr void f(_Tp&) { std::cout << "f(T&) is called!n"; }
static constexpr void f(_Tp&&) { std::cout << "f(T&&) is called!n"; }
};
When I was experimenting with strong types, I found out the first instance, T::f(_Tp&) was called when I attempted to create the strong type implicitly. Why is this? (See following)
using T_int = T<int>;
T_int::f(
typename strong_types::create_strong_type<int, struct tag>(5)()
); // calls f::(T&) (?)
using KG = typename strong_types::create_strong_type<double, struct KG_tag>;
T_int::f(KG(4.2)()); // calls f(T&&)
Note that operator() returns the value given through the constructor.
Feel free to ask if I need to elaborate.
EDIT: strong_types is a namespace. It exists among other things of the alias create_strong_type:
namespace strong_type {
template <typename T, typename tag>
using create_strong_type = Strong_Type<T, tag>;
...
}
...
template <typename T, typename tag>
struct Strong_Type {
constexpr explicit Strong_Type(const T& value) : _value(value) {}
constexpr explicit Strong_Type(T&& value) : _value(std::move(value)) {}
constexpr T& operator()() noexcept { return _value; }
private:
T _value;
};
c++ reference c++14 lvalue type-alias
asked Nov 19 at 16:57
Lourens Dijkstra
769
I created a struct with two static functions for testing purposes. The first instance of f is called when an l-value reference is passed. The second instance is called when an r-value is passed:
template <typename _Tp>
struct T {
static constexpr void f(_Tp&) { std::cout << "f(T&) is called!n"; }
static constexpr void f(_Tp&&) { std::cout << "f(T&&) is called!n"; }
};
When I was experimenting with strong types, I found out the first instance, T::f(_Tp&) was called when I attempted to create the strong type implicitly. Why is this? (See following)
using T_int = T<int>;
T_int::f(
typename strong_types::create_strong_type<int, struct tag>(5)()
); // calls f::(T&) (?)
using KG = typename strong_types::create_strong_type<double, struct KG_tag>;
T_int::f(KG(4.2)()); // calls f(T&&)
Note that operator() returns the value given through the constructor.
Feel free to ask if I need to elaborate.
EDIT: strong_types is a namespace. It exists among other things of the alias create_strong_type:
namespace strong_type {
template <typename T, typename tag>
using create_strong_type = Strong_Type<T, tag>;
...
}
...
template <typename T, typename tag>
struct Strong_Type {
constexpr explicit Strong_Type(const T& value) : _value(value) {}
constexpr explicit Strong_Type(T&& value) : _value(std::move(value)) {}
constexpr T& operator()() noexcept { return _value; }
private:
T _value;
};
c++ reference c++14 lvalue type-alias
c++ reference c++14 lvalue type-alias
asked Nov 19 at 16:57
Lourens Dijkstra
769
asked Nov 19 at 16:57
Lourens Dijkstra
769
edited Nov 19 at 17:57
asked Nov 19 at 16:57
Lourens Dijkstra
769
asked Nov 19 at 16:57
Lourens Dijkstra
769
asked Nov 19 at 16:57
Lourens Dijkstra
769
769
3
what isstrong_types?
– Piotr Skotnicki
Nov 19 at 17:32
Sure, but is it a third party library ?
– Piotr Skotnicki
Nov 19 at 17:40
Oo, I am sorry for my misunderstanding.strong_typesis a namespace I created myself.
– Lourens Dijkstra
Nov 19 at 17:41
1
Also unrelated: Identifiers beginning with underscore, followed by uppercase letter are reserved and you are not allowed to use them.
– user10605163
Nov 19 at 17:45
1
Please fix the typos in your edits, the identifiers do not match, e.g.strong_typevsstrong_typesandStrong_Typeand probablyStrong_Typeshould be part of the namespace and beforecreate_strong_type?. It doesn't compile the way it is even if generously moving around code.
– user10605163
Nov 19 at 17:51
|
show 3 more comments
3
what isstrong_types?
– Piotr Skotnicki
Nov 19 at 17:32
Sure, but is it a third party library ?
– Piotr Skotnicki
Nov 19 at 17:40
Oo, I am sorry for my misunderstanding.strong_typesis a namespace I created myself.
– Lourens Dijkstra
Nov 19 at 17:41
1
Also unrelated: Identifiers beginning with underscore, followed by uppercase letter are reserved and you are not allowed to use them.
– user10605163
Nov 19 at 17:45
1
Please fix the typos in your edits, the identifiers do not match, e.g.strong_typevsstrong_typesandStrong_Typeand probablyStrong_Typeshould be part of the namespace and beforecreate_strong_type?. It doesn't compile the way it is even if generously moving around code.
– user10605163
Nov 19 at 17:51
3
3
what is
strong_types ?– Piotr Skotnicki
Nov 19 at 17:32
what is
strong_types ?– Piotr Skotnicki
Nov 19 at 17:32
Sure, but is it a third party library ?
– Piotr Skotnicki
Nov 19 at 17:40
Sure, but is it a third party library ?
– Piotr Skotnicki
Nov 19 at 17:40
Oo, I am sorry for my misunderstanding.
strong_types is a namespace I created myself.– Lourens Dijkstra
Nov 19 at 17:41
Oo, I am sorry for my misunderstanding.
strong_types is a namespace I created myself.– Lourens Dijkstra
Nov 19 at 17:41
1
1
Also unrelated: Identifiers beginning with underscore, followed by uppercase letter are reserved and you are not allowed to use them.
– user10605163
Nov 19 at 17:45
Also unrelated: Identifiers beginning with underscore, followed by uppercase letter are reserved and you are not allowed to use them.
– user10605163
Nov 19 at 17:45
1
1
Please fix the typos in your edits, the identifiers do not match, e.g.
strong_type vs strong_types and Strong_Type and probably Strong_Type should be part of the namespace and before create_strong_type?. It doesn't compile the way it is even if generously moving around code.– user10605163
Nov 19 at 17:51
Please fix the typos in your edits, the identifiers do not match, e.g.
strong_type vs strong_types and Strong_Type and probably Strong_Type should be part of the namespace and before create_strong_type?. It doesn't compile the way it is even if generously moving around code.– user10605163
Nov 19 at 17:51
|
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The difference is not due to using an alias (using), but to the type you pass as first template argument to create_strong_type. In one case, it's an int, and in the other, a double.
Try T<double>::f(KG(4.2)()); and you will see the argument is passed as lvalue reference (because of the return type of Strong_Type::operator(), which is T&).
answered Nov 19 at 17:57
Nelfeal
4,170621
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The difference is not due to using an alias (using), but to the type you pass as first template argument to create_strong_type. In one case, it's an int, and in the other, a double.
Try T<double>::f(KG(4.2)()); and you will see the argument is passed as lvalue reference (because of the return type of Strong_Type::operator(), which is T&).
answered Nov 19 at 17:57
Nelfeal
4,170621
add a comment |
up vote
1
down vote
accepted
The difference is not due to using an alias (using), but to the type you pass as first template argument to create_strong_type. In one case, it's an int, and in the other, a double.
Try T<double>::f(KG(4.2)()); and you will see the argument is passed as lvalue reference (because of the return type of Strong_Type::operator(), which is T&).
answered Nov 19 at 17:57
Nelfeal
4,170621
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The difference is not due to using an alias (using), but to the type you pass as first template argument to create_strong_type. In one case, it's an int, and in the other, a double.
Try T<double>::f(KG(4.2)()); and you will see the argument is passed as lvalue reference (because of the return type of Strong_Type::operator(), which is T&).
answered Nov 19 at 17:57
Nelfeal
4,170621
The difference is not due to using an alias (using), but to the type you pass as first template argument to create_strong_type. In one case, it's an int, and in the other, a double.
Try T<double>::f(KG(4.2)()); and you will see the argument is passed as lvalue reference (because of the return type of Strong_Type::operator(), which is T&).
answered Nov 19 at 17:57
Nelfeal
4,170621
answered Nov 19 at 17:57
Nelfeal
4,170621
answered Nov 19 at 17:57
Nelfeal
4,170621
answered Nov 19 at 17:57
Nelfeal
4,170621
4,170621
add a comment |
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3
what is
strong_types?– Piotr Skotnicki
Nov 19 at 17:32
Sure, but is it a third party library ?
– Piotr Skotnicki
Nov 19 at 17:40
Oo, I am sorry for my misunderstanding.
strong_typesis a namespace I created myself.– Lourens Dijkstra
Nov 19 at 17:41
1
Also unrelated: Identifiers beginning with underscore, followed by uppercase letter are reserved and you are not allowed to use them.
– user10605163
Nov 19 at 17:45
1
Please fix the typos in your edits, the identifiers do not match, e.g.
strong_typevsstrong_typesandStrong_Typeand probablyStrong_Typeshould be part of the namespace and beforecreate_strong_type?. It doesn't compile the way it is even if generously moving around code.– user10605163
Nov 19 at 17:51