C ++ 1z中std :: make_pair和std :: make_tuple的用处 [英] Usefulness of std::make_pair and std::make_tuple in C++1z
问题描述
据我了解,存在 std :: make_pair
和 std :: make_tuple
的好处是您不必自己编写类型,因为它们是自动推导的。在C ++ 1z中,我们有模板参数推导对于类模板,它使我们可以简单地编写
std :: pair p(1,2.5); // C ++ 1z
而不是
auto p = std :: make_pair(1,2.5); // C ++ 11/14
std :: tuple
是类似的。这导致以下问题:在C ++ 1z中,是否存在使用 std :: make_pair
和 std的情况是有利的: :make_tuple
而不是使用 std :: pair
和 std :: tuple
的构造函数?
请考虑仅使用纯C ++ 1z代码(即无需与C ++ 14向后兼容),并假定每个人都熟悉此C ++。 1z功能。
在C ++ 1z中,是否存在有益的情况?使用
std :: make_pair
和std :: make_tuple
而不是使用std的构造函数:: pair
和std :: tuple
?
每条规则总是有有趣的例外。您想对 std :: reference_wrapper
做什么?
int我= 42;
auto r = std :: ref(i);
对p(i,r); // std :: pair< int,std :: reference_wrapper< int> >
自动q = std :: make_pair(i,r); // std :: pair< int,int&
如果需要后者,请 std :: make_pair
是您想要的。
在通用代码中出现了另一种情况。假设我有一个要转换为元组的模板参数包,我会写这个吗?
template< typename ... TS
auto foo(Ts ... ts){
return std :: tuple(ts ...);
}
现在,该代码的目的可能是获得 std :: tuple< Ts ...>
,但这不是不必要会发生的事情。这取决于 Ts ...
:
- 如果
Ts ...
是一个单独的对象,它本身就是一个tuple
,您可以得到它的副本。也就是说,foo(std :: tuple {1})
给了我tuple< int>
而不是tuple< tuple< int>>
。 - 如果
Ts ...
是一个对
对的单个对象,我得到了那些元素的元组
。也就是说,foo(std :: pair {1,2})
给了我tuple< int,int>
而不是tuple< pair< int,int>>
。
在通用代码中,我会避免将CTAD用于 tuple
之类的类型,因为目前尚不清楚要获得什么。 make_tuple
没有这个问题。 make_tuple(tuple {1})
是 tuple< tuple< int>>
和 make_tuple(pair {1,2})
是 tuple< pair< int,int>>
,因为这就是您要的。
另外,由于 std :: make_pair
是函数模板,您可以将其传递给可能想要执行某些操作的另一个函数模板:
foo(std :: make_pair< int,int> );
这似乎没什么用,但是有人在用它来解决问题-而您不能只在其中传递 std :: pair
。
In my understanding, the sole reason for the existence of std::make_pair
and std::make_tuple
is that you don't have to write the types by yourself as they are automatically deduced. In C++1z, we have template argument deduction for class templates, which allows us to simply write
std::pair p(1, 2.5); // C++1z
instead of
auto p = std::make_pair(1, 2.5); // C++11/14
The situation for std::tuple
is analogous. This leads to the following question: In C++1z, is there a situation in which it is beneficial to use std::make_pair
and std::make_tuple
instead of using the constructors of std::pair
and std::tuple
?
Please, consider just pure C++1z code (i.e. no need for backward compatibility with C++14) and assume that everyone is familiar with this C++1z feature.
In C++1z, is there a situation in which it is beneficial to use
std::make_pair
andstd::make_tuple
instead of using the constructors ofstd::pair
andstd::tuple
?
There are always fun exceptions to every rule. What do you want to happen to std::reference_wrapper
?
int i = 42;
auto r = std::ref(i);
pair p(i, r); // std::pair<int, std::reference_wrapper<int> >
auto q = std::make_pair(i,r); // std::pair<int, int&>
If you want the latter, std::make_pair
is what you want.
Another situation cames in generic code. Let's say I have a template parameter pack that I want to turn into a tuple, would I write this?
template <typename... Ts>
auto foo(Ts... ts) {
return std::tuple(ts...);
}
Now the intent of that code is to probably to get a std::tuple<Ts...>
out of it, but that's not necessarily what happens. It depends on Ts...
:
- if
Ts...
is a single object that is itself atuple
, you get a copy of it. That is,foo(std::tuple{1})
gives me atuple<int>
rather than atuple<tuple<int>>
. - if
Ts...
was a single object that is apair
, I get atuple
of those elements. That is,foo(std::pair{1, 2})
gives me atuple<int, int>
rather than atuple<pair<int, int>>
.
In generic code, I would stay away from using CTAD for types like tuple
because it's never clear what you're going to get. make_tuple
doesn't have this problem. make_tuple(tuple{1})
is a tuple<tuple<int>>
and make_tuple(pair{1, 2})
is a tuple<pair<int, int>>
because that's what you asked for.
Additionally, since std::make_pair
is a function template, you can pass that into another function template that maybe wants to do something:
foo(std::make_pair<int, int>);
This doesn't seem super useful, but somebody somewhere is using it to solve a problem - and you can't just pass std::pair
there.
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