C ++：std :: tie的返回类型，带std :: ignore [英] C++: Return type of std::tie with std::ignore

问题描述

我想知道C ++ 11标准是否对由<$ c>返回的 std :: tuple 的类型当一些参数 std :: ignore 时，$ c> std :: tie 。

I am wondering if the C++11 standard gives any requirement about the type of the std::tuple returned by std::tie when some arguments are std::ignore.

更具体地说，我可以假设：

More specifically, can I assume that:

1. decltype（std :: tie（42，std :: ignore ））不等于 decltype（std :: tie（std :: ignore，42））


2. decltype（std :: tie（42，std :: ignore））不同于 decltype（std :: tie ））


3. decltype（std :: tie（std :: ignore，42）） decltype（std :: tie（42））


4. std :: ignore（std :: ignore））不等于 decltype（std :: tie / li>


1. decltype(std::tie(42, std::ignore)) is not the same as decltype(std::tie(std::ignore, 42))
2. decltype(std::tie(42, std::ignore)) is not the same as decltype(std::tie(42))
3. decltype(std::tie(std::ignore, 42)) is not the same as decltype(std::tie(42))
4. decltype(std::tie(std::ignore, std::ignore)) is not the same as decltype(std::tie(std::ignore))

换句话说，从类型角度看，生成的元组是否具有类型 decltype对于所有符合 std :: ignore 的位置的模板参数

In other words, from the type perspective, does the generated tuple behaves as a tuple which has type decltype(std::ignore) for all template arguments that match std::ignore by position?

推荐答案

是的， std :: tie 返回 std :: tuple< T& ...> $ c> T ... 是赋予它的类型。

std :: ignore 有一个未指定类型，但它仍然会出现在 tuple 根据您在 std :: tie 中指定的位置。

Yes you can, std::tie returns a std::tuple<T&...> where T... are the types that are given to it.
std::ignore has an unspecified type, but it will still appear in the tuple according to where you specified it in std::tie.

如果您觉得自己感觉更好，您可以在您的代码中包括：

If if makes you feel better, you could include in your code somewhere:

    int n;
    auto i = std::tie(std::ignore, n);
    auto j = std::tie(n, std::ignore);
    auto k = std::tie(n);
    static_assert(!std::is_same<decltype(i), decltype(j)>::value, "");
    static_assert(!std::is_same<decltype(i), decltype(k)>::value, "");
    static_assert(!std::is_same<decltype(j), decltype(k)>::value, "");

等等。这样，如果您的假设无效，编译将失败。

and so on for whatever combinations you are explicitly using. This way compilation will fail if your assumption is invalid.

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