std :: ignore的要求 [英] Requirements for std::ignore

问题描述

C ++ 11引入了一个名为 std :: ignore 的对象：

C++11 introduces an object called std::ignore:

const /* unspecified */ ignore;

为了简洁，让

typedef decltype(std::ignore) T; 

从我可以知道， T 是因为 std :: tie 的规定，它是 CopyAssignable 20.4.2.4:7]。

From what I can tell, the only requirement for T is that it is CopyAssignable, due to the specification of std::tie [C++11, 20.4.2.4:7].

在g ++ - 4.8中，我发现 T 是另外 DefaultConstructible （例如， T x; 编译）。这是实现定义的行为吗？

In g++-4.8, I find that T is additionally DefaultConstructible (e.g., T x; compiles). Is this implementation-defined behavior?

（如果对 T 有其他要求， 。）

(If there are other requirements on T that I have missed, please elaborate.)

推荐答案

标准对类型有 ，除了它是一个与所有其他类型不同的类型。

The standard has no requirements on the type of ignore, besides the fact that it is a type that is distinct from all other types.

无论标准库容器允许 ignore 与 tie 配合使用时，可获得所需的行为。图书馆 可以为其提供模板< T&> operator =（const T&） overload，或者可以使用一些其他机制使其工作。标准没有说。因此，它甚至不需要 CopyAssignable 。

Whatever machinery that a standard library container does to allow ignore to gain the required behavior when used with tie is up to that standard library implementation. The library may give it a template<T&> operator=(const T&) overload, or it may use some other mechanism to make it work. The standard doesn't say. So it doesn't even have to be CopyAssignable.

注意 tie 只有特殊的行为，如果你特别使用忽略 。如果你使用一些其他值，你自己创建（因为类型没有要求，你不能保证能够做），你会得到未定义的行为。

Note that tie only has special behavior if you specifically use ignore. If you use some other value, created by yourself (which, since the type has no requirements, you are not guaranteed to be able to do), you will get undefined behavior.

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