在没有默认构造函数的情况下构造一个空对象 [英] Construct an empty object without the default constructor

问题描述

假设我有一个 F 类型.我知道 F 为空，但是 F 没有默认构造函数，所以我不能使用 F()来构造它.无论如何，有没有一种方法可以获取类型为 F 的有效对象?我似乎还记得有人提到过，有一种奥秘地使用工会的方法.理想情况下，它将是 constexpr 友好的.

Suppose I have a type F. I know that F is empty, but F has no default constructor, so I can't use F() to construct it. Is there a way to obtain a valid object of type F anyway? I seem to recall a mention that there was such a way with arcane usage of unions. Ideally, it would be constexpr friendly.

这很有用，因为无捕获的lambda仅在C ++ 20中获得了默认构造函数.在C ++ 17中，如果我想将lambda传递给模板"并在没有该lambda实例的情况下调用该lambda，则需要能够从该类型中重构它.

This can be useful because captureless lambdas only gained a default constructor in C++20. In C++17, if I want to "pass a lambda to a template" and call that lambda without having an instance of it, I need to be able to reconstruct it from the type.

auto const f = [](int x) { return x; };
using F = decltype(f);

static_assert(std::is_empty_v<F>);
static_assert(!std::is_default_constructible_v<F>);

magically-construct-an-F(42);

推荐答案

对于您自己的类型，您可以从自身复制或移动构造对象: F f = f .这本身并不会导致UB，请参见 CWG363 .但是，对于编译器提供的闭包类型，即使您知道它为空，也不清楚.

For your own types, you could copy- or move-construct an object from itself: F f = f. This does not lead to UB by itself, see CWG363. However, it is not so clear for a compiler-provided closure type, even if you know it is empty.

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