c++ - 即使复制构造函数不可用，返回仅移动类型也会编译

Question

以下编译没有错误:

#include <memory>

std::unique_ptr<int> f() {
    std::unique_ptr<int> x(new int(42));
    return x;
}

int main() {
    std::unique_ptr<int> y = f();
}

我认为 f() 的返回值是由 x 复制初始化的，但是 std::unique_ptr 是一个 move-只打字。因为复制构造函数不可用，这怎么不是格式错误的呢？标准中的相关条款是什么？有没有什么地方说如果 f() 是只移动类型而不是 return 语句变成移动构造而不是复制构造？

Answer 1

I thought that the return value of f() was copy-initialized by x, but std::unique_ptr is a move-only type

f() 的返回值确实是从表达式x 复制初始化的，但复制初始化并不总是意味着复制构造。如果表达式是右值，则移动构造函数将由重载决议选择(假设存在移动构造函数)。

虽然 return x; 语句中的表达式 x 确实是一个左值(这可能会让您认为我刚才写的内容不适用), 在返回具有自动存储持续时间的命名对象的情况下，编译器应首先尝试将 id-expression 视为重载决策的右值。

What is the relevant clause in the standard? Is there somewhere that says if f() is a move-only type than a return statement becomes a move construction instead of a copy construction?

根据 C++ 标准的第 12.8/32 段([class.copy]/32，草案 N4296):

When the criteria for elision of a copy/move operation are met, but not for an exception-declaration, and the object to be copied is designated by an lvalue, or when the expression in a return statement is a (possibly parenthesized) id-expression that names an object with automatic storage duration declared in the body or parameter-declaration-clause of the innermost enclosing function or lambda-expression, overload resolution to select the constructor for the copy is first performed as if the object were designated by an rvalue. [...]