c++ - 为什么 move 构造函数只被调用一次？

Question

我有以下功能:

SomeClass func()
{
    SomeClass someObject;
    someObject.mutate("some text");
    return someObject;
}

以及以下两种情况:

1

int main()
{
    func();
    return 0;
}

2

int main()
{
    SomeClass someObject = func();
    return 0;
}

我关闭了 NRVO，没有发生复制/move 省略。

在这两种情况下，我都有相同的输出:

"default constructor"
"move constructor"

为什么 move 构造函数在情况 2 中只被调用一次？我原以为会被调用一次以获得函数的返回值，然后被调用一次以初始化 someObject 变量。

更新:更清楚:输出用于调试构建。对于发布版本，我只有“默认构造函数”，由于复制/move 省略，这对我来说似乎很清楚。我想了解调试版本的不同输出。

Answer 1

这是由于 C++17 中的复制省略。来自 cppreference :

Under the following circumstances, the compilers are permitted, but not required to omit the copy and move (since C++11) construction of class objects even if the copy/move (since C++11) constructor and the destructor have observable side-effects.
[...]
In the initialization of an object, when the source object is a nameless temporary and is of the same class type (ignoring cv-qualification) as the target object. When the nameless temporary is the operand of a return statement, this variant of copy elision is known as RVO, "return value optimization".
This optimization is mandatory; see above. (since C++17)

另见复制初始化。同样来自 cppreference :

The effects of copy initialization are:
First, if T is a class type and the initializer is a prvalue expression whose cv-unqualified type is the same class as T, the initializer expression itself, rather than a temporary materialized from it, is used to initialize the destination object: see copy elision (since C++17)

请注意，这不是 NRVO(命名 返回值优化)。这是 RVO。

在 C++14 中，如果您不希望优化不会发生(参见 -fno-elide-constructors)。
Demo (使用 GCC，但 Clang 产生相同的结果)

在 C++17 中，它是强制性的，所以它确实发生了。
Demo (同样，GCC 但 Clang 同意)