C++只调用一个构造函数

Question

这个问题已经有答案了:

What are copy elision and return value optimization? (5 个回答)

已关闭 2 年前。

我正在尝试理解右值引用。这是我到目前为止编写的代码:

class A {
public:
    A(const char* str) {
        std::cout << str;
    }

    A(A&& other) {
        std::cout << "other";
    }
};

int main() {
    A m(A(A(A("hello"))));
}

输出仅为"hello" ，这让我很困惑。

自 A("hello")是传递给第二个构造函数的临时对象，但代码输出就像只调用了第一个构造函数。

我想这要么是编译器优化，要么我错过了有关构造函数和右值的一些细节。

Answer 1

您观察到的内容也称为 copy elision ，更准确地说:

Under the following circumstances, the compilers are required to omit the copy and move construction of class objects, even if the copy/move constructor and the destructor have observable side-effects. The objects are constructed directly into the storage where they would otherwise be copied/moved to. The copy/move constructors need not be present or accessible:

(...)

• In the initialization of an object, when the initializer expression is a prvalue of the same class type (ignoring cv-qualification) as the variable type:

T x = T(T(f())); // only one call to default constructor of T, to initialize x

这正是你的情况。

请注意，从 C++17 开始，优化是强制的。对于 C++11 来说它是可选的，但您的编译器似乎还是应用了它。