为什么不调用复制构造函数，但在此代码中调用缺省构造函数？ [英] Why copy-constructor is not called but default constructor is invoked in this code?

问题描述

我理解当从已经存在的对象创建对象时，以及当函数按值返回对象时，将调用复制构造函数。那么为什么在下面的代码中，复制构造函数不是被调用，而是默认构造函数？

I have an understanding that copy constructor will be called when an object is created from an already existing object and also when a function returns object by value. So why in the below code the copy constructor is not called but the default constructor?

class A {
public:
    A() { cout << "A" << endl; }
    A(A &) { cout << "A&" << endl; }
    A(const A &) { cout << "const A&" << endl; }
};
A fun() {
    class A a;
    return a;
}
int main() {
    class A a = fun(); // output: A
}

推荐答案

回答：编译器优化。

• 首先， a 函数直接在 main 函数的范围内创建，以避免必须将局部参数复制（或在C ++ 11中移动）函数通过函数的返回。这是返回值优化。

• First, the a object from your function is created directly in the scope of the main function, in order to avoid having to copy (or move in C++11) the local parameter out of the scope of the function via the function's return. This is return value optimization.

然后，在main中，语句等价于 class A a = A（），并且允许编译器创建 a 对象，而不从临时对象复制。这是复制。

Then, in main, the statement becomes equivalent to class A a = A() and again the compiler is allowed to the create the a object in place, without copying from a temporary object. This is copy elision.

这是允许的，即使复制构造函数旁路完全）具有副作用，如在您的示例中。

This is allowed even if the copy constructor (which is bypassed entirely) has side-effects, like in your example.

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