C ++中构造函数的返回类型 [英] return type of the constructor in C++

问题描述

我知道C ++中没有返回类型的构造函数

I know that there is no return type of the constructors in C++

但是，下面的代码编译正确。下面的代码中构造函数返回的是什么？

However, the code below compiles right. What is returned by the constructor in the code below?

class A{

public:
A() {}
}


A a = A();      //what is returned by A() here, why?

这里是否有冲突？

推荐答案

构造函数不返回任何内容。语法 A（）不是构造函数调用，它创建一个 A 类型的临时对象（并调用构造函数

Nothing is returned from the constructor. The syntax A() is not a constructor call, it creates a temporary object of type A (and calls the constructor in the process).

您不能直接调用构造函数，构造函数将作为对象构造的一部分进行调用。

You can't call a constructor directly, constructors are called as a part of object construction.

在代码中，在临时构造过程中，将调用默认构造函数（您定义的构造函数）。然后，在 a 的构造过程中，使用临时变量作为参数来调用复制构造函数（由编译器自动生成）。

In your code, during the construction of the temporary the default constructor (the one you defined) is called. Then, during the construction of a, the copy constructor (generated automatically by the compiler) is called with the temporary as an argument.

正如Greg正确指出的，在某些情况下（包括这一个），允许编译器避免复制构造和默认构造 a 复制构造函数必须可访问）。我知道没有编译器不会执行这样的优化。

As Greg correctly points out, in some circumstances (including this one), the compiler is allowed to avoid the copy-construction and default-construct a (the copy-constructor must be accessible however). I know of no compiler that wouldn't perform such optimization.

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