为什么在这种情况下不调用复制构造函数？ [英] Why copy constructor is not called in this case?

问题描述

以下是小代码片段：

class A
{
public:
    A(int value) : value_(value)
    {
        cout <<"Regular constructor" <<endl;
    }

    A(const A& other)   : value_(other.value_)  
    {
        cout <<"Copy constructor" <<endl;
    }

private:
    int value_;
};
int main()
{
    A a = A(5);
}

我假设输出将是Regular Constructor LHS的复制构造函数。所以我避免了这种风格，总是声明变量类 A a（5）; 。但是令我惊讶的是，在上面的代码中，复制构造函数从未被调用（Visual C ++ 2008）

I assumed that output would be "Regular Constructor" (for RHS) followed by "Copy constructor" for LHS. So I avoided this style and always declared variable of class as A a(5);. But to my surprise in the code above copy constructor is never called (Visual C++ 2008)

有人知道这种行为是编译器优化的结果，和便携）功能的C ++？谢谢。

Does anybody know if this behavior is a result of compiler optimization, or some documented (and portable) feature of C++? Thanks.

推荐答案

从另一个注释：所以默认情况下我不应该依赖它（因为它可能取决于编译器）

From another comment: "So by default I should not rely on it (as it may depend on the compiler)"

不，它不依赖于编译器，实际上无论如何。任何编译器都不会浪费时间构造一个A，然后复制它。

No, it does not depend on the compiler, practically anyway. Any compiler worth a grain of sand won't waste time constructing an A, then copying it over.

在标准中，它明确表示它完全可以接受 T = x; 等价于说 T（x）; 。 （§12.8.15，pg.211）这样做与 T（T（x））显然是多余的，所以它删除内部 / code>。

In the standard it explicitly says that it is completely acceptable for T = x; to be equivalent to saying T(x);. (§12.8.15, pg. 211) Doing this with T(T(x)) is obviously redundant, so it removes the inner T.

为了获得所需的行为，你必须强制编译器默认构造第一个A：

To get the desired behavior, you'd force the compiler to default construct the first A:

A a;
// A is now a fully constructed object,
// so it can't call constructors again:
a = A(5);

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