C++ 销毁表达式中的临时对象 [英] C++ destruction of temporary object in an expression

问题描述

给定以下代码:

#include <iostream>

struct implicit_t
{
    implicit_t(int x) :
        x_m(x)
    {
        std::cout << "ctor" << std::endl;
    }

    ~implicit_t()
    {
        std::cout << "dtor" << std::endl;
    }

    int x_m;
};

std::ostream& operator<<(std::ostream& s, const implicit_t& x)
{
    return s << x.x_m;
}

const implicit_t& f(const implicit_t& x)
{
    return x;
}

int main()
{
    std::cout << f(42) << std::endl;

    return 0;
}

我得到以下输出:

ctor
42
dtor

虽然我知道这是正确的，但我不确定为什么.有stdc++知识的可以给我解释一下吗?

While I know this is correct, I'm not certain why. Is there anyone with stdc++ knowledge who can explain it to me?

推荐答案

临时对象被销毁作为评估完整表达式 (1.9) 的最后一步，该表达式(词法上)包含它们的创建点.[12.2/3]

Temporary objects are destroyed as the last step in evaluating the full-expression (1.9) that (lexically) contains the point where they were created. [12.2/3]

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