C破坏表达式中的临时对象

给出以下代码：

#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.是否有任何有stdc知识的人可以向我解释一下？

解决方法:

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]

标签：temporary-objects,c,destructor,constructor
来源： https://codeday.me/bug/20190927/1822878.html