c++ - 另一个 C++ 临时生命周期困惑

Question

在下面的代码片段中...存储对 Quote::toXML 返回的临时字符串的引用是否安全？在ToXML::s_成员变量，至少只要与 << 一起使用即可仅运营商？ IE。是子表达式 q.toXML 的结果活着直到下一个; ？

这里的完整表达是什么？ q.toXML的返回值。整个std::cout或调用ToXML构造函数？

#include <iostream>
#include <string>

struct ToXML
{
    ToXML(char const * const tag, std::string const & s) : tag_(tag), s_(s)
    {
    }

    char const * tag_;
    std::string const & s_;
};

std::ostream & operator << (std::ostream & os, ToXML const & v)
{
    return os << "<" << v.tag_ << ">" << v.s_ << "</" << v.tag_ << ">";
}

struct Quote
{
    std::string toXML() const
    {
        return "<Quote/>";
    }
};

int main()
{
    Quote q;
    std::cout << ToXML("quote", q.toXML()) << std::endl;
    return 0;
}

Answer 1

是的，它很安全。

来自[class.temp]:

There are two contexts in which temporaries are destroyed at a different point than the end of the full-expression. [...]

The second context is when a reference is bound to a temporary.117 The temporary to which the reference is bound or the temporary that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference except:
— A temporary object bound to a reference parameter in a function call (5.2.2) persists until the completion of the full-expression containing the call.

我们正处于那个要点中。临时对象绑定(bind)到引用参数 (s) 并持续存在，直到包含调用的完整表达式完成为止。也就是说，它会一直持续到

std::cout << ToXML("quote", q.toXML()) << std::endl;
// --- here ---------------------------------------^

由于它在整个使用过程中持续存在，因此非常安全。但是，一旦您执行以下操作:

ToXML x("quote", q.toXML());

你被悬空引用困住了，所以我会谨慎使用这个模式。