另一个 C++ 临时生命周期混乱

Question

In the below code snippet... is it safe to store a reference to the temporary string returned by Quote::toXML in ToXML::s_ member variable, at least as long as it is used with the << operator only? Ie is the result of the sub-expression q.toXML alive until the next ; ?

And what is the full expression here wrt the return value of q.toXML . The entire std::cout or the call to ToXML constructor ?

#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

Yes, it is safe.

From [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.

We're in that bullet point. The temporary object is bound to a reference parameter ( s ) and persists until the completion of the full-expression containing the call. That is, it persists until

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

Since it lasts throughout its usage, it is perfectly safe. However, as soon as you do something like:

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

you're stuck with a dangling reference, so I would use this pattern sparingly.