c++ - 引用未命名的临时对象(生命周期)

Question

看完this answer来自 ildjarn ，我写了下面的例子，它看起来像一个未命名的临时对象与其引用具有相同的生命周期!

• 这怎么可能？
• 它是否在 C++ 标准中指定？
• 哪个版本？

源代码:

#include <iostream>  //cout
#include <sstream>   //ostringstream 

int main ()
{
        std::ostringstream oss;
        oss << 1234;

        std::string const& str = oss.str();
        char        const* ptr = str.c_str();

        // Change the stream content
        oss << "_more_stuff_";
        oss.str(""); //reset
        oss << "Beginning";
        std::cout << oss.str() <<'\n';

        // Fill the call stack
        // ... create many local variables, call functions...

        // Change again the stream content
        oss << "Again";
        oss.str(""); //reset
        oss << "Next should be '1234': ";
        std::cout << oss.str() <<'\n';

        // Check if the ptr is still unchanged
        std::cout << ptr << std::endl;
}

执行:

> g++ --version
g++ (GCC) 4.1.2 20080704 (Red Hat 4.1.2-54)
Copyright (C) 2006 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
> g++ main.cpp -O3
> ./a.out
Beginning
Next should be '1234':
1234

Answer 1

How come this is possible?

因为标准是这么说的，因为它被认为是有用的。右值引用和 const 左值引用延长了临时对象的生命周期:

[C++11: 12.2/5]: [..] 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 [..]

[C++11: 8.5.3/5] 中详尽的措辞要求我们不得将临时对象绑定(bind)到非const 左值引用。

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Is it specified in the C++ standard? Which version?

是的。所有这些。