问题:

#include <iostream>
int foo()
{
  return 0;
}

int main()
{
  const int& a = foo();
  std::cout << &a << std::endl;
}

In this code, a binds to a rvalue. Is it legal to take its address? (And by legal I mean: in the code ill-formed? Am I causing an undefined behaviour?)

回答1:

This is fine. In C++11 you can even do this:

int&& a = foo();
a = 123;

You can kind of think about temporaries like this (conceptually and in general):

x = func(); // translated as:

auto __temporary = func();
x = __temporary;
__destruct_value_now_and_not_later(__temporary);

Except if x is the definition of a reference type, the compiler notes that you're purposefully referring to the temporary value and extends its lifetime by removing the early destruction code, making it a normal variable.