问题描述

I understand that if a temporary is bound to a reference member in the constructor's initializer list, the object will be destroyed as the constructor returns.

However, consider the following code:

#include <functional>
#include <iostream>

using callback_func = std::function<int(void)>;

int
func(const callback_func& callback)
{
  struct wrapper
  {
    const callback_func& w_cb;
    wrapper(const callback_func& cb) : w_cb {cb} { }
    int call() { return this->w_cb() + this->w_cb(); }
  };
  wrapper wrp {callback};
  return wrp.call();
}

int
main()
{
  std::cout << func([](){ return 21; }) << std::endl;
  return 0;
}

This looks perfectly valid to me. The callback object will live during the whole execution of the func function and no temporary copy should be made for wrapper's constructor.

Indeed, GCC 4.9.0 compiles fine with all warnings enabled.

However, GCC 4.8.2 compiler gives me the following warning:

$ g++ -std=c++11 -W main.cpp 
main.cpp: In constructor ‘func(const callback_func&)::wrapper::wrapper(const callback_func&)’:
main.cpp:12:48: warning: a temporary bound to ‘func(const callback_func&)::wrapper::w_cb’ only persists until the constructor exits [-Wextra]
     wrapper(const callback_func& cb) : w_cb {cb} { }
                                            ^

Is this a false positive or am I misunderstanding the object lifetimes?

Here are my exact compiler versions tested:

$ g++ --version
g++ (GCC) 4.8.2
Copyright (C) 2013 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++ --version
g++ (GCC) 4.9.0 20140604 (prerelease)
Copyright (C) 2014 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.

This is a bug in gcc 4.8 that has been fixed in 4.9. Here is the bug report:

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=50025

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