C++ Returning reference to temporary [duplicate]

Answer 1

You're creating a temporary value on the stack AnotherObject() and returning it right before it gets destroyed. Your function's caller would get garbage, and so it's forbidden.

Maybe you want to allocate it on the heap and return a pointer to it instead?

return new AnotherObject();

Alternatively, declare your function to return a "copy" to your object, instead of a reference like I'm assuming you are returning right now:

AnotherObject f()
{
  return AnotherObject();  // return value optimization will kick in anyway!
}