I want to return an argument as a reference. Something like:
Foo& GetFoo(Foo f)
{
...
return f;
}
I know returning a local variable as a reference doesn't works:
Foo& GetFoo()
{
Foo f;
...
return f;
} // f is destroyed here
Also i know that returning a global variable as a reference works:
Foo f;
...
Foo& GetFoo()
{
...
return f;
}
But i don't know about arguments, and i didn't see anything about.
An argument as in your first example:
Foo& GetFoo(Foo f)is just the same as a temporary, at function scope. So if you try to return that by reference, it won't work.
What are you trying to do exactly? As is, it sounds like a code-smell. Try to explain what your actual problem is, and we can give you some possible solutions :)
If the argument is a value-type, it has the same restriction as a local variable - i.e. you can't return a reference to it.
If the argument is already a reference, you can return that reference back out safely, but it's kind of a weird thing to do.
Nonsense. It's a very common thing to do. It's all though the standard library.
std::ostream&
operator<<(std::ostream& ostr, std::string const& s)
{
s.streambuf.copyin(s);
return ostr;
}
