[FIXED] Why can a non-const reference parameter be bound to a temporary object?

Issue

char f1();
void f2(char&);

struct A {};

A    f3();
void f4(A&);

int main()
{
    f2(f1()); // error C2664. This is as expected.
    f4(f3()); // OK! Why???
}

error C2664: 'void f4(char &)' : cannot convert argument 1 from 'char' to 'char &'

I have been taught that in C++ a non-const reference parameter cannot be bound to a temporary object; and in the code above, f2(f1()); triggers an error as expected.

However, why does the same rule not apply to the code line f4(f3());?

PS: My compiler is VC++ 2013. Even if I comment the line f2(f1());, then the code containing f4(f3()); will be compiled without any errors or warnings.

Update:

MSDN says:

In previous releases of Visual C++, non-const references could be bound to temporary objects. Now, temporary objects can only be bound to const references.

So I think it is a bug of VC++. I have submitted a bug report to VC++ team

Solution

If you compile with the /Za option to disable language extensions, the compiler rejects both calls:

> cl /Za test.cpp
Microsoft (R) C/C++ Optimizing Compiler Version 18.00.21005.1 for x86
Copyright (C) Microsoft Corporation.  All rights reserved.

test.cpp
test.cpp(11): error C2664: 'void f2(char &)' : cannot convert argument 1 from 'char' to 'char &'
test.cpp(12): error C2664: 'void f4(A &)' : cannot convert argument 1 from 'A' to 'A &'
        A non-const reference may only be bound to an lvalue

There are several (very constrained) circumstances in which the compiler, with language extensions enabled, will still allow a non-const lvalue reference to bind to an rvalue expression. My understanding is that this is largely to avoid breaking several enormous legacy codebases that rely on this "extension."

(In general, use of /Za is not recommended for many reasons, but mostly because the Windows SDK headers cannot be #included with the /Za option.)

Answered By - James McNellis

Answer Checked By - Marie Seifert (PHPFixing Admin)