rvalue to lvalue conversion Visual Studio

Question

In Visual Studio 2012RC there are some non-standard extensions. For example this code compiles:

#include 

using namespace std;

void value(str         


        

Answer 1

Basically, VS will allocate the space somewhere and just let the reference point to it, as if it was a reference-to-const without the constness (or in C++11 an rvalue reference).

You can disable this behaviour with the /Za (disable language extensions) compiler switch under

Properties -> C/C++ -> Language

If I remember correctly.

Answer 2

In standard C++ you cannot bind a temporary (rvalue / string("nice")) to a non-const reference (lvalue), but the microsoft compiler allows it. The warning is telling you that the code is compiling due to an extension and will not compile with any other compiler.

Answer 3

A temporary object of class type is still an object. It lives somewhere in memory, which means that there's nothing unusual in the compiler being able to attach a reference to it. At physical level whether it is a const reference or non-const reference makes no difference. In other words, in cases like that the language restriction is purely conceptual, artificial. The compiler simply ignores that restriction. There's no need to "transform" anything here. The reference is simply attached directly to the object, wherever that object happens to reside.

Basically, for a class that provides the outside word with access to the value of its this pointer (or with lvalue access to *this) the behavior can be immediately and easily simulated

struct S {
  S& get_lvalue() { return *this; }
};

void foo(S& s);
...

foo(S().get_lvalue());

The above code is perfectly legal and it works around the aforementioned restriction. You can think of MSVC++ behavior as being equivalent to this.