Why does an lvalue cast work?

Question

I saw this kind of cast for the first time today, and I'm curious as to why this works. I thought casting in this manner would assign to the temporary, and not the class member. Using VC2010.

class A
{
public:

   A() :
      m_value(1.f)
   {
      ((float)m_value) = 10.f;
   }

   const float m_value;
};

Is that the exact snippet you tried? Normally this should error out because m_value is not initialized and should not compile the assignment. — PlasmaHH, Mar 11 '13 at 16:12
http://liveworkspace.org/code/nVBzK$0, apparently it doesn't work — Tony The Lion, Mar 11 '13 at 16:12
That **is not** allowed, and it should error out for multiple reasons (including the fact that `m_value` is not assigned a value in the initialization list) — David Rodríguez - dribeas, Mar 11 '13 at 16:13
Before your question, Didn't you need to initialize that `m_value` at A's constructor? — masoud, Mar 11 '13 at 16:14
I edited to be what I actually used. It still assigns it. Glad to know my suspicions were correct. — Hanoixan, Mar 11 '13 at 16:25
I believe that's a GCC extension. It does not work under Clang. See [C/C++ Language Compatibility, lvalue cast](http://clang.llvm.org/compatibility.html#lvalue-cast) — jww, Aug 02 '15 at 04:08

score 5 · Answer 1 · answered Mar 11 '13 at 16:30

Even after fixing all other problems to make the code compile, it only works in VC2010 because it uses a non-standard extension. And If you specify /Wall to see all warnings, you compiler will emit

warning C4213: nonstandard extension used : cast on l-value

score 4 · Accepted Answer · answered Mar 11 '13 at 16:17

It shouldn't work. An explicit type conversion to float with cast notation will be a prvalue (§5.4):

The result of the expression (T) cast-expression is of type T. The result is an lvalue if T is an lvalue reference type or an rvalue reference to function type and an xvalue if T is an rvalue reference to object type; otherwise the result is a prvalue.

My emphasis added.

The assignment operator requires an lvalue as its left operand (§5.17):

All require a modifiable lvalue as their left operand and return an lvalue referring to the left operand.

A prvalue is not an lvalue.

Why does an lvalue cast work?

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