cast(A)b is not an lvalue

[monarch_dodra]

On Wednesday, 26 December 2012 at 19:45:53 UTC, monarch_dodra 
wrote:
> On Wednesday, 26 December 2012 at 17:13:14 UTC, Ali Çehreli 
> wrote:
>> On 12/26/2012 09:05 AM, Namespace wrote:
>>>> I can answer the question in the subject line without 
>>>> looking at
>>>> dpaste: Yes, in many cases the result of a cast operation is 
>>>> an
>>>> rvalue. It is a temporary that is constructed at the spot 
>>>> for that
>>>> cast operation.
>>>>
>>>> Imagine casting an int to a double. The four bytes of the 
>>>> int is
>>>> nowhere close to what the bit representation of a double is, 
>>>> so a
>>>> double is made at the spot.
>>>>
>>>> Ali
>>>
>>> My question is: Should not work all three?
>>> IMO: yes.
>>
>> Here is the code:
>>
>> import std.stdio;
>>
>> static if (!is(typeof(writeln)))
>> 	alias writefln writeln;
>>
>> class A { }
>> class B : A { }
>>
>> void foo(ref A a) { }
>>
>> void main()
>> {
>> 	A a = new A();
>> 	A ab = new B();
>> 	B b = new B();
>> 	
>> 	foo(a);
>> 	foo(ab);
>> 	foo(b); // < compile error
>> }
>>
>> foo() takes a class _variable_ by reference (not a class 
>> _object_ by reference). Since b is not an A variable, one is 
>> constructed on the spot.
>>
>> Imagine foo() actually does what its signature suggest:
>>
>> void foo(ref A a) {
>>    a = new B();
>> }
>>
>> That line above is an attempt to modify the caller's rvalue.
>>
>> Ali
>
> The example is much better with a "new A();" actually ;)

Wait, never mind. Your example is better.

I actually fell in the "trap" thinking my variable got modified :)