Re: Type Inference Bug?

[Daniel Murphy via Digitalmars-d]

"Meta"  wrote in message news:wzczhiwokauvkkevt...@forum.dlang.org...
shared const int i;

static if (is(typeof(i) T == shared U, U))
{
     //Prints "shared(const(int))"
     pragma(msg, U);
}

This seems like subtly wrong behaviour to me. If T == shared U, for some 
U, then shouldn't U be unshared? If T is shared(const(int)), and T is the 
same as the type U with the 'shared' qualifier applied to it, then U 
should be of type const(int), not shared(const(int)).

I'm bringing this up partially because it seems wrong to me, and partially 
because we currently don't have a good why of "shaving" the outermost 
qualifier off a type, and this seemed the natural way to do it to me (I 
was surprised when it didn't work).

It doesn't print anything for me.  This code seems to have the desired 
effect:

shared const int i;

void main()
{
   static if (is(typeof(i) : shared(U), U))
   {
        //Prints "const(int)"
        pragma(msg, U);
   }
}