opUnary overloading

[Timon Gehr]

On 08/26/2012 10:33 PM, cal wrote:
>
> I have a struct wrapping a union of an int and a float, and want to
> overload opUnary for this struct:
>
> struct Value
> {
>      enum Type {
>          INT, FLOAT
>      }
>
>      Type type;
>
>      union {
>          int i;
>          float f;
>      }
>
>      this(int _i) {
>          i = _i;
>          type = Type.INT;
>      }
>
>      this(float _f) {
>          f = _f;
>          type = Type.FLOAT;
>      }
>
>      auto opUnary(string s)() if (s == "-") {
>          if (type == Type.INT)
>              return -i;
>          if (type == Type.FLOAT) // If i comment this out, I get an int
> back
>              return -f;          // as expected
>          assert(0);
>      }
> }
>
> Value v1 = Value(25);
> assert(v1.type == Value.Type.INT);
> auto neg = -v1;
> writeln(typeof(neg).stringof); // this returns float, when it should
> return int
>
> The part I don't understand is this: in the opUnary method, if I comment
> out the second if conditional (if (type == Type.FLOAT)) then when doing
> the negation on v1 I get an INT as expected. If I don't comment this
> code out, I get a float.
>
> When commented out, I get back -25 as an int, which i expect. When not
> commented out, I get back -nan, which suggests to me that it is
> returning the (uninitialized) float from the union, not simply giving me
> back the int as a float.
>
> Could someone explain this?
>
>
>

Compiler bug.

http://d.puremagic.com/issues/

Also see:
http://d.puremagic.com/issues/show_bug.cgi?id=8307

I assume the compiler fails to add an implicit conversion node into the
AST because at the point where -i is returned, the return type is still
assumed to be 'int'. What you are observing is not type coercion, but
an x87 FPU stack underflow.