On 2/25/11, Jonathan M Davis <jmdavisProg@gmx.com> wrote:
> However, it is true that in this particular case, it's annoying. Still, in
> the
> general case, I do believe that it's definitely the right behavior.
>
> - Jonathan M Davis
>

I do agree it's the right thing to do, and I wouldn't want to change it.

But I'd like the compiler to be smart in this case and see that there's no way 128 will ever be stored into a byte.

It's not some crucial piece of code that I need working. I just ran into it by accident really.

On 02/25/2011 10:52 AM, Andrej Mitrovic wrote:
> So I'd like to print all values storable in a byte in hex representation:
>
> import std.stdio;
> void main()
> {
>      int counter;
>      foreach (byte index; byte.min..byte.max)
>      {
>          if (!(counter % 4))
>              writeln();
>
>          writef("%#.2x, ", index);
>          counter++;
>      }
> }
>
> If you run this, you'll realize that it doesn't print the final 0x7F. This is because in a foreach range literal (is that the correct term?), the left side is inclusive, but the right side isn't.

I've seen the same "problem" with enums.

import std.stdio;

enum E { x, y, z }

void main()
{
    foreach (e; E.min .. E.max) {
        writeln(e);
    }
}

The value 2 is excluded. Of course it's rare to use enums like that in a foreach loop as their values are not always continuous.

I found out the better solution before sending this message:

    foreach (e; __traits(allMembers, E)) {
        writeln(e);
    }

The difference is, the type of 'e' is string above. Finally, the following produces integer values:

    foreach (e; __traits(allMembers, E)) {
        writeln(to!E(e));
    }

Ok, good... :)

Ali