Foreach with byte problems

[Steven Schveighoffer]

On Fri, 25 Feb 2011 13:52:53 -0500, Andrej Mitrovic <none at none.none> 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.
>
> Hence this will run the foreach from 0 to 9:
> foreach (index; 0..10) { }
>
> So I figured I'd just add a +1 at the end, but then I get an error:
>     foreach (byte index; byte.min..byte.max+1)
>     {
>     }
>
> Error: cannot implicitly convert expression (128) of type int to
> byte.
>
> Of course 128 can't fit in a byte. But how am I supposed to print out  
> the last value if the right hand side of a range literal is  
> non-inclusive?

foreach(int index; byte.min..byte.max + 1)

The truth is, you don't want to use byte to represent your comparisons,  
because byte.max + 1 isn't a valid byte.

And instead of counter, you can use a formula instead:

if((index - byte.min) % 4 == 0)
    writeln();

-Steve