When using multi-dimensional arrays I easily get confused as to the order of the notation. But why does args[i][] equal args[][i] (code below)? I assume args[i][] is the more correct version as only args[i][0] gives the correct results?

-Silv3r

-----------------------------------------------------------------------
import std.stdio;

void main(char[][] args)
{
 writefln("args.length = %d\n", args.length);
 for (int i = 0; i < args.length; i++)
 {
  if(args[i][] == args[][i]) // why does args[i][] equal args[][i]?
   writefln("Why does this work? [%d] = '%s'", i, args[i][]);
  writefln("<%s>,<%s>", args[i][0],args[0][i]);
 }
}
-----------------------------------------------------------------------

Silv3r wrote:

> When using multi-dimensional arrays I easily get confused as to the order of the notation. But why does args[i][] equal args[][i] (code below)? I assume args[i][] is the more correct version as only args[i][0] gives the correct results?
> 
> -Silv3r
> 
> -----------------------------------------------------------------------
> import std.stdio;
> 
> void main(char[][] args)
> {
>  writefln("args.length = %d\n", args.length);
>  for (int i = 0; i < args.length; i++)
>  {
>   if(args[i][] == args[][i]) // why does args[i][] equal args[][i]?
>    writefln("Why does this work? [%d] = '%s'", i, args[i][]);
>   writefln("<%s>,<%s>", args[i][0],args[0][i]);
>  }
> }
> -----------------------------------------------------------------------

ok, first args[] is the same as writing only args as a slice of an entire dynamic array is itself.

second think of the array brackets as a stack

(char[])[] args

so args is an array (of char arrays)
if you take an element of that you eliminate the outer brackets
char[] a=args[0];

hope this helps and if I misunderstood the question I'm sorry.

Reply to Silv3r,

> When using multi-dimensional arrays I easily get confused as to the
> order of the notation. But why does args[i][] equal args[][i] (code
> below)? I assume args[i][] is the more correct version as only
> args[i][0] gives the correct results?
> 
> -Silv3r
> 

> if(args[i][] == args[][i]) // why does args[i][] equal args[][i]?

args[i][] == get args[i], take all of it as a slice
args[][i] == get all of args as a slice, take [i] of it

in an expression [] is the same as [0..$]

Reply to BCS,

This reminds me of something I have been finding handy recently:

//take a slice starting at i and of length j:
arr[i..$][0..j]

the alternative is a bit uglier IMHO

arr[i..i+j]

Thanks, now I understand arrays a bit better!