modulus and array.length

[David Osborne]

Hi,

I've noticed that the modulus operator acts differently when the divisor is
the length of an array and the dividend is negative.
For instance, this code:

import std.stdio;
const int x = 4;
void main(){
    int[x] arr = [0, 1, 2, 3];

    writeln("Using arr.length");
    for(int i = -3; i <= 0; i++){
        writefln("%d mod %d = %d", i, arr.length, i % arr.length);
    }
    writeln("Using x");
    for(int i = -3; i <= 0; i++){
        writefln("%d mod %d = %d", i, x, i % x);
    }
}

Produces this output:

Using arr.length
-3 mod 4 = 1
-2 mod 4 = 2
-1 mod 4 = 3
0 mod 4 = 0
Using x
-3 mod 4 = -3
-2 mod 4 = -2
-1 mod 4 = -1
0 mod 4 = 0

 Which is a useful (but undocumented?) feature that lets you loop through
arrays backwards.
However, when the length of the array is odd...
const int x = 3;
...
int[x] arr = [0, 1, 2];
It looks like this:

Using arr.length
-3 mod 3 = 1  <-- this should be 0
-2 mod 3 = 2  <-- this should be 1
-1 mod 3 = 0  <-- this should be 2
0 mod 3 = 0
Using x
-3 mod 3 = 0
-2 mod 3 = -2
-1 mod 3 = -1
0 mod 3 = 0

Does anyone know of a reason for this? It doesn't seem like a bug, but I
don't know why it would do something like that.

Upon further investigation, I found that when arr.length is even, i acts
like (x*(abs(i)/x + 1) + i) , and when arr.length is odd, i acts like
(x*(abs(i)/x - 1) + i)

(I'm using dmd 2.049 on linux)


~Dave
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