digitalmars.D.learn - Multidimensional arrays, foreach loops and slices

```Hello all,

definitively before, but a search doesn't bring anything up: is
it possible to foreach() over every element of a multidimensional
array, i.e. something like,

int a;

foreach(ref int n; a)
n = uniform!"[]"(0, 1);

... or is it necessary to do a separate foreach over each
dimension?

The second query relates to slices.  Suppose I have an N*M 2d
array (let's call it a), and a 1d array of length N (let's call
it s) whose elements are unsigned integers whose values fall in
[0, M).  What I'd like is to take a slice of a such that I get
out a 1d array of length N such that the i'th element corresponds
to a[s[i]].

Is this possible?  And if so, how is it achieved?

Thanks and best wishes,

-- Joe
```
May 30 2012
```Joseph Rushton Wakeling:

A couple of queries.

I think you have to write two small generic functions to do those
things (or write inline code).

Bye,
bearophile
```
May 30 2012    Philippe Sigaud <philippe.sigaud gmail.com> writes:
```On Wed, May 30, 2012 at 6:15 PM, Joseph Rushton Wakeling
<joseph.wakeling webdrake.net> wrote:
Hello all,

definitively before, but a search doesn't bring anything up: is it possib=

le
to foreach() over every element of a multidimensional array, i.e. somethi=

ng
like,

=C2=A0 =C2=A0int a;

=C2=A0 =C2=A0foreach(ref int n; a)
=C2=A0 =C2=A0 =C2=A0 =C2=A0n =3D uniform!"[]"(0, 1);

... or is it necessary to do a separate foreach over each dimension?

When you use foreach on a, it will iterate on a's elements, in this
case the bidimensional subarrays. You'll have to foreach separately on
each dimension to act on individual, scalar, elements.

Of course, D can automate this for you:

module test;

import std.algorithm;
import std.random;
import std.range;
import std.stdio;

template rank(R)
{
static if (isInputRange!R)
enum size_t rank =3D 1 + rank!(ElementType!R);
else
enum size_t rank =3D 0;
}

/**
Maps fun at depth downToRank inside a range of ranges.
*/
auto depthMap(alias fun, int downToRank =3D 0, R)(R range)
{
static if (0<=3D downToRank && downToRank < rank!R)
return map!(depthMap!(fun, downToRank, ElementType!R))(range);
else // rank!R =3D=3D 0 -> range is a scalar, not a range
return fun(range);
}

int fill(int _) { return uniform!"[]"(0,1);}

void main()
{
auto r =3D [[[0,0,0],[0,0,0],[0,0,0]],[[0,0,0],[0,0,0],[0,0,0]],[[0,0,0=
],[0,0,0],[0,0,0]]];
auto d =3D depthMap!fill(r);
writeln(r);
writeln(d);
}

The second query relates to slices. =C2=A0Suppose I have an N*M 2d array =

(let's
call it a), and a 1d array of length N (let's call it s) whose elements a=

re
unsigned integers whose values fall in [0, M). =C2=A0What I'd like is to =

take a
slice of a such that I get out a 1d array of length N such that the i'th
element corresponds to a[s[i]].

Is this possible? =C2=A0And if so, how is it achieved?

You meant a[i][s[i]], right?

auto a =3D [[1,2,3,4,5],[6,7,8,9,10],[11,12,13,14,15]];
auto s =3D [4,3,2];
auto slice =3D map!( i =3D> a[i][s[i]])(iota(0,3));
writeln(slice);

This is a lazy range whose elements are a[i][s[i]]. If you're not
afraid of allocating everything at once, call std.range.array() on it
to transform it into an array:

auto slice2 =3D array(slice);
writeln(typeof(slice2).stringof);

Philippe
```
May 30 2012