It is common in a recursive function to amend a global array using the function parameter to refer to the element eg

int[10];

void foo (int i) {
  foreach (x; 0 .. 9) {
     t[i] = x;
     foo ();

C in a for loop allows use of t[i] directly as the iterating variable so you don't need the dummy variable x which has no real function.

D does not.  The error generated is "no identifier for declarator t[i]". For a long time, I thought that was specific to foreach but the same thing happens with for.

It looks like an unnecessary restriction.  Am I missing something?

Of course, in the first line I meant to say
int[10] t;

On Mon, 02 Jun 2014 15:21:06 -0400, Logesh Pillay <lp.court.jester@gmail.com> wrote:

> It is common in a recursive function to amend a global array using the function parameter to refer to the element eg
>
> int[10];
>
> void foo (int i) {
>    foreach (x; 0 .. 9) {
>       t[i] = x;
>       foo ();
>
> C in a for loop allows use of t[i] directly as the iterating variable so you don't need the dummy variable x which has no real function.
>
> D does not.  The error generated is "no identifier for declarator t[i]". For a long time, I thought that was specific to foreach but the same thing happens with for.
>
> It looks like an unnecessary restriction.  Am I missing something?

Can you post a full compiling example? I can't figure out what you are trying to do with this code.

-Steve

On Monday, 2 June 2014 at 19:21:07 UTC, Logesh Pillay wrote:
> It is common in a recursive function to amend a global array using the function parameter to refer to the element eg
>
> int[10];
>
> void foo (int i) {
>   foreach (x; 0 .. 9) {
>      t[i] = x;
>      foo ();
>
> C in a for loop allows use of t[i] directly as the iterating variable so you don't need the dummy variable x which has no real function.
>
> D does not.  The error generated is "no identifier for declarator t[i]". For a long time, I thought that was specific to foreach but the same thing happens with for.
>
> It looks like an unnecessary restriction.  Am I missing something?

works fine for me: http://dpaste.dzfl.pl/6ff9a3909fde

Issue has nothing to do with recursion.  That's only where I keep seeing it.

eg a function to generate combinations.

import std.stdio;

int[3] t;

void foo (int i) {
  if (i == 3)
    writef("%s\n", t);
  else foreach (x; 0 .. 3) {
    t[i] = x;
    foo(i+1);
  }
}

void main() {
  foo(0);
}

In C, I could put in the equivalent for statement for foreach, t[i] as the iterating variable. I won't need x which exists as a middleman only and save myself two lines.

On Mon, 02 Jun 2014 15:47:02 -0400, Logesh Pillay <lp.court.jester@gmail.com> wrote:

> Issue has nothing to do with recursion.  That's only where I keep seeing it.
>
> eg a function to generate combinations.
>
> import std.stdio;
>
> int[3] t;
>
> void foo (int i) {
>    if (i == 3)
>      writef("%s\n", t);
>    else foreach (x; 0 .. 3) {
>      t[i] = x;
>      foo(i+1);
>    }
> }
>
> void main() {
>    foo(0);
> }
>
> In C, I could put in the equivalent for statement for foreach, t[i] as the iterating variable. I won't need x which exists as a middleman only and save myself two lines.

OK, I get it. You want to do:

foreach(t[i]; 0 .. 3)

But this doesn't work.

This should (the equivalent for C):

for(t[i] = 0; t[i] < 3; ++t[i])

I'm trying to think of a way to do this without loops, but not sure. Note that foreach is expected to be given a new variable to declare, so you can't foreach with an existing variable on the left.

-Steve

Sorry, I am embarassed to say I've just tried the for equivalent and it works with t[i] as the iterating variable.

import std.stdio;

int[3] t;

void foo (int i) {
  if (i == 3)
    writef("%s\n", t);
  else for (t[i] = 0; t[i] < 3; t[i]++)
    foo(i+1);
}

void main() {
  foo(0);
}

Sorry, yet another overhasty post

On Mon, 02 Jun 2014 15:58:01 -0400, Steven Schveighoffer <schveiguy@yahoo.com> wrote:


> I'm trying to think of a way to do this without loops, but not sure.

I'm surprised, I looked for some kind of "apply" function like map, but just calls some function with each element in the range.

Something like this would make this a 1 (2?) liner:

if(i == t.length) writeln(t) else each!((x) => {t[i] = x; foo(i+1);})(iota(x.length));

But I can't find a phobos primitive for each. Would have expected it in std.algorithm or std.functional?

-Steve

On Monday, 2 June 2014 at 20:23:12 UTC, Steven Schveighoffer wrote:
> On Mon, 02 Jun 2014 15:58:01 -0400, Steven Schveighoffer <schveiguy@yahoo.com> wrote:
>
>
>> I'm trying to think of a way to do this without loops, but not sure.
>
> I'm surprised, I looked for some kind of "apply" function like map, but just calls some function with each element in the range.
>
> Something like this would make this a 1 (2?) liner:
>
> if(i == t.length) writeln(t) else each!((x) => {t[i] = x; foo(i+1);})(iota(x.length));
>
> But I can't find a phobos primitive for each. Would have expected it in std.algorithm or std.functional?
>
> -Steve

Its been discussed a few times. There were some objections (IIRC Walter thought that there was no significant advantage over plain foreach).

On Mon, 02 Jun 2014 17:25:24 -0400, John Colvin <john.loughran.colvin@gmail.com> wrote:

> On Monday, 2 June 2014 at 20:23:12 UTC, Steven Schveighoffer wrote:
>> On Mon, 02 Jun 2014 15:58:01 -0400, Steven Schveighoffer <schveiguy@yahoo.com> wrote:
>>
>>
>>> I'm trying to think of a way to do this without loops, but not sure.
>>
>> I'm surprised, I looked for some kind of "apply" function like map, but just calls some function with each element in the range.
>>
>> Something like this would make this a 1 (2?) liner:
>>
>> if(i == t.length) writeln(t) else each!((x) => {t[i] = x; foo(i+1);})(iota(x.length));
>>
>> But I can't find a phobos primitive for each. Would have expected it in std.algorithm or std.functional?
>>
>> -Steve
>
> Its been discussed a few times. There were some objections (IIRC Walter thought that there was no significant advantage over plain foreach).

Indeed, foreach is like such a construct:

... else each!((x) {t[i] = x; foo(i+1);})(iota(t.length));
... else foreach(x; 0 .. t.length) {t[i] = x; foo(i+1);}

It's even shorter and clearer.

I agree with Walter. Since such a construct by definition wouldn't return anything, you can't chain it. There really is little reason to have it.

-Steve