Thread overview
Why does compose from std.functional return a templated function
Sep 16, 2020
Jan Hönig
Sep 16, 2020
Daniel Kozak
Sep 16, 2020
Jan Hönig
Sep 16, 2020
Daniel Kozak
Sep 16, 2020
Simen Kjærås
September 16, 2020
I have toyed with the compose template in std.functional and ran into some problems.
rikki_cattermole on discord helped me a lot to solve my problem.

However, what still remains (for me) to understand is why.

Source code for `compose`: https://github.com/dlang/phobos/blob/master/std/functional.d#L1161



`compose` pipes together given functions. And returns a TEMPLATED function.
Why does it return a templated function? At compile-time, compose definitly knows, what kinds of function it composes together. So with std.traits, it could put there a definitve type, depending on the given function(s).

I somewhat see, that inside compose itself, this probably solves some issues with casting (maybe).
However, the last composition, i.e. the one which is returned, does not need to be templated, since it is known, what parameter has the last function.

In my case, I failed to understand, that it returns a non-initialized template function, which lead into compile problems.

In general I can imagine that this leads to weird compile errors, which are hard to understand. (types, casting, etc.)


My main question is why? Is there something, which I am missing, that explains, why it is beneficial to return a templated function?

(maybe, because I might want to compose together templated non-initialized functions?)
September 16, 2020
On Wed, Sep 16, 2020 at 12:00 PM Jan Hönig via Digitalmars-d-learn < digitalmars-d-learn@puremagic.com> wrote:

> ...
>
> My main question is why? Is there something, which I am missing, that explains, why it is beneficial to return a templated function?
>
> (maybe, because I might want to compose together templated
> non-initialized functions?)
>

It has to be templated because than you can alias it and use it many times something like

import std.stdio;
import std.functional : compose;
import std.algorithm.comparison : equal;
import std.algorithm.iteration : map;
import std.array : split, array;
import std.conv : to;

alias StrArrToIntArr = compose!(array,map!(to!int), split);
void main()
{
    auto str1 = "2 4 8 9";
    int[] intArr = StrArrToIntArr(str1);
}


If compose would not be template it would need to store functions addresses so it would need to have some array of functions, this would be ineffective and need to use GC


September 16, 2020
On Wed, Sep 16, 2020 at 12:50 PM Daniel Kozak <kozzi11@gmail.com> wrote:

>
>
> On Wed, Sep 16, 2020 at 12:00 PM Jan Hönig via Digitalmars-d-learn < digitalmars-d-learn@puremagic.com> wrote:
>
>> ...
>>
>> My main question is why? Is there something, which I am missing, that explains, why it is beneficial to return a templated function?
>>
>> (maybe, because I might want to compose together templated
>> non-initialized functions?)
>
>
(maybe, because I might want to compose together templated
non-initialized functions?)

Yes


September 16, 2020
On Wednesday, 16 September 2020 at 10:50:06 UTC, Daniel Kozak wrote:
> On Wed, Sep 16, 2020 at 12:00 PM Jan Hönig via Digitalmars-d-learn < digitalmars-d-learn@puremagic.com> wrote:
>
>> ...
>>
>> My main question is why? Is there something, which I am missing, that explains, why it is beneficial to return a templated function?
>>
>> (maybe, because I might want to compose together templated
>> non-initialized functions?)
>>
>
> It has to be templated because than you can alias it and use it many times something like
>
> import std.stdio;
> import std.functional : compose;
> import std.algorithm.comparison : equal;
> import std.algorithm.iteration : map;
> import std.array : split, array;
> import std.conv : to;
>
> alias StrArrToIntArr = compose!(array,map!(to!int), split);
> void main()
> {
>     auto str1 = "2 4 8 9";
>     int[] intArr = StrArrToIntArr(str1);
> }
>
>
> If compose would not be template it would need to store functions addresses so it would need to have some array of functions, this would be ineffective and need to use GC

Right, if i give it non-initialized templated functions, it makes a lot of sense to return a template function as well.
But for functions without templates? Probably not a frequent usecase I guess.
September 16, 2020
On Wednesday, 16 September 2020 at 09:59:59 UTC, Jan Hönig wrote:
> I have toyed with the compose template in std.functional and ran into some problems.
> rikki_cattermole on discord helped me a lot to solve my problem.
>
> However, what still remains (for me) to understand is why.
>
> Source code for `compose`: https://github.com/dlang/phobos/blob/master/std/functional.d#L1161
>
>
>
> `compose` pipes together given functions. And returns a TEMPLATED function.
> Why does it return a templated function? At compile-time, compose definitly knows, what kinds of function it composes together. So with std.traits, it could put there a definitve type, depending on the given function(s).
>
> I somewhat see, that inside compose itself, this probably solves some issues with casting (maybe).
> However, the last composition, i.e. the one which is returned, does not need to be templated, since it is known, what parameter has the last function.
>
> In my case, I failed to understand, that it returns a non-initialized template function, which lead into compile problems.
>
> In general I can imagine that this leads to weird compile errors, which are hard to understand. (types, casting, etc.)
>
>
> My main question is why? Is there something, which I am missing, that explains, why it is beneficial to return a templated function?
>
> (maybe, because I might want to compose together templated non-initialized functions?)

It's perfectly possible to compose templated functions without wanting to specify the template parameters, and not allowing this would significantly hamper compose's usability.

The other complication is overloads. Here's a version of compose that generates the correct overloads:

template getOverloads(alias fn) {
    static if (__traits(compiles, __traits(getOverloads, __traits(parent, fn), __traits(identifier, fn), true))) {
        alias getOverloads = __traits(getOverloads, __traits(parent, fn), __traits(identifier, fn), true);
    } else {
        alias getOverloads = fn;
    }
}

template compose(funs...) if (funs.length > 0) {
    static foreach (overload; getOverloads!(funs[$-1])) {
        static if (__traits(isTemplate, overload)) {
            auto compose(Args...)(Args args) {
                static if (funs.length == 1) {
                    return overload(args);
                } else {
                    return funs[0](.compose!(funs[1..$])(args));
                }
            }
        } else {
            import std.traits : Parameters;
            auto compose(Parameters!overload args) {
                static if (funs.length == 1) {
                    return overload(args);
                } else {
                    return funs[0](.compose!(funs[1..$])(args));
                }
            }
        }
    }
}

As you can see, this is *a lot* more complex than the version in std.functional. The benefit is you can take the address of it easily. Here's how you can do the same with std.functional.compose:

import std.functional : compose;
import std.meta : Instantiate;

unittest {
    auto sun = &Instantiate!(compose!(fun, gun), int);
    sun(3);
}

void fun(T)(T t) {
}

int gun(int t) {
    return t;
}

I will argue the latter is an acceptable cost of avoiding the dense, bug-prone monstrosity of the former.

--
  Simen