Re: Why does compose from std.functional return a templated function
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 = !(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
Re: Why does compose from std.functional return a templated function
On Wed, Sep 16, 2020 at 12:50 PM 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?) > > (maybe, because I might want to compose together templated non-initialized functions?) Yes
Re: Why does compose from std.functional return a templated function
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.
Re: Why does compose from std.functional return a templated function
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
Why does compose from std.functional return a templated function
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?)
