Thread overview
Append to 'map' result
Jul 04, 2017
Jean-Louis Leroy
Jul 04, 2017
H. S. Teoh
Jul 04, 2017
Jean-Louis Leroy
Jul 05, 2017
Ali Çehreli
Jul 05, 2017
Jean-Louis Leroy
Jul 05, 2017
Ali Çehreli
Jul 05, 2017
Jean-Louis Leroy
Jul 04, 2017
Nicholas Wilson
Jul 05, 2017
Jonathan M Davis
July 04, 2017
On Tue, Jul 04, 2017 at 11:27:25PM +0000, Jean-Louis Leroy via Digitalmars-d-learn wrote:
> I want to create a range that consists of the result of a map()
> followed by a value, e.g.:
> 
>   int[] x = [ 1, 2, 3];
>   auto y = map!(x => x * x)(x);
>   auto z = y ~ 99; // how???
> 
> I have tried several variations: convert 99 to a dynamic array, to a range, convert range to dynamic array (couldn't even figure that one); to no avail.
[...]

Try:

	auto z = chain(y, only(99));


T

-- 
I don't trust computers, I've spent too long programming to think that they can get anything right. -- James Miller
July 04, 2017
I want to create a range that consists of the result of a map() followed by a value, e.g.:

  int[] x = [ 1, 2, 3];
  auto y = map!(x => x * x)(x);
  auto z = y ~ 99; // how???

I have tried several variations: convert 99 to a dynamic array, to a range, convert range to dynamic array (couldn't even figure that one); to no avail.

Help please...


July 04, 2017
On Tuesday, 4 July 2017 at 23:27:25 UTC, Jean-Louis Leroy wrote:
> I want to create a range that consists of the result of a map() followed by a value, e.g.:
>
>   int[] x = [ 1, 2, 3];
>   auto y = map!(x => x * x)(x);
>   auto z = y ~ 99; // how???
>
> I have tried several variations: convert 99 to a dynamic array, to a range, convert range to dynamic array (couldn't even figure that one); to no avail.
>
> Help please...

using
   auto y = x.map!(x => x * x).array;
will work.
July 04, 2017
On Tuesday, 4 July 2017 at 23:26:28 UTC, H. S. Teoh wrote:
> On Tue, Jul 04, 2017 at 11:27:25PM +0000, Jean-Louis Leroy via Digitalmars-d-learn wrote:
>> I want to create a range that consists of the result of a map()
>> followed by a value, e.g.:
>> 
>>   int[] x = [ 1, 2, 3];
>>   auto y = map!(x => x * x)(x);
>>   auto z = y ~ 99; // how???
>> 
>> I have tried several variations: convert 99 to a dynamic array, to a range, convert range to dynamic array (couldn't even figure that one); to no avail.
> [...]
>
> Try:
>
> 	auto z = chain(y, only(99));

Thanks!

However, when I try to use this construct in my real code, I still have a problem:

    // k is a ClassInfo
    chain(map!(i => i.classinfo)(k.interfaces), only(k.base))

meth.d(311): Error: template std.range.chain cannot deduce function from argument types !()(MapResult!(__lambda3, Interface[]), OnlyResult!(TypeInfo_Class, 1LU)), candidates are:
/usr/include/d/std/range/package.d(795):        std.range.chain(Ranges...)(Ranges rs) if (Ranges.length > 0 && allSatisfy!(isInputRange, staticMap!(Unqual, Ranges)) && !is(CommonType!(staticMap!(ElementType, staticMap!(Unqual, Ranges))) == void))



July 04, 2017
On Tuesday, July 04, 2017 23:50:57 Nicholas Wilson via Digitalmars-d-learn wrote:
> On Tuesday, 4 July 2017 at 23:27:25 UTC, Jean-Louis Leroy wrote:
> > I want to create a range that consists of the result of a map()
> >
> > followed by a value, e.g.:
> >   int[] x = [ 1, 2, 3];
> >   auto y = map!(x => x * x)(x);
> >   auto z = y ~ 99; // how???
> >
> > I have tried several variations: convert 99 to a dynamic array, to a range, convert range to dynamic array (couldn't even figure that one); to no avail.
> >
> > Help please...
>
> using
>     auto y = x.map!(x => x * x).array;
> will work.

Yeah, that will convert the range to a dynamic array, but if you're looking to do is append onto a range, then using chain like H.S. Teoh showed is better. And even if you ultimately want a dynamic array, using chain and then calling array on the result than allocating the array and then appending to it as an array means that you're only going to allocate the array once, whereas if you append to the array, you might end up causing it to be reallocated (though if you're just appending a single element to it it'll have the space unless in std.array.array allocates a dynamic array with exactly the space required for its elements, which I doubt).

- Jonathan M Davis

July 04, 2017
On 07/04/2017 04:57 PM, Jean-Louis Leroy wrote:
> On Tuesday, 4 July 2017 at 23:26:28 UTC, H. S. Teoh wrote:
>> On Tue, Jul 04, 2017 at 11:27:25PM +0000, Jean-Louis Leroy via
>> Digitalmars-d-learn wrote:
>>> I want to create a range that consists of the result of a map()
>>> followed by a value, e.g.:
>>>
>>>   int[] x = [ 1, 2, 3];
>>>   auto y = map!(x => x * x)(x);
>>>   auto z = y ~ 99; // how???
>>>
>>> I have tried several variations: convert 99 to a dynamic array, to a
>>> range, convert range to dynamic array (couldn't even figure that
>>> one); to no avail.
>> [...]
>>
>> Try:
>>
>>     auto z = chain(y, only(99));
>
> Thanks!
>
> However, when I try to use this construct in my real code, I still have
> a problem:
>
>     // k is a ClassInfo
>     chain(map!(i => i.classinfo)(k.interfaces), only(k.base))
>
> meth.d(311): Error: template std.range.chain cannot deduce function from
> argument types !()(MapResult!(__lambda3, Interface[]),
> OnlyResult!(TypeInfo_Class, 1LU)), candidates are:
> /usr/include/d/std/range/package.d(795):
> std.range.chain(Ranges...)(Ranges rs) if (Ranges.length > 0 &&
> allSatisfy!(isInputRange, staticMap!(Unqual, Ranges)) &&
> !is(CommonType!(staticMap!(ElementType, staticMap!(Unqual, Ranges))) ==
> void))
>
>
>
No time to dig deeper but this is because the two ranges that chain() receives do not have a common type. (Rather, that type is 'void'):

import std.algorithm;
import std.range;
import std.traits;

class C {
}

void main() {
    auto k = C.classinfo;
    auto m = map!(i => i.classinfo)(k.interfaces);
    auto b = only(k.base);

    // Types
    alias M = typeof(m);
    alias B = typeof(b);

    // Element types
    alias EM = ElementType!M;
    alias EB = ElementType!B;

    // Two of chain's constraints are satisfied:
    static assert(isInputRange!M);
    static assert(isInputRange!B);
    // Now it needs to see a CommonType of range elements

    pragma(msg, EM);    // const(void)[]()
    pragma(msg, EB);    // const(void)[]()

    pragma(msg, CommonType!(EM, EB)); // void

    // For that reason, the following does not work:
    // chain(m, b);
}

Ali


July 05, 2017
On Wednesday, 5 July 2017 at 00:28:01 UTC, Ali Çehreli wrote:
> On 07/04/2017 04:57 PM, Jean-Louis Leroy wrote:
>>[...]
> No time to dig deeper but this is because the two ranges that chain() receives do not have a common type. (Rather, that type is 'void'):
>
> [...]

I suspect that that is the reason, although I checked in several ways that I had ClassInfo elements on both sides. Ok, I'll keep looking tomorrow with a fresh eye...
July 04, 2017
On 07/04/2017 05:52 PM, Jean-Louis Leroy wrote:
> On Wednesday, 5 July 2017 at 00:28:01 UTC, Ali Çehreli wrote:
>> On 07/04/2017 04:57 PM, Jean-Louis Leroy wrote:
>>> [...]
>> No time to dig deeper but this is because the two ranges that chain()
>> receives do not have a common type. (Rather, that type is 'void'):
>>
>> [...]
>
> I suspect that that is the reason, although I checked in several ways
> that I had ClassInfo elements on both sides. Ok, I'll keep looking
> tomorrow with a fresh eye...

This is related to TypeInfo.init, which unfortunately is *not* the .init property in this case. :( Luckily, it will be fixed in 2.075. This is what object.d has:

class TypeInfo
{
// ...

    /**
     * Return default initializer.  If the type should be initialized to all
     * zeros, an array with a null ptr and a length equal to the type size will
     * be returned. For static arrays, this returns the default initializer for
     * a single element of the array, use `tsize` to get the correct size.
     */
    abstract const(void)[] initializer() nothrow pure const @safe @nogc;

    /// $(RED Removed.) Please use `initializer` instead.
    @disable static const(void)[] init(); // since 2.074
    /* Planned for 2.075: Remove init, making way for the init type property,
    fixing issue 12233. */
}

The problem is, ElementType relies on .init, which works on most other type but not TypeInfo:

template ElementType(R)
{
    static if (is(typeof(R.init.front.init) T))
        alias ElementType = T;
    else
        alias ElementType = void;
}

So, unfortunately, most Phobos range functions cannot be used with TypeInfo as they would invariably touch ElementType. Here is a halfway workaround that uses each() instead of map(): :/

import std.algorithm;

class C {
}

void main() {
    auto k = C.classinfo;

    TypeInfo_Class[] arr;

    each!(i => arr ~= i.classinfo)(k.interfaces);
    arr ~= k.base;
}

Ali

July 05, 2017
On Wednesday, 5 July 2017 at 01:43:46 UTC, Ali Çehreli wrote:
> On 07/04/2017 05:52 PM, Jean-Louis Leroy wrote:
>> On Wednesday, 5 July 2017 at 00:28:01 UTC, Ali Çehreli wrote:
>>> On 07/04/2017 04:57 PM, Jean-Louis Leroy wrote:
>>>> [...]
>>> No time to dig deeper but this is because the two ranges that chain()
>>> receives do not have a common type. (Rather, that type is 'void'):
>>>
>>> [...]
>>
>
> This is related to TypeInfo.init, which unfortunately is *not* the .init property in this case. :( Luckily, it will be fixed in 2.075. This is what object.d has:
> [...]
> So, unfortunately, most Phobos range functions cannot be used with TypeInfo as they would invariably touch ElementType. Here is a halfway workaround that uses each() instead of map(): :/

Aaah, so I was not doing anything wrong ;-)

I was trying to factor this code:

      foreach (i; k.interfaces) {
        if (i.classinfo in classMap) {
          v.bases ~= classMap[i.classinfo];
        }
      }
      if (k.base in classMap) {
        v.bases ~= classMap[k.base];
      }

i.e. process the class base along with the interfaces in the same loop. I dwelled on it because my goal with this project is to learn (and evaluate) D.

Anyway, I am building my own graph of enriched ClassInfo objects; once that is done, I won't be impacted with the bug anymore.

Thanks for the diagnostic...

J-L