October 12, 2015 Re: how to do iota(0,256) with ubytes ? (cf need for iotaInclusive) | ||||
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Posted in reply to Andrei Alexandrescu | On Monday, 12 October 2015 at 16:34:09 UTC, Andrei Alexandrescu wrote:
> On 10/12/15 11:20 AM, Per Nordlöw wrote:
>> On Friday, 9 October 2015 at 02:41:50 UTC, Timothee Cour wrote:
>>> of course this doesn't work:
>>> auto b=iota(ubyte(0), ubyte(256));
>>> //cannot implicitly convert expression (256) of type int to ubyte
>>
>> What about adding an overload supporting
>>
>> iota!ubyte(0, 256)
>
> We can add iota!T() with no arguments that spans the entire range of T (integral). -- Andrei
Nice, that'll also be consistent with uniform!T().
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October 12, 2015 Re: how to do iota(0,256) with ubytes ? (cf need for iotaInclusive) | ||||
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Posted in reply to Nordlöw | On 10/12/15 11:51 PM, Nordlöw wrote:
> On Monday, 12 October 2015 at 20:39:11 UTC, Andrei Alexandrescu wrote:
>>> Alexandrescu wrote:
>>>> We can add iota!T() with no arguments that spans the entire range of T
>>>> (integral). -- Andrei
>
> From a quick glance I couldn't find a way to reuse the existing
> overloads. Can anybody come with a reusing solution?
One possibility (can't look at the code now) is to change the implementation to use a closed interval for state, then use that. But then you still need to mind overflow. Whole range is just a bit special like that. -- Andrei
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October 12, 2015 Re: how to do iota(0,256) with ubytes ? (cf need for iotaInclusive) | ||||
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Posted in reply to Andrei Alexandrescu Attachments:
| On Mon, Oct 12, 2015 at 1:41 PM, Andrei Alexandrescu via Digitalmars-d < digitalmars-d@puremagic.com> wrote: > On 10/12/15 10:55 PM, Timothee Cour via Digitalmars-d wrote: > >> that's only a partial fix: >> I also would like to express: >> >> iotaInclusive(1,256) >> > > iota!ubyte.drop(1) > > iotaInclusive(1,256,3) >> > > iota!ubyte.drop(1).stride(3) > That's not a good workaround; it's error-prone in more general cases: auto fun(ubyte a, ubyte stride){ // return iotaInclusive(a,256, stride);// simple // error prone with your suggestion: auto b=some function of a, stride; return iota!ubyte.drop(b).stride(stride); } > > Just playing devil's advocate. > > > Andrei > > |
October 12, 2015 Re: how to do iota(0,256) with ubytes ? (cf need for iotaInclusive) | ||||
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Posted in reply to Nordlöw | On Monday, 12 October 2015 at 20:51:40 UTC, Nordlöw wrote:
> On Monday, 12 October 2015 at 20:39:11 UTC, Andrei Alexandrescu wrote:
>>> Alexandrescu wrote:
>>>> We can add iota!T() with no arguments that spans the entire range of T
>>>> (integral). -- Andrei
>
> From a quick glance I couldn't find a way to reuse the existing overloads. Can anybody come with a reusing solution?
Wouldn't it be enough changing the overload
auto iota(E)(E end)
to
auto iota(E)(E end = E.max)
?
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October 12, 2015 Re: how to do iota(0,256) with ubytes ? (cf need for iotaInclusive) | ||||
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Posted in reply to Nordlöw | On 10/12/2015 10:51 PM, Nordlöw wrote: > On Monday, 12 October 2015 at 20:39:11 UTC, Andrei Alexandrescu wrote: >>> Alexandrescu wrote: >>>> We can add iota!T() with no arguments that spans the entire range of T >>>> (integral). -- Andrei > > From a quick glance I couldn't find a way to reuse the existing > overloads. Can anybody come with a reusing solution? auto iota(T)(){ import std.range; return chain(iota(T.min,T.max),only(T.max)); } void main(){ import std.stdio; writeln(iota!byte()); writeln(iota!dchar()); } |
October 12, 2015 Re: how to do iota(0,256) with ubytes ? (cf need for iotaInclusive) | ||||
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Posted in reply to Jacques Müller | On 10/12/2015 11:52 PM, Jacques Müller wrote:
> On Monday, 12 October 2015 at 20:51:40 UTC, Nordlöw wrote:
>> On Monday, 12 October 2015 at 20:39:11 UTC, Andrei Alexandrescu wrote:
>>>> Alexandrescu wrote:
>>>>> We can add iota!T() with no arguments that spans the entire range of T
>>>>> (integral). -- Andrei
>>
>> From a quick glance I couldn't find a way to reuse the existing
>> overloads. Can anybody come with a reusing solution?
>
> Wouldn't it be enough changing the overload
>
> auto iota(E)(E end)
>
> to
>
> auto iota(E)(E end = E.max)
>
> ?
>
No.
import std.range;
writeln(iota(byte.max));
starts at 0
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v
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19,
20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37,
38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55,
56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73,
74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91,
92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107,
108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121,
122, 123, 124, 125, 126]
^
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ends before 127
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October 12, 2015 Re: how to do iota(0,256) with ubytes ? (cf need for iotaInclusive) | ||||
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Posted in reply to Vladimir Panteleev | On 10/12/2015 11:02 PM, Vladimir Panteleev wrote:
> On Monday, 12 October 2015 at 16:34:09 UTC, Andrei Alexandrescu wrote:
>> On 10/12/15 11:20 AM, Per Nordlöw wrote:
>>> On Friday, 9 October 2015 at 02:41:50 UTC, Timothee Cour wrote:
>>>> of course this doesn't work:
>>>> auto b=iota(ubyte(0), ubyte(256));
>>>> //cannot implicitly convert expression (256) of type int to ubyte
>>>
>>> What about adding an overload supporting
>>>
>>> iota!ubyte(0, 256)
>>
>> We can add iota!T() with no arguments that spans the entire range of T
>> (integral). -- Andrei
>
> Nice, that'll also be consistent with uniform!T().
As will iota!"[]".
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October 13, 2015 Re: how to do iota(0,256) with ubytes ? (cf need for iotaInclusive) | ||||
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Posted in reply to Timon Gehr | On Monday, 12 October 2015 at 13:17:32 UTC, Timon Gehr wrote:
> On 10/09/2015 04:41 AM, Timothee Cour via Digitalmars-d wrote:
>>
>> Could we have a function with iota_inclusive that has inclusive bounds
>> for 'end' parameter ?
>>
>
> iota!"[]" ?
Yes please.
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