In Python -1%3 == 2 however in D -1%3 == -1
Is there a standard library function or something that gives me the Python version of modulo?

On 1/20/19 1:28 PM, faissaloo wrote:
> In Python -1%3 == 2 however in D -1%3 == -1
> Is there a standard library function or something that gives me the Python version of modulo?

Hm... (n%3+3)%3 should work.

-Steve

On Sunday, 20 January 2019 at 18:51:54 UTC, Steven Schveighoffer wrote:
> On 1/20/19 1:28 PM, faissaloo wrote:
>> In Python -1%3 == 2 however in D -1%3 == -1
>> Is there a standard library function or something that gives me the Python version of modulo?
>
> Hm... (n%3+3)%3 should work.
>
> -Steve

You only need the

(n % 3) + 3

On Monday, 21 January 2019 at 04:52:53 UTC, NaN wrote:
> On Sunday, 20 January 2019 at 18:51:54 UTC, Steven Schveighoffer wrote:
>> On 1/20/19 1:28 PM, faissaloo wrote:
>>> In Python -1%3 == 2 however in D -1%3 == -1
>>> Is there a standard library function or something that gives me the Python version of modulo?
>>
>> Hm... (n%3+3)%3 should work.
>>
>> -Steve
>
> You only need the
>
> (n % 3) + 3

You need both:

(-3 % 3) + 3 == 3
((-3 % 3) + 3) % 3 == 0

On 1/21/19 2:33 AM, Paul Backus wrote:
> On Monday, 21 January 2019 at 04:52:53 UTC, NaN wrote:
>> On Sunday, 20 January 2019 at 18:51:54 UTC, Steven Schveighoffer wrote:
>>> On 1/20/19 1:28 PM, faissaloo wrote:
>>>> In Python -1%3 == 2 however in D -1%3 == -1
>>>> Is there a standard library function or something that gives me the Python version of modulo?
>>>
>>> Hm... (n%3+3)%3 should work.
>>>
>>> -Steve
>>
>> You only need the
>>
>> (n % 3) + 3
> 
> You need both:
> 
> (-3 % 3) + 3 == 3
> ((-3 % 3) + 3) % 3 == 0

Probably, this optimizes into better code, but maybe the optimizer already does this with the expression above:

auto tmp = n % 3;
if(tmp < 0)
   tmp += 3;

It's just not a nice single expression.

-Steve

On Monday, 21 January 2019 at 15:01:27 UTC, Steven Schveighoffer wrote:
>
> Probably, this optimizes into better code, but maybe the optimizer already does this with the expression above:
>
> auto tmp = n % 3;
> if(tmp < 0)
>    tmp += 3;
>
> It's just not a nice single expression.
>
> -Steve

I don't think you can do this, imagine this case:

auto tmp = -1 % -3; // Note divisor in negative too.

tmp will be "-1" which already matches the Python way, so you can't add divisor anymore.

Matheus.

On 1/21/19 10:54 AM, Matheus wrote:
> On Monday, 21 January 2019 at 15:01:27 UTC, Steven Schveighoffer wrote:
>>
>> Probably, this optimizes into better code, but maybe the optimizer already does this with the expression above:
>>
>> auto tmp = n % 3;
>> if(tmp < 0)
>>    tmp += 3;
>>
>> It's just not a nice single expression.
>>
> 
> I don't think you can do this, imagine this case:
> 
> auto tmp = -1 % -3; // Note divisor in negative too.
> 
> tmp will be "-1" which already matches the Python way, so you can't add divisor anymore.

If the divisor is unknown, I hadn't considered the case. I was only considering the case where the divisor is a constant.

In that case, one can probably determine if the divisor is less than 0 before starting.

Not a Python user, just hoping to help answer questions :)

-Steve

On Monday, 21 January 2019 at 18:39:27 UTC, Steven Schveighoffer wrote:
> Not a Python user, just hoping to help answer questions :)

Yes I know in fact I'm not the OP but from what I understood from his post, he want to replicate, but I may be wrong.

If it's the case, this code may help him:

//DMD64 D Compiler 2.072.2
import std.stdio;
import std.math;

int mod(int i,int j){
    if(abs(j)==abs(i)||!j||!i||abs(j)==1){return 0;};
    auto m = (i%j);
    return (!m||sgn(i)==sgn(j))?(m):(m+j);
}

void main(){
    int j,i;
    for(j=1;j<9;++j){
        for(i=1;i<9;++i){
          writeln(-i, " % ", +j, " = ", (-i).mod(+j));
          writeln(+i, " % ", -j, " = ", (+i).mod(-j));
        }
    }
}


At least the code above gave me the same results as the Python version.

Matheus.