• faissaloo (3/3) Jan 20 2019 In Python -1%3 == 2 however in D -1%3 == -1
• Steven Schveighoffer (3/6) Jan 20 2019 Hm... (n%3+3)%3 should work.
• NaN (4/10) Jan 20 2019 You only need the
• Paul Backus (4/16) Jan 20 2019 You need both:
• Steven Schveighoffer (8/27) Jan 21 2019 Probably, this optimizes into better code, but maybe the optimizer
• Matheus (7/14) Jan 21 2019 I don't think you can do this, imagine this case:
• Steven Schveighoffer (7/25) Jan 21 2019 If the divisor is unknown, I hadn't considered the case. I was only
• Matheus (25/26) Jan 21 2019 Yes I know in fact I'm not the OP but from what I understood from

faissaloo <faissaloo gmail.com> writes:

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?

Steven Schveighoffer <schveiguy gmail.com> writes:

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

NaN <divide by.zero> writes:

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

Paul Backus <snarwin gmail.com> writes:

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

Steven Schveighoffer <schveiguy gmail.com> writes:

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

Matheus <stop spam.com> writes:

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.

Steven Schveighoffer <schveiguy gmail.com> writes:

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

Matheus <stop spam.com> writes:

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.