www.digitalmars.com         C & C++   DMDScript  

digitalmars.D.learn - Modulo that 'wraps' the number?

reply 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?
Jan 20
parent reply 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
Jan 20
parent reply 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
Jan 20
parent reply 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
Jan 20
parent reply 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
Jan 21
parent reply 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.
Jan 21
parent reply 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
Jan 21
parent 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.
Jan 21