Re: Modulo that 'wraps' the number?
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.
Re: Modulo that 'wraps' the number?
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
Re: Modulo that 'wraps' the number?
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.
Re: Modulo that 'wraps' the number?
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
Re: Modulo that 'wraps' the number?
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
Re: Modulo that 'wraps' the number?
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
Re: Modulo that 'wraps' the number?
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
Modulo that 'wraps' the number?
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?
