Re: mod of negative number
On Monday, 23 September 2024 at 19:52:02 UTC, Craig Dillabaugh
wrote:
Why does the following program:
\
import std.stdio;
int main(string[] args) {
uint Q = 7681;
writeln("Val = ", -1 % Q);
return 0;
}
\
Print
Val = 5568
Was hoping for 1.
I assume it is an integer promotion issue, but I am unsure how
to resolve. I tried replacing the Q with to!int(Q) but this
gave me -1, which is closer but not right either.
Check section "In Programming languages" -
https://en.wikipedia.org/wiki/Modulo
There are different operations (remainder and modulus) and
operators (modulo) which could be combined in a different ways
Split by top level comma
I *strongly* believe the answer is not reasonable using phoboes
tools; but Im going to leave it as an open question before I get
started on something from scratch.
Given well compliant phoboes-style ranges and a `ref string or
retro!string` that starts with '(',')','[',']','{','}', modify
that string so that you a) find the matching pair, b) return a
list of indexs for where the top level commas are
"(foo([1,2]),bar!"\","foobar" =>
"(foo([1,2]),bar!"\","",[12]
retro("abc[1,2,3,4,"❤️❤️❤️❤️","\","]") => "[1,2,3,4,"❤️❤️❤️❤️","\","]",
[2,4,6,8,35]
Re: mod of negative number
On 9/23/24 21:52, Craig Dillabaugh wrote:
Why does the following program:
\
import std.stdio;
int main(string[] args) {
uint Q = 7681;
writeln("Val = ", -1 % Q);
return 0;
}
\
Print
Val = 5568
Was hoping for 1.
I assume it is an integer promotion issue, but I am unsure how to
resolve. I tried replacing the Q with to!int(Q) but this gave me -1,
which is closer but not right either.
Yes, what this does is to promote `-1` to `uint.max` and the result you
are getting is `uint.max % 7686`.
`-1 % 7686` does not work because division rounds towards zero, so the
result of modulo may be negative.
In general `a%b` will give you a value between `-abs(b)+1` and
`abs(b)-1`, matching the sign of `a` (and ignoring the sign of `b`).
You can use something like `auto r=a%b; if(r<0) r+=abs(b);` or `auto
r=(a%b+b)%b;`, depending an your needs (note that those two are not the
same for negative `b`, but they match for positive `b`).
This implementation is equivalent to `(a%b+b)%b`, if you want to avoid
the second modulo and the compiler is not smart enough:
```d
int floormod(int a,int b){
bool sign=(a<0)^(b<0);
auto r=a%b;
if(sign&&r!=0) r+=b;
return r;
}
```
The interpretation of this is to compute the remainder for a division
that rounds to negative infinity.
Sometimes one might also want `a%0 = a`, but this is a special case that
has to be checked as the language will give you a division by zero error.
