digitalmars.D.learn - mod of negative number
- Craig Dillabaugh (15/15) Sep 23 Why does the following program:
- Craig Dillabaugh (3/18) Sep 23 Opps, sorry. I was expecting 7680 (not -1 or 5568).
- Craig Dillabaugh (5/9) Sep 23 snip
- Jonathan M Davis (5/20) Sep 23 Well, this is what the spec says:
- Sergey (6/21) Sep 24 Check section "In Programming languages" -
- Timon Gehr (24/45) Sep 24 Yes, what this does is to promote `-1` to `uint.max` and the result you
Why does the following program:
\<code>
import std.stdio;
int main(string[] args) {
uint Q = 7681;
writeln("Val = ", -1 % Q);
return 0;
}
\</code>
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.
Sep 23
On Monday, 23 September 2024 at 19:52:02 UTC, Craig Dillabaugh
wrote:
Why does the following program:
\<code>
import std.stdio;
int main(string[] args) {
uint Q = 7681;
writeln("Val = ", -1 % Q);
return 0;
}
\</code>
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.
Opps, sorry. I was expecting 7680 (not -1 or 5568).
Sep 23
Craig Dillabaugh <craig.dillabaugh gmail.com> On Monday, 23 September 2024 at 20:02:25 UTC, Craig Dillabaugh wrote:On Monday, 23 September 2024 at 19:52:02 UTC, Craig Dillabaugh wrote:snipWhy does the following program:Opps, sorry. I was expecting 7680 (not -1 or 5568).After a bit of research I see this is the same behavior as C, so I guess that explains the result.
Sep 23
On Monday, September 23, 2024 1:52:02 PM MDT Craig Dillabaugh via Digitalmars-
d-learn wrote:
Why does the following program:
\<code>
import std.stdio;
int main(string[] args) {
uint Q = 7681;
writeln("Val = ", -1 % Q);
return 0;
}
\</code>
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.
Well, this is what the spec says:
https://dlang.org/spec/expression.html#division
- Jonathan M Davis
Sep 23
On Monday, 23 September 2024 at 19:52:02 UTC, Craig Dillabaugh
wrote:
Why does the following program:
\<code>
import std.stdio;
int main(string[] args) {
uint Q = 7681;
writeln("Val = ", -1 % Q);
return 0;
}
\</code>
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
Sep 24
On 9/23/24 21:52, Craig Dillabaugh wrote:
Why does the following program:
\<code>
import std.stdio;
int main(string[] args) {
uint Q = 7681;
writeln("Val = ", -1 % Q);
return 0;
}
\</code>
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.
Sep 24