Re: 9999999999999999.0 - 9999999999999998.0
On Sunday, 6 January 2019 at 01:05:08 UTC, Adam D. Ruppe wrote: That's because it is done at compile time, since both are compile-time constants. The compiler will evaluate it using the maximum precision available to the compiler, ignoring your request to cast it to double (which annoys some people who value predictability over precision btw). At different precisions, you get different results. Thanks for that explanation, Adam! Very helpful. On Sunday, 6 January 2019 at 03:33:45 UTC, Jesse Phillips wrote: Since you got your answer you may also like http://dconf.org/2016/talks/clugston.html Thank you for pointing out that talk, Jesse. I will set aside some time to go through that! Samir
Re: 9999999999999999.0 - 9999999999999998.0
On Sunday, 6 January 2019 at 00:20:40 UTC, Samir wrote: [1] https://news.ycombinator.com/item?id=18832155 [2] https://en.wikipedia.org/wiki/IEEE_754 Since you got your answer you may also like http://dconf.org/2016/talks/clugston.html
Re: 9999999999999999.0 - 9999999999999998.0
On Sunday, 6 January 2019 at 00:20:40 UTC, Samir wrote:
import std.stdio: writeln;
void main(){
writeln(cast(double).0-9998.0);
}
That's because it is done at compile time, since both are
compile-time constants. The compiler will evaluate it using the
maximum precision available to the compiler, ignoring your
request to cast it to double (which annoys some people who value
predictability over precision btw). At different precisions, you
get different results.
I suggest breaking it up into a different variable to force a
runtime evaluation instead of using the compiler's constant
folding.
import std.stdio: writeln;
void main(){
double d = .0;
writeln(d-9998.0);
}
This gives 1. Making it float instead of double, you get
something different. With real (which btw is higher precision,
but terrible speed), you get 1 - this is what the compiler
happened to use at compile time.
