Re: Why pow() won't go beyond 2^31?
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Re: Why pow() won't go beyond 2^31?
On Thu, Nov 29, 2018 at 8:10 AM Murilo via Digitalmars-d-learn < digitalmars-d-learn@puremagic.com> wrote: > I am using the function pow() from std.math but if I try pow(2, > 32) it returns 0, it doesn't compute beyond the maximum value of > an int(2^31) and I am working with long. What should I do? > if you look at doc: https://dlang.org/phobos/std_math.html#.pow.2 you will see that return type is infered from pow arguments, so if both arguments are int for example the return value would be int too https://run.dlang.io/is/FMVJhY so if you want to have long as output one of your args should be (u)long or you can enforce that by pow!long(2,32); https://run.dlang.io/is/WlDfsE
Re: Why pow() won't go beyond 2^31?
On Thursday, 29 November 2018 at 07:07:06 UTC, Murilo wrote:
I am using the function pow() from std.math but if I try pow(2,
32) it returns 0, it doesn't compute beyond the maximum value
of an int(2^31) and I am working with long. What should I do?
what exactly is your input?
´´´
import std.stdio;
import std.experimental.all;
void main()
{
long u = 2;
assert(pow(u,32) == 4294967296);
assert(pow(2UL,32) == 4294967296);
}
´´´
Why pow() won't go beyond 2^31?
I am using the function pow() from std.math but if I try pow(2, 32) it returns 0, it doesn't compute beyond the maximum value of an int(2^31) and I am working with long. What should I do?
