On Tuesday, 17 January 2017 at 18:25:46 UTC, kinke wrote:
It should and I looked into that as well, but I didn't like the
implementation as loop:
https://github.com/dlang/phobos/blob/master/std/math.d#L5988
A special case for base x == 2 wouldn't hurt.
That seems strange. Why isn't that a bu
On Tuesday, 17 January 2017 at 17:56:13 UTC, Nordlöw wrote:
On Tuesday, 17 January 2017 at 16:40:57 UTC, kinke wrote:
If it doesn't have to be D ;), it can be as simple as
`core.stdc.math.ldexp(1, exponent)`. No CTFE though.
Isn't it a simple as
2.0^^exponent
?
It should and I looked i
On Tuesday, 17 January 2017 at 16:40:57 UTC, kinke wrote:
If it doesn't have to be D ;), it can be as simple as
`core.stdc.math.ldexp(1, exponent)`. No CTFE though.
Isn't it a simple as
2.0^^exponent
?
On Tuesday, 17 January 2017 at 00:08:24 UTC, Nordlöw wrote:
How do I best initialize a D double to an exact mantissa and
exponent representation?
I'm specifically interested in
2^^i for all i in [min_exp, max_exp]
If it doesn't have to be D ;), it can be as simple as
`core.stdc.math.lde
On Tuesday, 17 January 2017 at 00:08:24 UTC, Nordlöw wrote:
How do I best initialize a D double to an exact mantissa and
exponent representation?
I'm specifically interested in
2^^i for all i in [min_exp, max_exp]
See
std.bitmanip : FloatRep , DoubleRep;
On Tuesday, 17 January 2017 at 00:08:24 UTC, Nordlöw wrote:
How do I best initialize a D double to an exact mantissa and
exponent representation?
I'm specifically interested in
2^^i for all i in [min_exp, max_exp]
This mach module can do the job:
https://github.com/pineapplemachine/mach
How do I best initialize a D double to an exact mantissa and
exponent representation?
I'm specifically interested in
2^^i for all i in [min_exp, max_exp]
