Want a function that determines a double or float given its 80-bit IEEE 754 SANE (big endian) representation

[z]

On Tuesday, 22 August 2023 at 22:38:23 UTC, dan wrote:
> Hi,
>
> I'm parsing some files, each containing (among other things) 10 
> bytes said to represent an IEEE 754 extended floating point 
> number, in SANE (Standard Apple Numerical Environment) form, as 
> SANE existed in the early 1990s (so, big endian).
>
> Note that the number actually stored will probably be a 
> positive even integer less than 100,000, so a better format 
> would have been to store a two-byte ushort rather than a 
> 10-byte float.  However the spec chose to have an encoded float 
> there.
>
> I would like to have a function of the form
>
>     public bool ubytes_to_double( ubytes[10] u, out double d ) 
> { /* stuff */ }
>
> which would set d to the value encoded provided that the value 
> is a number and is sane, and otherwise just return false.
>
> So my plan is just to do this: examine the first 2 bytes to 
> check the sign and see how big the number is, and if it is 
> reasonable, convert the remaining 8 bytes to a fractional part, 
> perhaps ignoring the last 2 or 3 as not being significant.
>
> But --- it seems like this kind of task may be something that d 
> already does, maybe with some constructor of a double or 
> something.
>
> Thanks in advance for any suggestions.
>
> dan

On 32bit x86 an endianness swap and pointer cast to `real` should 
be enough.(seems to be the same format but i could be wrong.)
Else(afaik `real` on 64 bit x86 is just `double`?) you can always 
isolate sign mantissa and exponent to three isolated `double` 
values(cast from integer to `double`) and 
recalculate(`sign*mantissa*(2^^exponent)` according to wikipedia) 
the floating point number, since they mostly contain integers 
precision loss probably won't be a problem.