integral to floating point conversion

[Andrei Alexandrescu via Digitalmars-d]

On 07/03/2016 05:52 AM, Guillaume Boucher wrote:
> On Sunday, 3 July 2016 at 09:08:14 UTC, Ola Fosheim Grøstad wrote:
>> On Saturday, 2 July 2016 at 20:17:59 UTC, Andrei Alexandrescu wrote:
>>> So what's the fastest way to figure that an integral is convertible
>>> to a floating point value precisely (i.e. no other integral converts
>>> to the same floating point value)? Thanks! -- Andrei
>>
>> If it is within what the mantissa can represent then it is easy. But
>> you also have to consider the cases where the mantissa is shifted.
>>
>> So the real answer is:
>>
>> n is an unsigned 64 bit integer
>>
>> mbits = representation bits for mantissa +1
>>
>> tz = trailing_zero_bits(n)
>> lz = leading_zero_bits(n)
>>
>> assert(mbits >= (64 - tz - lz))
>
> This is the correct answer for another definition of "precisely
> convertible", not the one Andrei gave.

Well to be more precise here's what I'm looking for. When you compare an 
integral with a floating point number, the integral is first converted 
to floating point format. I.e. for long x and double y, x == y is the 
same as double(x) == y.

Now, say we want to eliminate the "bad" cases of this comparison. Those 
would make it non-transitive. Consider two distinct longs x1 and x2. If 
they convert to the same double y, then x1 == y and x2 == y are true, 
which is contradictory with x1 != x2.


Andrei