I am guessing there is a bug in b formatting. Note, even base 10:

   10b123456789010234567
123456789010234560

   10b123456789010234562
123456789010234560

Almost seems like it went "float" internally or some such.

On 4/25/2018 11:12 AM, Thomas Hickey wrote:
Maybe a better question is why are all of these true?

81985529216486896 = 16b123456789abcdf0

1

81985529216486896 = 16b123456789abcdef

1

81985529216486880 = 16b123456789abcde8

1

I'm running j806/j64/windows.

--Th

On Wed, Apr 25, 2018 at 10:40 AM, Raul Miller <[email protected]> wrote:

P.S.

    240 205 171 137 103 69 35 1 p. 256x
81985529216486896

--
Raul


On Wed, Apr 25, 2018 at 10:39 AM, Raul Miller <[email protected]>
wrote:
    _3 ic 240 205 171 137 103 69 35 1 { a.
81985529216486896

The byte order the machine is using here is little endian

https://en.wikipedia.org/wiki/Endianness#Little-endian

That means the least significant byte here was 239 (your example) or
240 (my example).

But your number was even and 239 is odd...

--
Raul

On Wed, Apr 25, 2018 at 10:29 AM, Thomas Hickey <[email protected]>
wrote:
I have a 64 bit number:
16b123456789abcdef = 81985529216486896

encoded in 8 bytes in a file:
239 205 171 137 103 69 35 1

but

_3 ic 239 205 171 137 103 69 35 1 { a. returns 81985529216486895 (1 less
than I expected)

16 #.inv 81985529216486895 returns 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

16 #.inv 81985529216486896 returns 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0

I suppose this has something to do with signed 64 bit integers, but I
don't
understand it. I'm running on a Intel machine (Surface laptop).

--Th
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