On Wednesday, 23 October 2013 at 16:58:21 UTC, Dicebot wrote:
By simply printing all bytes of a union as %b (binary) and
checking bit pattern as per http://en.wikipedia.org/wiki/NaN :)
I am not really proficient in IEEE stuff so checking bits is
only thing I am capable of :P
I'm not sure I follow. The values I have are:
src.l: 7ff7a50200000000
dst1.l: 7ff7a50200000000
(0111111111110111101001010000001000000000000000000000000000000000)
dst2.l: 7fffa50200000000
(0111111111111111101001010000001000000000000000000000000000000000)
dst1.d: nan
dst2.d: nan
dst1.payload: 3d2810
dst2.payload: 3d2810
Making the bit pattern more explicit:
S E F P
0 11111111111 0
111101001010000001000000000000000000000000000000000
0 11111111111 1
111101001010000001000000000000000000000000000000000
Is the only different bit not the quiet/signaling NaN flag? Why
do you say that the payload is different?
