boole 9 (ex-nor) bug

[Jef!]

Exclusive NOR the output should be 1 if both inputs are the same, 0/0 (operator bit 8 ) + 1/1 (operator bit 1). So operator bit 9.

(boole 9 0 0) and (boole 9 1 1) should return 1, while (boole 9 0 1) and (boole 9 1 0) should return 0

I just tested in Cad 2013/2014/2015 and 2016 and...

(boole 9 0 0) and (boole 9 1 1) return -1 while (boole 9 0 1) and (boole 9 1 0) return -2. (!)

 

I worked around that bug using "not XOR", ie: (- 1 (boole 6 0 0)), but I would still be curious if other and/or newer versions of Autocad than the above-mentioned (2012- / 2017+) had/still have that (boole 9 bug. I would also be curious to know if BricsCad have the same bug. What cad software do you use, what version and what does (boole 9 return?

 

Thanks and have a great week end!

[Lee Mac]

This is not a bug. Signed integers are stored in 32-bit 2's complement representation and so we have the following when performing the bitwise operations:

(boole 9 0 1) =

0 0 0 0 0 0 0 0  0 0 0 0 0 0 0 0  0 0 0 0 0 0 0 0  0 0 0 0 0 0 0 0 =  0
0 0 0 0 0 0 0 0  0 0 0 0 0 0 0 0  0 0 0 0 0 0 0 0  0 0 0 0 0 0 0 1 =  1
----------------------------------------------------------------------- NOT XOR
1 1 1 1 1 1 1 1  1 1 1 1 1 1 1 1  1 1 1 1 1 1 1 1  1 1 1 1 1 1 1 0 = -2

 

[Jef!]

36 minutes ago, Lee Mac said:

This is not a bug. Signed integers are stored in 32-bit 2's complement representation and so we have the following when performing the bitwise operations:

(boole 9 0 1) =

0 0 0 0 0 0 0 0  0 0 0 0 0 0 0 0  0 0 0 0 0 0 0 0  0 0 0 0 0 0 0 0 =  0
0 0 0 0 0 0 0 0  0 0 0 0 0 0 0 0  0 0 0 0 0 0 0 0  0 0 0 0 0 0 0 1 =  1
----------------------------------------------------------------------- NOT XOR
1 1 1 1 1 1 1 1  1 1 1 1 1 1 1 1  1 1 1 1 1 1 1 1  1 1 1 1 1 1 1 0 = -2

 

Oh! Right. Each bit of int1 is paired with the corresponding bit of int2. I was checking with 2 input logic gate literature. My bad.

As long as I'm comparing only 1's and 0's, (- 1 (boole 6 0 0)) does what I need but a more robust solution to get 1 if and only if both int inputs bit x are the same would be as follow

(boole 1 x (boole 9 int1 int2))

 

(boole 1 1 (boole 9 0 0)) & (boole 1 1 (boole 9 1 1)) return 1

(boole 1 1 (boole 9 0 1)) & (boole 1 1 (boole 9 1 0)) return 0

 

Does it make sense? (or maybe I should ask is there anything that would make more sense?)