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tib
Switzerland
105 Posts |
 Posted - 09/06/2006 : 12:47:46 PM
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Hi,
might be a stupid question... but could not find a help nor forum topic...
AND & and OR | operators exist in Labtalk.
Is there an easy way to realize NOT, XOR and NAND in LabTalk?
e.g.
bin.(dec) --> bin. (dec.)
NOT 1100 (12) --> 0011 (3)
1101 (13) XOR 0100 (4) --> 1011 (11)
...
Thanks for hints,
Tilman.
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tib
Switzerland
105 Posts |
Posted - 09/07/2006 : 04:28:11 AM
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Hi,
in the meantime I found a possible solution.
Key is the NOT function. NOT can be realized in the following way:
NOT z = MOD(2^n-1-z,2^n)
where
n is the number of digits of the binary number
z is the number you want to apply NOT on
with this you can realize NAND:
x NAND y = NOT (x & y)
with this you can realize XOR:
x XOR y = (x OR y) AND NOT(x AND y)
= (x | y) & (MOD(2^n-1-(x & y),2^n))
Maybe there is even a shorter way...
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Mike Buess
USA
3037 Posts |
Posted - 09/07/2006 : 10:25:46 AM
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Hi Tilman,
I question your definition of the bitwise NOT (or complement) operator. My C reference uses all 16 bits of unsigned integer to represent a decimal number in bitwise operations. Therefore...
NOT x = 65535 - x
x NAND y = NOT (x&y) = 65535 - (x&y)
x XOR y = (x|y) & (65535 - (x&y))
In Origin C you can use ^ for XOR by including a pragma in your code. Therefore you can create your own bitwise operators as shown below. Note that bw_xor(x,y) returns the same integer as does my LabTalk XOR expression above.
#pragma xor(push, FALSE)
uint bw_xor(uint x, uint y)
{
return x^y;
}
uint bw_not(uint x)
{
return 65535 - x;
}
uint bw_nand(uint x, uint y)
{
return 65535 - (x&y);
}
Mike Buess
Origin WebRing Member
Edited by - Mike Buess on 09/07/2006 11:37:11 AM |
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Topic |
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bitwise NOT, XOR, NAND ?