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Is this the best way to flip a bit?


Random Terrain

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Either I didn't know or I forgot that XOR could flip a bit. Can somebody post how you'd use XOR to flip each of the 8 bits individually? A made a little chart to fill in to make it easier:

 

a{0} = !a{0}	same as	a ^

a{1} = !a{1}	same as	a ^ 1

a{2} = !a{2}	same as	a ^

a{3} = !a{3}	same as	a ^

a{4} = !a{4}	same as	a ^

a{5} = !a{5}	same as	a ^

a{6} = !a{6}	same as	a ^

a{7} = !a{7}	same as	a ^

 

Thanks.

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Can someone also post a little example of how this works? It doesn't seem to be working correctly in my program. I must not be doing it right.

You would do

 

   a = a ^ %xxxxxxxx

where %xxxxxxxx is a bit pattern you want to flip a with. Anywhere there's a 0, the bit will be left as is. Anywhere there's a 1, the bit will be flipped. So for example if you wanted to flip bit 4, you would do

 

   a = a ^ %00010000

Bits 0-3, as well as bits 5-7, will stay as they are, and bit 4 will be flipped. You can also use decimal or hex values that represent the same bit pattern. For example

 

   dim frame = a
  frame = 0

loop

  drawscreen
  frame = frame ^ 1

  goto loop

will flip frame between 0 and 1, so you can tell whether you're on an odd frame or an even frame-- which would be useful if you were doing things that need to be split across two frames, and needed a way to tell if it's time to do stuff1 or stuff2.

 

Michael

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Thanks. Now I know why it wasn't working. I was using the wrong number with '^'.

 

a{0} = !a{0}	same as	a = a ^ 1

a{1} = !a{1}	same as	a = a ^ 2

a{2} = !a{2}	same as	a = a ^ 4

a{3} = !a{3}	same as	a = a ^ 8

a{4} = !a{4}	same as	a = a ^ 16

a{5} = !a{5}	same as	a = a ^ 32

a{6} = !a{6}	same as	a = a ^ 64

a{7} = !a{7}	same as	a = a ^ 128

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