[DONE] Swap 2 bits in an integer

[rsts]

I'm actually attempting to "flip" the two end bits in a 3 bit combo while keeing the middle bit constant: The higher order bits may stay the same.

011 becomes 110 or 001 becomes 100. I can NOT it and reverse it all, but then how do I NOT the middle bit by itself?

Or what's the better way?

cheers

[Kaeru Gaman]

what about XOR?

Code: Select all

value ! %101

[rsts]

I almost addressed the topic to you since I knew you were one of the experts in bitwise

Modulo? I never even considered that!

Take me a few mins to wrap my brain around this. Might be back with a question if I can't

Once again, thanks, Mr Gaman.

cheers

[Arctic Fox]

Modulo? I never even considered that!

Isn't it a binary number (like $ for hexadecimal numbers)

[rsts]

My mistake

Anyway, doesn't work for %000, which should remain 000

[Arctic Fox]

How about "the string way"?

Code: Select all

Val(ReverseString(RSet(Bin(value), 3, "0")))

Edit Updated with the use of RSet()

[netmaestro]

Anyway, doesn't work for %000, which should remain 000

I don't get that from your post. You said you wanted to flip bits 0 and 2 from whatever they were, didn't you? Then 000 would become 101.

[rsts]

Anyway, doesn't work for %000, which should remain 000

I don't get that from your post. You said you wanted to flip bits 0 and 2 from whatever they were, didn't you? Then 000 would become 101.

Of course it would. Like I said - I've been at this too long

cheers

[Kaeru Gaman]

indeed, that's what I meant...

Code: Select all

For n=0 To 7
  t = n ! %101
  Debug RSet(Bin(n),3,"0") + " -> " + RSet(Bin(t),3,"0")
Next

but this does nit really flip bit 0 and 2, it just reverses the two each.

the ReverseString solution should work better...

[Arctic Fox]

OK, I got it now, too - forget my previous post

[Michael Vogel]

Code: Select all

For n=0 To 7
	t = (n&%1111010) | ((n&1)<<2) | ((n>>2)&1)
	Debug RSet(Bin(n),3,"0") + " -> " + RSet(Bin(t),3,"0")
Next

[rsts]

Code: Select all

For n=0 To 7
	t = (n&%1111010) | ((n&1)<<2) | ((n>>2)&1)
	Debug RSet(Bin(n),3,"0") + " -> " + RSet(Bin(t),3,"0")
Next

You give this to a guy who's still trying to figure out XOR?

This forum is the best Thanks, Mr Vogel.

cheers

[netmaestro]

For readability, here's my two cents:

Code: Select all

#bit2 = 4
#bit0 = 1

Macro FlipBits_0_and_2(number)
  If number & #bit2
    number &~ #bit2
  Else
    number | #bit2
  EndIf
  
  If number & #bit0
    number &~ #bit0
  Else
    number | #bit0
  EndIf
EndMacro

number = %000

FlipBits_0_and_2(number)

Debug RSet(Bin(number),3,"0")

FlipBits_0_and_2(number)

Debug RSet(Bin(number),3,"0")

[rsts]

For readability, here's my two cents:

Code: Select all

#bit2 = 4
#bit0 = 1

Macro FlipBits_0_and_2(number)
  If number & #bit2
    number &~ #bit2
  Else
    number | #bit2
  EndIf
  
  If number & #bit0
    number &~ #bit0
  Else
    number | #bit0
  EndIf
EndMacro

number = %000

FlipBits_0_and_2(number)

Debug RSet(Bin(number),3,"0")

FlipBits_0_and_2(number)

Debug RSet(Bin(number),3,"0")

AHA

Now it begins to sink in.

You would never believe I was a math major. But that was back in the days of sliderules.

The best to all of you

[Deeem2031]

I don't see why you don't use the binary XOR operator:

Code: Select all

#bit2 = 4
#bit0 = 1

Macro FlipBits_0_and_2(number)
  number ! #bit2
  
  number ! #bit0
EndMacro

number = %000

FlipBits_0_and_2(number)

Debug RSet(Bin(number),3,"0")

FlipBits_0_and_2(number)

Debug RSet(Bin(number),3,"0")