Efficient bitwise operations

Chandrashekhar Mullaparthi Chandrashekhar.Mullaparthi@REDACTED
Fri Jun 20 11:52:15 CEST 2003

You can treat each byte as an integer and do

bsl - bit shift left
bsr - bit shift right

These operators are described in the first chapter of the red book.

11> 1 bsl 1.
12> 1 bsl 2.
13> 4 bsr 1.
14> 4 bsr 2.

To swap nibbles

swap(Int) ->
    ((Int band 15) bsl 4) + ((Int band 240) bsr 4).


-----Original Message-----
From: Rudolph van Graan [mailto:rvg@REDACTED]
Sent: 20 June 2003 09:34
To: erlang-questions@REDACTED
Subject: Efficient bitwise operations

Hi there again,

Thanks again for all the answers previously - helped me a lot! 

I've had to write some bitwise operations yesterday, because I couldn't
locate any bifs or operations that do the same:

Bitwise roll left:

	lroll1(<<B7:1,B6:1,B5:1,B4:1,B3:1,B2:1,B1:1,B0:1>>) ->

	rroll(<<Byte>>,Rolls) when Rolls > 0 ->
	    Output = rroll1(<<Byte>>),

	rroll(Byte,0) ->

Bitwise roll right:

	lroll(<<Byte>>,Rolls) when Rolls > 0 ->
	    Output = lroll1(<<Byte>>),

	lroll(Byte,0) ->

	rroll1(<<B7:1,B6:1,B5:1,B4:1,B3:1,B2:1,B1:1,B0:1>>) ->

Bitwise high <-> low nibble swap:

	swap(<<B7:1,B6:1,B5:1,B4:1,B3:1,B2:1,B1:1,B0:1>>) ->     

My question here has to do with the efficiency of the above code.
Normally, I would have used a single assembly instruction to achieve the
same result, but even though the code works, how efficient can it really
be? Are there any obvious things I've missed in the documentation?

Thanks again!


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