问题
I need to perform a bitwise equality between two bytes. That means that for instance if I have two bytes: 00011011 and 00011110 the result is 11111010 The only fast way I see is to use the following statement
byte a, b;//set input bytes
byte c = ~(a^b);//output bytes
But I wonder if there is a faster solution for this. After these equality operation I want to mask the bits I need. So I need to use an AND-operation. So the code becomes:
byte a, b;//set input bytes
byte m;//mask, intresting bits are set to 1, others to 0
byte c = (~(a^b))&m;//output bytes
aren't there any faster and more simple methods that don't need to use all those bitwise operations because this part of the code will be called very often.
回答1:
I doubt it can be done in fewer operations. That looks optimal. Perhaps you could store ~(a^b) in a lookup table (256*256 entries)? I doubt you would get much benefit and possibly even make things worse, but you could try it.
回答2:
You are looking in the wrong place for this optimization; you won't end up finding any better bitwise operation here. Even if you did, it's hardly going to speed anything up. The real win will come from processing more than just a byte at a time. The processor is already having to do a bunch of bit shifting and masking operations just so that it can pretend for you that you are working with bytes. Process your arrays of bytes 1 word at a time, or use vector instructions if they are available.
回答3:
These operations seem fast enough to be honest. I think you shouldn't try to optimize them further, but finish your software first, see if you are happy with the overall performance and use a profiler if you are not. I am fairly sure the problem will be elsewhere.
回答4:
What you want is an XNOR operation. Unfortunately this is not supported by C#/Mono. I think your solution is optimal.
来源：https://stackoverflow.com/questions/2200856/bitwise-equality