Get MSB / LSB in register

问题

This might seem like a stupid question, but I cant really tell what I'm missing

On ARM 7:

I have a 8 digit number in register 0, say

10110111

I want to 'loop' through this and do something with the current bit until the 8 bits are up, but I'm having alot of trouble with this simple issue..

My logic is:

- get MSB / LSB of number in r0
- shift it to r1
- lsl / lsr r0

But from this logic, I dont know how you would get the MSB / LSB. Could anyone help me out? Or is there a better way of looping through this?

Thank you!

回答1:

Check out the rbit and clz instructions. These allow you to compute the MSB and LSB index.

回答2:

I'm having a hard time understanding your issue. But if you shift with flags LSLS you can read the carrybit and it from there.

回答3:

I've got a workaround to this issue, but I'm not sure it's the best method.

Solution:

Use a 2nd register with value 1, with every iteration, AND (r0,r1) thats your bit. Then just LSL r1, and continue.

eg

mov r0, #0b10101010 @original
mov r1, #1

loop:
    and r2,r0,r1 @this is your bit
    lsl r1
    b loop

来源：https://stackoverflow.com/questions/44069659/get-msb-lsb-in-register