Sign in to follow this  

implementing SHR in ASM

This topic is 4779 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

If you intended to correct an error in the post then please contact us.

Recommended Posts

how can implement SHR in assembly?( without the SHR instructor, of course :-D) I thought I seperate each bit with AND and then get a new number, but this is way too long...

Share this post


Link to post
Share on other sites


;stdcall dword shiftRight (dword number, dword shiftCount)
shiftRight:
push ebp
mov ebp, esp

mov ebx, 2
mov ecx, [ebp + 12]
mov eax, [ebp + 8]

.divide:
xor edx, edx
div ebx
loop .divide

mov esp, ebp
pop ebp
ret 8






[smile]

This must be slow as hell [smile]

Oxyd

Share this post


Link to post
Share on other sites
Can’t think why anyone would want to implement a SHR in assembler without actually using the SHR instruction! That’s why it’s there isn’t it?

However, the SHR shifts all the bits in a source operand by the number in the second operand, zeroising the top bits.

Repeated DIV by 2 of the source operand will do it as has been shown already.

Another option, if the number of shifts > 1, is to do a single DIV by 2 and then a SAR by 1 less than the number of desired shifts.

The SAR shifts to the right, but retains the existing value of the top bit. If you know your source operand has a zero top bit, a SAR has the same effect as a SHR.


Share this post


Link to post
Share on other sites

This topic is 4779 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

If you intended to correct an error in the post then please contact us.

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

Sign in to follow this