Adding al to edi?

[ThoughtCriminal]

add edi,al wont work. I tried.

Any ideas?

Thanks.

[QvasiModo]

The problem is EDI is 32 bits in size, and AL is 8 bits in size. But you can try this:

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movzx eax, al
add edi, eax

The first instruction extends the 8-bit value of AL into EAX, the second adds EAX (32 bits) to EDI (32 bits).

[ThoughtCriminal]

Thanks. I forgot about those instructions.

I need to use movsx.

[gabor]

Hi!

You have to use movsx? Is this a misspelling or you really have to use a negative number (a value with sign bit set)?

As far as I know movsx and movzx are slow. Instead of movzx eax,al (the upper 12 bits are set to 0) use

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and eax,0FFh
to clear the upper part of eax.
Or the best way is to clear the whole eax before getting a byte value into al:

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xor eax,eax
...
; get value into al
add edi,eax


If you need to use movsx eax,al (the upper 12 bits are set to 1) You could make a

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or eax,0FFFFFF00h
to set those bits.
Another way is to use cbw/cwd but these instructions are slow again.

Greets,Gábor

[zooba]

If you need to use movsx eax,al (the upper 12 bits are set to 1) You could make a

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or eax,0FFFFFF00h
to set those bits.

This method only works if you include a test (and al, 80h) and a jump. From my own testing, movsx is the cheapest way to sign-extend 8/16-bits to 32.

You are correct with your other points, but simply setting the upper 12 bits to 1 may make a negative out of a positive.

[QvasiModo]

I'd also like to add that, if I'm not wrong:

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xor eax, eax
mov al, value

would actually be slower than movzx.