Adding values without sign-extension?

Asked Sep 26 '15 at 11:40

Active Sep 26 '15 at 11:56

Viewed 405 times

I know that the CPU doesn't care about the sign bit, because of two's complement. But when using different operand sizes, then the source value gets sign extended to the destination's size.

xor rax, rax
add rax, byte 0xAD ; adds 0xFFFFFFFFFFFFFFAD to RAX

Is there an add instruction that adds values without sign-extension (and avoids null bytes)?

(I'm working with a x86_64 architecture)

edited Sep 26 '15 at 11:56

Jester

56,577
4
81
125

asked Sep 26 '15 at 11:40

Cubi73

1,891
3
31
52

Not with an immediate. You can go through a register. – Jester Sep 26 '15 at 11:50
So `add r64, r8` is not a sign-extended addition? – Cubi73 Sep 26 '15 at 11:52
There is no `add r64, r8`. – Jester Sep 26 '15 at 11:52
Sorry. `r64` and `r8` should mean 64-bit register and 8-bit register – Cubi73 Sep 26 '15 at 11:53
3

There is no such instruction. You will use `add r64, r64` after you load your immediate into a 64 bit register zero extended. `mov cl, 0xad; movzx ecx, cl; add rax, rcx` (note that 32 bit ops zero extend automatically). – Jester Sep 26 '15 at 11:53
Ahhh, ok, now it works :D Thank you – Cubi73 Sep 26 '15 at 11:56

Adding values without sign-extension?

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