CODEWITHSUNDEEP
assemblyx86
CodeLikeBananas
CodeLikeBananas

Reputation: 13

x86 linux - GAS - AT&T - Stosb?

I am currently trying to teach myself x86 under the circumstances posted in my title. Please note I do not just want an answer, I want to learn this and be fluent in it.

I am having a lot of trouble storing data in memory addresses other than the main registers (eax, ebx, ecx, edx). 

.global main

.text

str: .string "data: %d\n"

main:

pushl   $3 
call    malloc
popl    %ecx
movl    $4, %esi          #source 
movl    %eax, %edi       #destination
stosb
movl    %eax, %esi
lodsb
pushl   %edi
pushl   $str
call    printf
popl    %ecx
popl    %ecx

After the stosb, shouldn't movl put the destination address (memory address produced from malloc, and original source address) into the destination address, and move the $4 in %esi back into %edi?

Upvotes: 0

Views: 1334

Answers (2)

gsg
gsg

Reputation: 9377

It's not quite clear what you are trying to do. However, note that stosb does not make any use of esi (or ecx). rep stosb will make use of ecx: read the descriptions of both forms of that instruction carefully and make sure you are doing the right thing.

You can download Intel's reference manual from this page.

Upvotes: 0

C. K. Young
C. K. Young

Reputation: 223193

stosb stores a single byte. In x86, an address is a dword, so you needed to use stosd instead. (Remember that stosb and lodsb write to/read from %al, and stosd and lodsd write to/read from %eax.)

Also, your $4 was initially stored in %esi, but that was later overwritten by the movl %eax, %esi instruction.

Upvotes: 1

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