x86 Assembler: shl and other instructions

Question

I'm trying to understand some assembler code inside a loop. The loop runs from 1 to 255 and does the following inside the loop:

mov    eax,DWORD PTR [ebp-0x4]
shl    eax,0x2
add    eax,DWORD PTR [ebp+0x8]
mov    DWORD PTR [eax],0x0

Here the DWORD PTR [ebp-0x4] refers to the number going from 1 to 255.

Can someone figure out what is going on here? Thanks.

score 5 · Accepted Answer · answered May 15 '11 at 19:30

5

It's just zeroing an array apparently:

mov    eax,DWORD PTR [ebp-0x4] ; load index
shl    eax,0x2                 ; multiply index by 4 to get byte offset
add    eax,DWORD PTR [ebp+0x8] ; add byte offset to array base address
mov    DWORD PTR [eax],0x0     ; zero value at array[index]

answered May 15 '11 at 19:30

Paul R

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Great, of course! Thanks. Been looking at it for a while, but just couldn't see it. – danielhc May 15 '11 at 19:37
1

why don't just use LEA for address calculation? – phuclv Sep 10 '14 at 03:41

x86 Assembler: shl and other instructions

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