Masm32 - Addressing modes and arrays

Question

I am a bit confused by addressing modes.

 array1 DWORD 200 DUP(?)

If i have this statement:

mov EAX, [EBX + EDI + 10]

EBX is the base, EDI is the index, and + 10 is saying add ten more bytes to EDI? so, EDI would then return 14 bytes long? and return the address and put it into EAX?

Similiarly,

mov EAX, [EBX + EDI * 4]

Is this saying that at what ever index in the array, returns 4 bytes long? Because the array is 4 bytes long for each index right? I guess im not understanding what * 4 does!

in addition,

  mov ECX, 100 ; loop counter
  mov ESI, 0

  top1:

  mov [array1 + ESI * 4], ECX

  inc ESI

  loop top1

This would be putting the number 100 into the array starting at index zero 4 bytes big? So, index[0] = 100, index[1]= 101, index[2]= 102?

Thank you

A mov like that doesn't put the address in the destination but the value at that address. If you want the address, that's what lea is for. — harold

twalberg twalberg · Accepted Answer · 2012-07-19T21:25:02

mov EAX, [EBX + EDI + 10]

Take the value in EBX and add the value in EDI and add 10, then use this result as an address to load a 32-bit value into EAX.

mov EAX, [EBX + EDI * 4]

Take the value in EBX and add to it 4 times the value in EDI, then use the result as an address to load a 32-bit value into EAX.

The last one stores a decreasing sequence of numbers starting at 100 into the memory locations starting at array1. ECX is decremented every time the loop instruction executes, until it reaches 0. The address each progressive value of ECX is stored at is array1 plus 4 times ESI, which is incremented on every iteration.

Masm32 - Addressing modes and arrays

1 Answers