Question

I am trying to learn assembly language. Can someone explain and/or give an example of how to use addressing modes to access elements in each of the following array types?

1. array of DWORD

2. array of structures, where each structure contains two DWORDs

3. array of arrays, where each array has 10 DWORDs

Answer 1

You don’t mention which processor specifically you are targeting, but in 386+ this would work. I don’t have MASM, but MASM uses Intel syntax.

1. Let’s assume that ebx is the base register and esi is the index register of the element. Standard 32-bit registers (eax, ebx, ecx, edx, ebp, esi, edi) are valid for the addressing mode used here. esp can be used as base register but not as index register. The value of index register can be optionally scaled with 2, 4 or 8. In this example we can use a scaling factor 4 for a dword array (in 386 legal scaling factors are 1, 2, 4 and 8).

   to read the value from array into eax: mov eax, [ebx+4*esi]
   to store the value of eax into array: mov [ebx+4*esi], eax

2. Let’s keep still ebx as the base address. We can use a scaling factor 8 for esi for a struct array in which each struct consists of 2 dwords.

   to read the value of the first dword into eax: mov eax, [ebx+8*esi]
   to store the value of eax into first dword: mov [ebx+8*esi], eax

   to read the value of the second dword into eax: mov eax, [ebx+8*esi+4]
   to store the value of eax into second dword: mov [ebx+8*esi+4], eax

   If the index can’t be hardcoded, you can just add 4 to ebx (or whatever registers you use to store the base address). And if you have hardcoded base-address, then you could use eg. esi to address the struct and eg. ebx to address the dword you want. Note that in 386+ you can’t scale more than one register (the index register) in indirect addressing.

3. Let’s still assume you don’t know the base address of your array beforehand, and you’ll have it in ebx, in esi you have the index of the struct and in edx you have the index of the dword.

   To get the address of the struct you can use lea multiplication optimization (10 = 8 + 2):

   Edit: Fix: lea esi,[4*esi] (dword is 4 bytes)

   lea edi,[ebx+8*esi]

   lea edi,[edi+2*esi]

   Now you have the address of the struct in edi. You just need to add the index of the dword (stored in edx in this example) multiplied by 4 (because each dword is 4 bytes).

   To read value of dword to eax: mov eax,[edi+4*edx].

   To store value of eax into dword: mov [edi+4*edx],eax.