x86 Assembly pointers

Question

I am trying to wrap my mind around pointers in Assembly.

What exactly is the difference between:

mov eax, ebx 

and

mov [eax], ebx 

and when should dword ptr [eax] should be used?

Also when I try to do mov eax, [ebx] I get a compile error, why is this?

Answer 1

As has already been stated, wrapping brackets around an operand means that that operand is to be dereferenced, as if it were a pointer in C. In other words, the brackets mean that you are reading a value from (or storing a value into) that memory location, rather than reading that value directly.

So, this:

mov  eax, ebx 

simply copies the value in ebx into eax. In a pseudo-C notation, this would be: eax = ebx.

Whereas this:

mov  eax, [ebx] 

dereferences the contents of ebx and stores the pointed-to value in eax. In a pseudo-C notation, this would be: eax = *ebx.

Finally, this:

mov  [eax], ebx 

stores the value in ebx into the memory location pointed to by eax. Again, in pseudo-C notation: *eax = ebx.

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The registers here could also be replaced with memory operands, such as symbolic variable names. So this:

mov  eax, [myVar] 

dereferences the address of the variable myVar and stores the contents of that variable in eax, like eax = myVar.

By contrast, this:

mov  eax, myVar 

stores the address of the variable myVar into eax, like eax = &myVar.

At least, that's how most assemblers work. Microsoft's assembler (called MASM), and the Microsoft C/C++ compiler's inline assembly, is a bit different. It treats the above two instructions as equivalent, essentially ignoring the brackets around memory operands.

To get the address of a variable in MASM, you would use the OFFSET keyword:

mov  eax, OFFSET myVar 

However, even though MASM has this forgiving syntax and allows you to be sloppy, you shouldn't. Always include the brackets when you want to dereference a variable and get its actual value. You will never get the wrong result if you write the code explicitly using the proper syntax, and it'll make it easier for others to understand. Plus, it'll force you to get into the habit of writing the code the way that other assemblers will expect it to be written, rather than relying on MASM's "do what I mean, not what I write" crutch.

Speaking of that "do what I mean, not what I write" crutch, MASM also generally allows you to get away with omitting the operand-size specifier, since it knows the size of the variable. But again, I recommend writing it for clarity and consistency. Therefore, if myVar is an int, you would do:

mov  eax, DWORD PTR [myVar]    ; eax = myVar 

or

mov  DWORD PTR [myVar], eax    ; myVar = eax 

This notation is necessary in other assemblers like NASM that are not strongly-typed and don't remember that myVar is a DWORD-sized memory location.

You don't need this at all when dereferencing register operands, since the name of the register indicates its size. al and ah are always BYTE-sized, ax is always WORD-sized, eax is always DWORD-sized, and rax is always QWORD-sized. But it doesn't hurt to include it anyway, if you like, for consistency with the way you notate memory operands.

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Also when I try to do mov eax, [ebx] I get a compile error, why is this?

Um…you shouldn't. This assembles fine for me in MSVC's inline assembly. As we have already seen, it is equivalent to:

mov  eax, DWORD PTR [ebx] 

and means that the memory location pointed to by ebx will be dereferenced and that DWORD-sized value will be loaded into eax.

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why I cant do mov a, [eax] Should that not make "a" a pointer to wherever eax is pointing?

No. This combination of operands is not allowed. As you can see from the documentation for the MOV instruction, there are essentially five possibilities (ignoring alternate encodings and segments):

mov  register, register     ; copy one register to another mov  register, memory       ; load value from memory into register mov  memory,   register     ; store value from register into memory mov  register, immediate    ; move immediate value (constant) into register mov  memory,   immediate    ; store immediate value (constant) in memory 

Notice that there is no mov memory, memory, which is what you were trying.

However, you can make a point to what eax is pointing to by simply coding:

mov  DWORD PTR [a], eax 

Now a and eax have the same value. If eax was a pointer, then a is now a pointer to that same memory location.

If you want to set a to the value that eax is pointing to, then you will need to do:

mov  eax, DWORD PTR [eax]    ; eax = *eax mov  DWORD PTR [a], eax      ; a   = eax 

Of course, this clobbers the pointer and replaces it with the dereferenced value. If you don't want to lose the pointer, then you will have to use a second "scratch" register; something like:

mov  edx, DWORD PTR [eax]    ; edx = *eax mov  DWORD PTR [a], edx      ; a   = edx 

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I realize this is all somewhat confusing. The mov instruction is overloaded with a large number of potential meanings in the x86 ISA. This is due to x86's roots as a CISC architecture. By contrast, modern RISC architectures do a better job of separating register-register moves, memory loads, and memory stores. x86 crams them all into a single mov instruction. It's too late to go back and fix it now; you just have to get comfortable with the syntax, and sometimes it takes a second glance.