x86 Instruction help: MOV [edx], eax

Question

I'm reading through a book that is describing C linked lists and how they are represented within x86 ASM. I having difficulty understanding the instruction MOV [edx], eax.

If the instruction was reversed, MOV eax, [edx], I would understand it to mean copy the 4 bytes represented by the memory address stored in edx and store it in eax.

What does MOV [edx], eax represent?

If using the [] with the MOV instruction, I thought it meant to copy the data residing at the memory address to it's destination. If that is true, how can you copy whatever is in eax to a data value in edx?

Answer 1

It's Intel assembly syntax. In Intel assembly syntax the destination is always the first operand, the rest operands are source operands. The other commonly used assembly syntax for x86 is AT&T, but as Intel and AT&T syntaxes look very different, they are easy to distinguish.

mov [edx],eax stores the value of eax in memory, to the address given in edx (in little-endian byte order).

mov eax,[edx] does exactly the reverse, reads a value stored from the memory, from the address given in edx, and stores it in eax.

[reg] always means indirect addressing, it's just like a pointer *reg in C.

To copy the contents of eax to edx, all you need is mov edx,eax. Destination is first operand, source is the second operand.