Understanding mov in assembly

Question

address and data manipulation, click to see the image

mov $0x100, %eax
mov $0x105, (%eax)

In a 32bit little-endian architecture

Could someone explain me how these mov's are working?

I understand that the second mov is informing the processor to write on the memory address corresponding to the aex register the value 0x105, but I fail to understand why it is being written on the 0x100 address

Thanks for any help

Answer 1

mov $0x100, %eax

it moves the immediate value 0x100 into the eax register

mov $0x105, (%eax)

it moves the immediate value 0x105 into the dereferenced address 0x100 from eax because you are using (%eax). It has the value 0x100 you put in the previous step.

Think about the concept pointer in C which saves the address as its value - it uses *ptr to dereference the value while here it uses (%register).

To help you get something into assembly, check out here: CS107 Guide to x86-64