what is the difference between these movw to AX and movb to AL load instructions, from an odd offset-CodePudding

.global main

.data
arr: .short 1, 0xEA, 0x2, 0x3, 0b1010

.text
main:

      lea (arr), %rbx
      movb 3(%rbx),%al #1

      movw arr 3, %ax #2

when I try this code al gets the value 0x0 however ax gets the value 0x200 can you tell me why is that? what is the difference between #1 and #2 ?

CodePudding user response：

Since x86 is a little-endian architecture, the data in arr looks like:

arr 0    1    2    3    4    5    6    7    8    9
 0x01 0x00 0xEA 0x00 0x02 0x00 0x03 0x00 0x0A 0x00

#1 reads data to al. al is a one-byte register, so 0x00 at the 3 is loaded and the value becomes 0x0.

#2 reads data to ax. ax is a two-byte register, so 0x00 0x02 is loaded and the value becomes 0x200.

The difference is the size of the destination registers.