My understanding of x86 registers say that each register can be accessed by the entire 32 bit code and it is broken into multiple accessible registers.
In this example EAX being a 32 bit register, if we call AX it should return the first 16 bits, and if we call AH or AL it should return the next 8 bits after the 16 bits and AL should return the last 8 bits.
So my question, because I don't truly believe is this is how it operates. If we store the 32 bit value aka EAX storing:
0000 0100 0000 1000 0110 0000 0000 0111
So if we access AX it should return
0000 0100 0000 1000
if we read AH it should return
0000 0100
and when we read AL it should return
0000 0111
Is this correct? and if it is what value does AH truly hold?
This question is related to
assembly
x86
cpu-registers
No -- AL is the 8 least significant bits of AX. AX is the 16 least significant bits of EAX.
Perhaps it's easiest to deal with if we start with 04030201h in eax. In this case, AX will contain 0201h, AH wil contain 02h and AL will contain 01h.
| 0000 0001 0010 0011 0100 0101 0110 0111 | ------> EAX
| 0100 0101 0110 0111 | ------> AX
| 0110 0111 | ------> AL
| 0100 0101 | ------> AH
AX is the 16 lower bits of EAX. AH is the 8 high bits of AX (i.e. the bits 8-15 of EAX) and AL is the least significant byte (bits 0-7) of EAX as well as AX.
Example (Hexadecimal digits):
EAX: 12 34 56 78
AX: 56 78
AH: 56
AL: 78
The below snippet examines EAX using GDB.
(gdb) info register eax
eax 0xaa55 43605
(gdb) info register ax
ax 0xaa55 -21931
(gdb) info register ah
ah 0xaa -86
(gdb) info register al
al 0x55 85
no your ans is Wrong
Selection of Al and Ah is from AX not from EAX
e.g
EAX=0000 0000 0000 0000 0000 0000 0000 0111
So if we call AX it should return
0000 0000 0000 0111
if we call AH it should return
0000 0000
and when we call AL it should return
0000 0111
Example number 2
EAX: 22 33 55 77
AX: 55 77
AH: 55
AL: 77
example 3
EAX: 1111 0000 0000 0000 0000 0000 0000 0111
AX= 0000 0000 0000 0111
AH= 0000 0000
AL= 0000 0111
Source: Stackoverflow.com