How do AX AH AL map onto EAX

Question

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

User · Answer

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

User · Answer

No  that s not quite right  EAX is the full 32-bit value AX is the lower 16-bits AL is the lower 8 bits AH is the bits 8 through 15  zero-based   So AX is composed of AH AL halves  and is itself the low half of EAX    The upper half of EAX isn t directly accessible as a 16-bit register  you can shift or rotate EAX if you want to get at it   For completeness  in addition to the above  which was based on a 32-bit CPU  64-bit Intel AMD CPUs have RAX  which hold a 64-bit value  and where EAX is mapped to the lower 32 bits   All of this also applies to EBX RBX  ECX RCX  and EDX RDX   The other registers like EDI RDI have a DI low 16-bit partial register  but no high-8 part  and the low-8 DIL is only accessible in 64-bit mode  Assembly registers in 64-bit architecture  Writing AL  AH  or AX merges into the full AX EAX RAX  leaving other bytes unmodified for historical reasons    In 32 or 64-bit code  prefer a movzx eax  byte  mem  or movzx eax  word  mem  load if you don t specifically want this merging  Why doesn  39 t GCC use partial registers   Writing EAX zero-extends into RAX    Why do x86-64 instructions on 32-bit registers zero the upper part of the full 64-bit register

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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    EAX - Full 32 bit value  AX  - lower 16 bit value  AH  - Bits from 8 to 15  AL  - lower 8 bits of EAX AX

User · Answer

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

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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

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0000 0001 0010 0011 0100 0101 0110 0111   ------ gt  EAX                        0100 0101 0110 0111   ------ gt  AX                                  0110 0111   ------ gt  AL                        0100 0101             ------ gt  AH

[assembly] How do AX, AH, AL map onto EAX?

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