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How to parse movb instruction

I am writing a disassembler, and I was looking at the format of the instruction (and doing some sorting manually), and I came across an instruction that I seem to be unable to decode.

The output for this particular command (from objdump):

c6 05 14 a0 04 08 01    movb   $0x1,0x804a014

However, I do not understand how the command is decoded, since the operation code c6 must be MOV Eb Ib (Mod R / M to imm8).

Can anyone tell me how it is decoded?

Thank!

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3 answers

This is due (in part) to the answer of Alex Frunze, but it’s a little bit, so I’ll give a few explanations here:

  • - c6/0, , 2 . r/m 8, , mod/rm, . 8 .
  • /0 , mod/rm. 3-5 mod/rm . c6, mod/rm, 3-5 0, mov.
  • 5 (, c6), r/m 00 000 101 ( ).
  • " " ( 0-2) r/m r/m. r/m, 101 (5), " dword", 4 mod/rm .
  • 14 a0 04 08 - 0x0804a014
  • 1 -

, .

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, . , , - 1 , 0x804a014. - C:

*(unsigned char *)0x804a014 = 1;

opcode c6, . MOV 2A .

05 - ModR/M. , 2-2 2A, "32- ModR/M Byte". 05 " ModR/M Byte (in Hexadecimal)" . , , ModR/M "disp32". : " disp32 32- , ModR/M". : 14 a0 04 08.

, 8- 01, .

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c6 - ( Mod/RM, /digit (reg))
05 - Mod/RM byte (mod = 00b, r/m = 101b,/digit (reg) = 0 - )
14 a0 04 08 - disp32
01 - imm8

And this is a movfrom Ibto Eb. You are probably confusing the AT & T syntax in which objdump shows disassembly with Intel / AMD documentation. The order of the operands in the AT & T syntax is different from the order in the x86 processor manuals.

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