EMU8086 adds value to wrong register












0















var1 db 1h, 2h
var2 db 2h
MOV AL, [var1 + 1]
MOV [var2], AL


If I emulate this code, I see EMU8086 set up AH (not AL) part of register AX. Why?










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    0















    var1 db 1h, 2h
    var2 db 2h
    MOV AL, [var1 + 1]
    MOV [var2], AL


    If I emulate this code, I see EMU8086 set up AH (not AL) part of register AX. Why?










    share|improve this question

























      0












      0








      0








      var1 db 1h, 2h
      var2 db 2h
      MOV AL, [var1 + 1]
      MOV [var2], AL


      If I emulate this code, I see EMU8086 set up AH (not AL) part of register AX. Why?










      share|improve this question














      var1 db 1h, 2h
      var2 db 2h
      MOV AL, [var1 + 1]
      MOV [var2], AL


      If I emulate this code, I see EMU8086 set up AH (not AL) part of register AX. Why?







      nasm emu8086






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 25 '18 at 12:52









      Bartłomiej StasiakBartłomiej Stasiak

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          EMU8086 begins to process at the beginning (var1) and interpretes the bytes as instruction and following bytes accordingly. So, the code looks for EMU8086 like



          ADD [BP+SI], AX
          ADD AH, [BX + SI] + 00001
          MOV [00002], AL


          As you can see, this code changes AH but not AL. In short: EMU8086 requires a proper DOS program.






          share|improve this answer























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            EMU8086 begins to process at the beginning (var1) and interpretes the bytes as instruction and following bytes accordingly. So, the code looks for EMU8086 like



            ADD [BP+SI], AX
            ADD AH, [BX + SI] + 00001
            MOV [00002], AL


            As you can see, this code changes AH but not AL. In short: EMU8086 requires a proper DOS program.






            share|improve this answer




























              0














              EMU8086 begins to process at the beginning (var1) and interpretes the bytes as instruction and following bytes accordingly. So, the code looks for EMU8086 like



              ADD [BP+SI], AX
              ADD AH, [BX + SI] + 00001
              MOV [00002], AL


              As you can see, this code changes AH but not AL. In short: EMU8086 requires a proper DOS program.






              share|improve this answer


























                0












                0








                0







                EMU8086 begins to process at the beginning (var1) and interpretes the bytes as instruction and following bytes accordingly. So, the code looks for EMU8086 like



                ADD [BP+SI], AX
                ADD AH, [BX + SI] + 00001
                MOV [00002], AL


                As you can see, this code changes AH but not AL. In short: EMU8086 requires a proper DOS program.






                share|improve this answer













                EMU8086 begins to process at the beginning (var1) and interpretes the bytes as instruction and following bytes accordingly. So, the code looks for EMU8086 like



                ADD [BP+SI], AX
                ADD AH, [BX + SI] + 00001
                MOV [00002], AL


                As you can see, this code changes AH but not AL. In short: EMU8086 requires a proper DOS program.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 25 '18 at 16:07









                rkhbrkhb

                11.4k72142




                11.4k72142
































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