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I am trying to write a program which takes the binary input from a text file and sends it as a parameter to an assembly function. That assembly function must print this binary input to the screen. The input is sent from c code to assembly code by its address.

When I try to assemble my asm file, I get an "invalid combination of opcode and operands" error on the mov msg, [esp+8] line. I want to copy my char arg from the stack to my static variable. Why isn't that a valid instruction?

The full code is:

segment .data
        len equ 31
segment .bss
        msg resb 0
segment .text
global sequence_generator

sequence_generator:

       push ebp
       mov ebp, esp
       mov msg, [esp+8]

       mov eax,4
       mov ebx,1
       mov ecx,msg
       mov edx,len
       int 80h

       pop ebp
       ret
Peter Cordes
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Nope
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    `mov msg, [esp+8]` what you think this is doing? And this `msg resb 0`? And having `edx` equal to fixed 31 is intended? So the C code will always send 31+ characters? – Ped7g Oct 10 '16 at 12:44
  • BTW, if it's truly binary data, it will contain non-printable characters, which doesn't work very well in Linux, so probably you may want to do some processing over them to change unprintable values to `'.'`, or to print hexadecimal values instead of bytes themselves. (so from data `" "` (triple space) you will print `"20 20 20"`) – Ped7g Oct 10 '16 at 12:47

1 Answers1

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I wonder what have you tried to do in this line:

mov msg, [esp+8]

But you are not allowed to move from memory to memory. Refer to this page, for instance.

If you want to move something from memory to memory, use a register as a temporary storage. For example:

mov eax, [var1]
mov [var2], eax
Ruslan Osmanov
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