Assembly Language nasm error

Question

I have written the following assembly code as prescribed by my text book in the intel 64 bit syntax

Section .text
global _short
_start:
        jmp short    Gotocall
shellcode:
        pop             rcx
        xor             eax,eax
        mov byte        [rcx+8], al
        lea             rdi, [rax]
        mov long        [rcx+8], rdi
        mov long        [rcx+12], eax
        mov byte        al, 0x3b
        mov             rsi, rax
        lea             rdi, [esi+8]
        lea             edx, [esi+12]
        int             0x80

Gotocall:

        call            shellcode
        db              '/bin/shJAAAAKKKK'

but i get a nasm error in line 10 like this asmshell.asm:10: error: mismatch in operand sizes Can anybody tell me what mistake is their in my code.

And can anybody please tell me some good references to the 64 bit intel assembly instructions.

And how do you invoke `nasm`, how do you attempt to build your program? — Some programmer dude, Aug 15 '17 at 18:18
@Someprogrammerdude i use this command nasm -f elf64 asmshell.asm — Sourav Rai, Aug 15 '17 at 18:29
If your textbook advises `mov byte al, 0x3b` is it any good? What other size can `al` be? — Weather Vane, Aug 15 '17 at 19:07
@WeatherVane yes that was unnecesary. I copied the code as it was in my book. — Sourav Rai, Aug 15 '17 at 19:11
Note that this can't work as actual shellcode injected into stack memory; the int 0x80 ABI is 32-bit and will thus truncate pointers. [What happens if you use the 32-bit int 0x80 Linux ABI in 64-bit code?](https://stackoverflow.com/q/46087730) — Peter Cordes, Dec 14 '20 at 14:42

score 1 · Accepted Answer · answered Aug 15 '17 at 18:59

If you mean the error is on line 10

mov long [rcx+8], rdi

I was about to ask you what size long qualifier is, but the next line

mov long [rcx+12], eax

shows that you are moving two different sizes of register to the same size destination. In the first case the 64-bit register rdi, in the second case the 32-bit register eax, and long cannot satisfy them both.

Why not just drop the long since by specifying the register, the assembler knows the size of the destination? But sadly, you have only allowed 4 bytes memory to store a 64-bit register, given away by the [rcx+8] followed by [rcx+12].

Perhaps you intended

mov long [rcx+8], edi

thanks @WeatherVane i corrected it and now it compiles smoothly. Assembly language can be frustating sometimes. — Sourav Rai, Aug 15 '17 at 19:13

Assembly Language nasm error

1 Answers1