66 lines
1.8 KiB
ArmAsm
66 lines
1.8 KiB
ArmAsm
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;; Hello World Program #1
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;; Compile with: nasm -f elf hello.s
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;; Link with: ld -m elf_i386 -o hello hello.o
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;; Run with: ./hello
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;; sys/unistd_32.h
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%define SYS_write 1
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%define SYS_exit 60
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;; unistd.h
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%define STDOUT 1
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section .data
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msg db "Okay, so we're doing 64-bit subroutines now, huh?", 0Ah, 00h
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section .text
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global _start
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_start:
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mov rsi, msg ; Put the address of our message into rax.
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call strlen ; call the function strlen. We're using RAX as our argument.
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;; At this point, RDX is length and RSI still points to the source.
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;; The differences here make me wonder if my understanding of the 32-bit
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;; version are incorrect, since now the two fields that matter are
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;; already populated.
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;;
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;; Note that this is non-standard. RAX is usually assumed to hold
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;; the return value, but I'm cheating by knowing that printit
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;; wants the result of strlen in RDX. Whether this is exploitable
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;; in a compiler (you know, that whole "choosing registers"
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;; question) is something to ponder.
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call printit
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call exit
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strlen:
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mov rdx, rsi
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strlen_next:
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cmp byte [rdx], 0
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jz strlen_done
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inc rdx
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jmp strlen_next
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strlen_done:
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sub rdx, rsi
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ret
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;; Takes RDI as the address of the message and RDX as the length,
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;; and prints them. Restores all used registers when finished.
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printit:
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push rdi
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push rax
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mov rdi, STDOUT ; using STDOUT (see definition above)
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mov rax, SYS_write ; Using WRITE in 32-bit mode?
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syscall
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pop rax
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pop rdi
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;; Since this terminates the program, I'm not worried about
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;; managing the stack correctly.
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exit:
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mov rdi, 0
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mov rax, SYS_exit
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syscall
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