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## Example showing arg order on the stack.
# Show difference between ascii codes of first letter of first arg and first
# letter of second arg.
#
# To run:
# $ subx translate ex9.subx ex9
# $ subx run ex9 z x
# Expected result:
# $ echo $?
# 2
#
# At the start of a SubX program:
# argc: *ESP
# argv[0]: *(ESP+4)
# argv[1]: *(ESP+8)
# ...
# Locals start from ESP-4 downwards.
== code
# instruction effective address operand displacement immediate
# op subop mod rm32 base index scale r32
# 1-3 bytes 3 bits 2 bits 3 bits 3 bits 3 bits 2 bits 2 bits 0/1/2/4 bytes 0/1/2/4 bytes
# s1 = argv[1] (EAX)
8b/copy 1/mod/*+disp8 4/rm32/sib 4/base/ESP 4/index/none 0/r32/EAX 8/disp8 . # copy *(ESP+8) to EAX
# s2 = argv[2] (EBX)
8b/copy 1/mod/*+disp8 4/rm32/sib 4/base/ESP 4/index/none 3/r32/EBX 0xc/disp8 . # copy *(ESP+12) to EBX
# call string_equal(s1, s2)
# push args
50/push-EAX
53/push-EBX
# call
e8/call ascii_difference/disp32
# discard args
81 0/subop/add 3/mod/direct 4/rm32/ESP . . . . . 8/imm32 # add 8 to ESP
# exit(EAX)
89/copy 3/mod/direct 3/rm32/EBX . . . 0/r32/EAX . . # copy EAX to EBX
b8/copy . . . . . . . 1/imm32/exit # copy 1 to EAX
cd/syscall 0x80/imm8
ascii_difference: # (s1, s2) : null-terminated ascii strings
# a = first letter of s1 (ECX)
8b/copy 1/mod/*+disp8 4/rm32/sib 4/base/ESP 4/index/none 0/r32/EAX 8/disp8 . # copy *(ESP+8) to EAX
8b/copy 0/mod/indirect 0/rm32/EAX . . . 0/r32/EAX . . # copy *EAX to EAX
# b = first letter of s2 (EDX)
8b/copy 1/mod/*+disp8 4/rm32/sib 4/base/ESP 4/index/none 1/r32/ECX 4/disp8 # copy *(ESP+4) to ECX
8b/copy 0/mod/indirect 1/rm32/ECX . . . 1/r32/ECX . . # copy *ECX to ECX
# a-b
29/subtract 3/mod/direct 0/rm32/EAX . . . 1/r32/ECX . . # subtract ECX from EAX
c3/return
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