解析器：移动指令预期的指令

Question

我正在尝试制作一个简单的汇编程序，即将两个数字相加并显示它们，然后将两个数字相减并显示它们。 但是我遇到了错误：

oppgave3.asm:28: error: parser: instruction expected
oppgave3.asm:29: error: comma, colon, decorator or end of line expected after operand
oppgave3.asm:30: error: symbol `move' redefined
oppgave3.asm:30: error: parser: instruction expected
oppgave3.asm:31: error: comma, colon, decorator or end of line expected after operand
oppgave3.asm:32: error: comma, colon, decorator or end of line expected after operand
oppgave3.asm:33: error: comma, colon, decorator or end of line expected after operand
oppgave3.asm:37: error: symbol `move' redefined
oppgave3.asm:37: error: parser: instruction expected
oppgave3.asm:38: error: comma, colon, decorator or end of line expected after operand
oppgave3.asm:39: error: symbol `move' redefined
oppgave3.asm:39: error: parser: instruction expected
oppgave3.asm:40: error: comma, colon, decorator or end of line expected after operand
oppgave3.asm:41: error: comma, colon, decorator or end of line expected after operand
oppgave3.asm:42: error: comma, colon, decorator or end of line expected after operand

这就是我想要做的：我有两个子程序，一个用于加法，一个用于减法。

section .data
a dw 4
b dw 2


section .bss
c resb 1

section .text
global_start:
_start:

call addition
mov eax,4
mov ebx,1
mov ecx,c
mov edx,1
int 0x80

call subtraction
mov eax,4
mov ebx,1
mov ecx,c
mov edx,1
int 0x80

addition:
move eax,[a]
sub eax '0'
move ebx,[b]
sub ebx '0'
add eax and ebx
add eax '0'
mov [c],eax
ret

subtraction:
move eax,[a]
sub eax '0'
move ebx,[b]
sub ebx '0'
sub eax and ebx
add eax '0'
mov [c],eax
ret

Answer 1

你写的是“move”而不是“mov”

未被识别为指令助记符的标记被视为标签。 就像move nop等价于move: nop 。 这就是为什么您会在一些后来的用途中symbol 'move' redefined " symbol 'move' redefined的原因。

还有其他各种语法错误，例如sub eax and ebx而不是sub eax, ebx和sub ebx '0'缺少逗号

Answer 2

我想你有一个错字。 你有一个“移动”指令，而我的猜测是它应该是“移动”，最后没有额外的 e 。 我不是装配专家，所以我在这里可能是错的。