有没有办法再重复一次？
Is there a way to repeat this one more time?
我正在编写 nasm x86 汇编代码，执行 +-_ 其中空白是用户输入。现在我已经在它执行此操作的地方拥有它，但我需要它来阅读下一行并在那里执行此操作。我是汇编的新手，我不想使用循环或指针或任何东西。也许是时代的一种方式？
这是我的代码：
segment .data
newline db 0xA, 0xD
newlinelen equ $-newline
segment .bss
num1 resb 2
num2 resb 2
num3 resb 2
;num4 resb 2
;num5 resb 2
;num6 resb 2
res resb 1
;res2 resb 1
section .text
global main ;must be declared for using gcc
main: ;tell linker entry point
;reading num 1
mov eax, 3
mov ebx, 0
mov ecx, num1
mov edx, 2
int 0x80
;reading num 2
mov eax, 3
mov ebx, 0
mov ecx, num2
mov edx, 2
int 0x80
;reading num 3
mov eax, 3
mov ebx, 0
mov ecx, num3
mov edx, 2
int 0x80
;new line for next equation
mov edx, newline
mov ecx, newlinelen
mov ebx, 1
mov eax, 4
int 0x80
; moving the first number to eax register and second number to ebx
; and subtracting ascii '0' to convert it into a decimal number
mov eax, [num1]
sub eax, '0'
mov ebx, [num2]
sub ebx, '0'
mov ecx, [num3]
sub ecx, '0'
; add ebx to eax
add eax, ebx
; add '0' to to convert the sum from decimal to ASCII
add eax, '0'
; storing the sum in memory location res
mov [res], eax
;subtract ecx from eax
sub eax, ecx
mov [res], eax
; print the sum
mov eax, 4
mov ebx, 1
mov ecx, res
mov edx, 1
int 0x80
exit:
int 0x80
我的输入是1+2-1，输出是2，没错。我不需要在任何大于 4 的数字或负数上使用它。
我试过用数字 4、5、6 重复整个过程，但结果只给我一个空白的 ascii 字符，这是我得到的最接近结果的结果。
;;new equation?
;reading num 4
mov eax, 3
mov ebx, 0
mov ecx, num4
mov edx, 2
int 0x80
;reading num 5
mov eax, 3
mov ebx, 0
mov ecx, num5
mov edx, 2
int 0x80
;reading num 6
mov eax, 3
mov ebx, 0
mov ecx, num6
mov edx, 2
int 0x80
; moving the first number to eax register and second number to ebx
; and subtracting ascii '0' to convert it into a decimal number
mov eax, [num4]
sub eax, '0'
mov ebx, [num5]
sub ebx, '0'
mov ecx, [num6]
sub ecx, '0'
; add ebx to eax
add eax, ebx
; add '0' to to convert the sum from decimal to ASCII
add eax, '0'
; storing the sum in memory location res
mov [res2], eax
;subtract ecx from eax
sub eax, ecx
mov [res2], eax
; print the sum
mov eax, 4
mov ebx, 1
mov ecx, res2
mov edx, 1
int 0x80
我也试过用 AL 替换 eax，用 AH 替换另一个等式，但它并不像我也不理解的那样工作。然后我还尝试制作每个字节并放入段符号 resb 1，用于 + 符号，但没有用。我只是想让它在之后再读一行。
mov edx, newline
mov ecx, newlinelen
你在这里反转了寄存器。
mov eax, [num1]
...
mov [res], eax
这些指令之间的所有内容都应改用字节大小的寄存器。使用 AL、BL 和 CL。当前您正在覆盖不属于 res 变量的内存！
; add '0' to to convert the sum from decimal to ASCII
add eax, '0'
您应该先做减法，然后再转换回 ASCII，然后使用 mov [res], al.[=25= 将字符一次存储在 res 变量中]
exit:
int 0x80
您的出口缺少 eax 寄存器中的函数编号！
Is there a way to repeat this one more time?
即使您不喜欢，循环也是可行的方法。 times 运算符是一个汇编时功能。你可以把整个东西变成一个宏并调用它两次，但是你也可以像现在这样写第二次。
我只是重复了代码，而不是添加值 4、5 和 6，但将它们放在高位字符而不是低位字符中。还为新行添加了一个读取垃圾字符。
segment .data
newline db 0xA, 0xD
newlinelen equ $-newline
segment .bss
;defining all variables
num1 resb 2
num2 resb 2
num3 resb 2
;variables for next equation
num4 resb 2
num5 resb 2
num6 resb 2
;newlin for garbage character
newlin resb 1
;result 1 and result 2
res resb 1
res2 resb 1
section .text
global main ;must be declared for using gcc
main: ;tell linker entry point
;reading num 1
mov eax, 3
mov ebx, 0
mov ecx, num1
mov edx, 2
int 0x80
;reading num 2
mov eax, 3
mov ebx, 0
mov ecx, num2
mov edx, 2
int 0x80
;reading num 3
mov eax, 3
mov ebx, 0
mov ecx, num3
mov edx, 2
int 0x80
; moving the first number to eax register and second number to ebx
; and subtracting ascii '0' to convert it into a decimal number
;moving variables in lower halfs
mov AL, [num1]
sub AL, '0'
mov BL, [num2]
sub BL, '0'
mov CL, [num3]
sub CL, '0'
; add ebx to eax
add AL, BL
; storing the sum in memory location res
mov [res], AL
;subtract ecx from eax
sub AL, CL
; add '0' to to convert the sum from decimal to ASCII
add AL, '0'
mov [res], AL
; print the sum
mov eax, 4
mov ebx, 1
mov ecx, res
mov edx, 1
int 0x80
;new line for next equation
mov eax, 4
mov ebx, 1
mov ecx, newline
mov edx, newlinelen
int 0x80
;;new equation?
;;skip garbage character
mov eax, 3
mov ebx, 0
mov ecx, newlin
mov edx, 1
int 0x80
;reading num 4
mov eax, 3
mov ebx, 0
mov ecx, num4
mov edx, 2
int 0x80
;reading num 5
mov eax, 3
mov ebx, 0
mov ecx, num5
mov edx, 2
int 0x80
;reading num 6
mov eax, 3
mov ebx, 0
mov ecx, num6
mov edx, 2
int 0x80
; moving the first number to eax register and second number to ebx
; and subtracting ascii '0' to convert it into a decimal number
;moving variables in higher halfs
mov AH, [newlin]
mov BH, [num4]
sub BH, '0'
mov CH, [num5]
sub CH, '0'
mov DH, [num6]
sub DH, '0'
; add ebx to eax
add BH, CH
;subtract ecx from eax
sub BH, DH
; add '0' to to convert the sum from decimal to ASCII
add BH, '0'
;move ah into memory location res 2
mov [res2], BH
; print the sum
mov eax, 4
mov ebx, 1
mov ecx, res2
mov edx, 1
int 0x80
exit:
mov eax, 1
int 0x80
我正在编写 nasm x86 汇编代码，执行 +-_ 其中空白是用户输入。现在我已经在它执行此操作的地方拥有它，但我需要它来阅读下一行并在那里执行此操作。我是汇编的新手，我不想使用循环或指针或任何东西。也许是时代的一种方式？ 这是我的代码：
segment .data
newline db 0xA, 0xD
newlinelen equ $-newline
segment .bss
num1 resb 2
num2 resb 2
num3 resb 2
;num4 resb 2
;num5 resb 2
;num6 resb 2
res resb 1
;res2 resb 1
section .text
global main ;must be declared for using gcc
main: ;tell linker entry point
;reading num 1
mov eax, 3
mov ebx, 0
mov ecx, num1
mov edx, 2
int 0x80
;reading num 2
mov eax, 3
mov ebx, 0
mov ecx, num2
mov edx, 2
int 0x80
;reading num 3
mov eax, 3
mov ebx, 0
mov ecx, num3
mov edx, 2
int 0x80
;new line for next equation
mov edx, newline
mov ecx, newlinelen
mov ebx, 1
mov eax, 4
int 0x80
; moving the first number to eax register and second number to ebx
; and subtracting ascii '0' to convert it into a decimal number
mov eax, [num1]
sub eax, '0'
mov ebx, [num2]
sub ebx, '0'
mov ecx, [num3]
sub ecx, '0'
; add ebx to eax
add eax, ebx
; add '0' to to convert the sum from decimal to ASCII
add eax, '0'
; storing the sum in memory location res
mov [res], eax
;subtract ecx from eax
sub eax, ecx
mov [res], eax
; print the sum
mov eax, 4
mov ebx, 1
mov ecx, res
mov edx, 1
int 0x80
exit:
int 0x80
我的输入是1+2-1，输出是2，没错。我不需要在任何大于 4 的数字或负数上使用它。 我试过用数字 4、5、6 重复整个过程，但结果只给我一个空白的 ascii 字符，这是我得到的最接近结果的结果。
;;new equation?
;reading num 4
mov eax, 3
mov ebx, 0
mov ecx, num4
mov edx, 2
int 0x80
;reading num 5
mov eax, 3
mov ebx, 0
mov ecx, num5
mov edx, 2
int 0x80
;reading num 6
mov eax, 3
mov ebx, 0
mov ecx, num6
mov edx, 2
int 0x80
; moving the first number to eax register and second number to ebx
; and subtracting ascii '0' to convert it into a decimal number
mov eax, [num4]
sub eax, '0'
mov ebx, [num5]
sub ebx, '0'
mov ecx, [num6]
sub ecx, '0'
; add ebx to eax
add eax, ebx
; add '0' to to convert the sum from decimal to ASCII
add eax, '0'
; storing the sum in memory location res
mov [res2], eax
;subtract ecx from eax
sub eax, ecx
mov [res2], eax
; print the sum
mov eax, 4
mov ebx, 1
mov ecx, res2
mov edx, 1
int 0x80
我也试过用 AL 替换 eax，用 AH 替换另一个等式，但它并不像我也不理解的那样工作。然后我还尝试制作每个字节并放入段符号 resb 1，用于 + 符号，但没有用。我只是想让它在之后再读一行。
mov edx, newline mov ecx, newlinelen
你在这里反转了寄存器。
mov eax, [num1] ... mov [res], eax
这些指令之间的所有内容都应改用字节大小的寄存器。使用 AL、BL 和 CL。当前您正在覆盖不属于 res 变量的内存！
; add '0' to to convert the sum from decimal to ASCII add eax, '0'
您应该先做减法，然后再转换回 ASCII，然后使用 mov [res], al.[=25= 将字符一次存储在 res 变量中]
exit: int 0x80
您的出口缺少 eax 寄存器中的函数编号！
Is there a way to repeat this one more time?
即使您不喜欢，循环也是可行的方法。 times 运算符是一个汇编时功能。你可以把整个东西变成一个宏并调用它两次，但是你也可以像现在这样写第二次。
我只是重复了代码，而不是添加值 4、5 和 6，但将它们放在高位字符而不是低位字符中。还为新行添加了一个读取垃圾字符。
segment .data
newline db 0xA, 0xD
newlinelen equ $-newline
segment .bss
;defining all variables
num1 resb 2
num2 resb 2
num3 resb 2
;variables for next equation
num4 resb 2
num5 resb 2
num6 resb 2
;newlin for garbage character
newlin resb 1
;result 1 and result 2
res resb 1
res2 resb 1
section .text
global main ;must be declared for using gcc
main: ;tell linker entry point
;reading num 1
mov eax, 3
mov ebx, 0
mov ecx, num1
mov edx, 2
int 0x80
;reading num 2
mov eax, 3
mov ebx, 0
mov ecx, num2
mov edx, 2
int 0x80
;reading num 3
mov eax, 3
mov ebx, 0
mov ecx, num3
mov edx, 2
int 0x80
; moving the first number to eax register and second number to ebx
; and subtracting ascii '0' to convert it into a decimal number
;moving variables in lower halfs
mov AL, [num1]
sub AL, '0'
mov BL, [num2]
sub BL, '0'
mov CL, [num3]
sub CL, '0'
; add ebx to eax
add AL, BL
; storing the sum in memory location res
mov [res], AL
;subtract ecx from eax
sub AL, CL
; add '0' to to convert the sum from decimal to ASCII
add AL, '0'
mov [res], AL
; print the sum
mov eax, 4
mov ebx, 1
mov ecx, res
mov edx, 1
int 0x80
;new line for next equation
mov eax, 4
mov ebx, 1
mov ecx, newline
mov edx, newlinelen
int 0x80
;;new equation?
;;skip garbage character
mov eax, 3
mov ebx, 0
mov ecx, newlin
mov edx, 1
int 0x80
;reading num 4
mov eax, 3
mov ebx, 0
mov ecx, num4
mov edx, 2
int 0x80
;reading num 5
mov eax, 3
mov ebx, 0
mov ecx, num5
mov edx, 2
int 0x80
;reading num 6
mov eax, 3
mov ebx, 0
mov ecx, num6
mov edx, 2
int 0x80
; moving the first number to eax register and second number to ebx
; and subtracting ascii '0' to convert it into a decimal number
;moving variables in higher halfs
mov AH, [newlin]
mov BH, [num4]
sub BH, '0'
mov CH, [num5]
sub CH, '0'
mov DH, [num6]
sub DH, '0'
; add ebx to eax
add BH, CH
;subtract ecx from eax
sub BH, DH
; add '0' to to convert the sum from decimal to ASCII
add BH, '0'
;move ah into memory location res 2
mov [res2], BH
; print the sum
mov eax, 4
mov ebx, 1
mov ecx, res2
mov edx, 1
int 0x80
exit:
mov eax, 1
int 0x80