组装中的算术运算

本文介绍了组装中的算术运算的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 SYS_EXIT equ 1 SYS_READ equ 3 SYS_WRITE equ 4 STDIN equ 0 STDOUT equ 1 segment .data msg db "Please enter a digit ", 0xA,0xD len equ $- msg segment .bss number1 resb 2 number2 resb 2 result resb 1 result2 resb 1 segment .text msg2 db "Please enter a second digit", 0xA,0xD len2 equ $- msg2 msg3 db "The sum is: " len3 equ $- msg3 msg4 db "The minus is: " len4 equ $- msg4 global _start _start: mov eax, SYS_WRITE ; System write mov ebx, STDOUT ; System output mov ecx, msg ; What to write mov edx, len ; Length to write int 0x80 ; Interupt Kernel mov eax, SYS_READ ; System read mov ebx, STDIN ; mov ecx, number1 mov edx, 2 int 0x80 mov eax, SYS_WRITE mov ebx, STDOUT mov ecx, msg2 mov edx, len2 int 0x80 mov eax, SYS_READ mov ebx, STDIN mov ecx, number2 mov edx, 2 int 0x80 call add add: mov eax, SYS_WRITE mov ebx, STDOUT mov ecx, msg3 mov edx, len3 int 0x80 ;load number1 into eax and subtract '0' to convert from ASCII to decimal mov eax, [number1] sub eax, '0' ; do the same for number2 mov ebx, [number2] sub ebx, '0' ; add eax and ebx, storing the result in eax add eax, ebx ; add '0' to eax to convert the digit from decimal to ASCII add eax, '0' ; store the result in result mov [result], eax ; print the result digit mov eax, SYS_WRITE mov ebx, STDOUT mov ecx, result mov edx, 1 int 0x80 ret minus: mov eax, SYS_WRITE mov ebx, STDOUT mov ecx, msg4 mov edx, len4 int 0x80 ;load number1 into eax and subtract '0' to convert from ASCII to decimal mov eax, [number1] sub eax, '0' ; do the same for number2 mov ebx, [number2] sub ebx, '0' ; add eax and ebx, storing the result in eax sub eax, ebx ; add '0' to eax to convert the digit from decimal to ASCII add eax, '0' ; store the result in result mov [result2], eax ; print the result digit mov eax, SYS_WRITE mov ebx, STDOUT mov ecx, result mov edx, 1 int 0x80 ret mul: ;load number1 into eax and subtract '0' to convert from ASCII to decimal mov al, [number1] sub al, '0' ; do the same for number2 mov bl, [number2] sub bl, '0' ; add eax and ebx, storing the result in eax mul bl ; add '0' to eax to convert the digit from decimal to ASCII add al, '0' ; store the result in result mov [result], al ; print the result digit mov eax, SYS_WRITE mov ebx, STDOUT mov ecx, result mov edx, 1 int 0x80 ret exit: mov eax, SYS_EXIT xor ebx, ebx int 0x80

上面的代码是我到目前为止所做的，我试图添加数字，但是添加函数似乎被调用了两次，如下图所示.我想做的是将两个数字相加，相减，相乘和相除.我该怎么办?当我尝试加2个数字时，有时会出现此输出细分错误.

The above code is what i have done so far, I tried to add the numbers but add function seems to be called twice as shown in the picture below. What am i trying to do is to add, subtract, multiply and divide the two numbers. How can i do that and when i try to add 2 numbers sometimes it gave me this output Segmentation fault.

推荐答案

您在_start块的中间定义了函数.

You defined your functions in the middle of the _start block.

在调试器中单步执行代码，请注意，在返回call add之后，执行将继续执行从add标签开始的指令.

Single-step your code in a debugger and notice that after call add returns, execution continues into the instructions starting at the add label.

最终，当您到达ret时会崩溃，即，当您尝试从_start进入ret时(因为堆栈的顶部是argc，而不是地址)，在System V ABI中的进程启动时).也许您甚至更早地犯了错；您没有发布调试器输出，我也没有在脑海中模拟它以查看会发生什么.

Eventually you crash when you reach the ret, i.e. when you try to ret from _start (because the top of the stack is argc, not an address, at process startup in the System V ABI). Or maybe you even fault earlier; you didn't post debugger output and I didn't simulate it in my head to see what happens.

这就是为什么The sum is: ?打印两次的原因.

This is why The sum is: ? is printed twice.

我想您的代码也无法处理多位数字，因此您得到了ASCII代码'0' + 8+9或其他任何代码.参见如何在不使用c库中的printf的情况下，在汇编级编程中打印整数?，或该数字的简化2位数最大版本.或者，如果您动态链接libc，请使用printf，以便它可以调用自己的init函数.

I guess your code also fails to handle multi-digit numbers, so you got the ASCII code '0' + 8+9 or whatever. See How do I print an integer in Assembly Level Programming without printf from the c library?, or a simplified 2-digit-max version of that. Or just use printf if you dynamically link libc, so it can call its own init functions.