Linux Intel 64bit 汇编分部 [英] Linux Intel 64bit Assembly Division

问题描述

我正在努力理解为什么我的除法不起作用，下面是我当前的代码，它只取两个个位数并试图将它们相除:

I am battling to understand why my division is not working, below is my current code, which simply takes in two single digits and attempts to divide them:

STDIN equ 0
SYS_READ equ 0

STDOUT equ 1
SYS_WRITE equ 1

segment .data
    num1 dq 0
    num2 dq 0
    quot dq 0
    rem dq 0

segment .text
    global _start
_start:
    mov rax, SYS_READ
    mov rdi, STDIN
    mov rsi, num1
    mov rdx, 2
    syscall

    mov rax, SYS_READ
    mov rdi, STDIN
    mov rsi, num2
    mov rdx, 2
    syscall

    mov rax, [num1]
    sub rax, '0'

    mov rbx, [num2]
    sub rbx, '0'

    xor rdx, rdx
    div rbx

    add rax, '0'

    mov [quot], rax
    mov [rem], rdx

    mov rax, SYS_WRITE
    mov rdi, STDOUT
    mov rsi, quot
    mov rdx, 1
    syscall

    mov rax, 60
    xor rdi, rdi
    syscall

现在据我所知，汇编程序将RDX:RAX除以操作数RBX.我只能假设这是问题所在，事实上我将 128 位值除以 64 位值.每当我输入诸如 8/2 或类似的东西时，我都会收到值 1 作为商.我在这里错过了什么?任何帮助将不胜感激.

Now as far as I understand when dividing the assembler will divide RDX:RAX by the operand RBX. I can only assume this is where the problem is coming in, the fact that I am dividing a 128bit value by a 64bit value. Whenever I enter something such as 8 / 2 or something similar, I receive the value 1 as the quotient. What am I missing here? Any help would be greatly appreciated.

推荐答案

您为操作数读取了 2 个字节，但似乎您忽略了第二个字节，而您不应该这样做.
假设您输入 8 和 2 各一行，您将看到8 "和2 ".然后减去 '0'，但保留 ' '，因此您的操作数将是 0x08 0x0A 和 0x02 0x0A，即 2568 和 2562.和 2568/2562 = 1.

You read 2 bytes for the operands, but it seems you ignore the 2nd, when you shouldn't.
Assuming you type 8 and 2 and one line each, you will read "8 " and "2 ". You then subtract '0', but you leave the ' ', so your operands will be 0x08 0x0A and 0x02 0x0A, which are 2568 and 2562. And 2568 / 2562 = 1.

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