Nasm x86 检查下溢

Question

考虑 16 位架构：

在下面的代码中，

test.s

bits 16

section .text


test:
    aam 7       ; divide n by 7
    mov cl, al  ; cl = n % 7
    dec ecx     ; <--- here if n was power of 7 we get the underflow

    ; ... more things where I make use of ah = n / 7

    add eax, ecx ; return that number
    ret

我想知道如何检查 under 流程。 jo、adc 和 setc（进位标志正常）都没有用。

如果不可能，则仅当 ecx 为负时才将其转换为 0 的汇编代码

Answer 1

感谢@PeterCordes 解决了这个问题，帮助我为 code golf 创造了答案：

0:  d4 07                   aam    0x7          ; eax contains n, with aam 0x7 -> al = n % 7 and ah = n / 7 
2:  88 c1                   mov    cl,al        ; keep result of n%7 in ecx
4:  66 c1 e8 08             shr    ax,0x8       ; by shifting 8 times ax, ax=al=n/7;
8:  66 6b c0 05             imul   ax,ax,0x5    ; we multiply by 5
c:  49                      dec    ecx          ; ecx = ecx - 1 -> give 65535 if n was power of 7
d:  78 02                   js     11 <end>     ; if it happened we don't care about the remainder so we can just return n/7*5, the Signed Flag will be set so we can jump 2 bytes forward.
f:  01 c8                   add    eax,ecx      ; if not then we add (n%7)-1 to n/7*5
00000011 <end>:
11: c3                      ret

Nasm x86 检查下溢

Nasm x86 Check for UNDERflow

assembly

nasm

x86-16