assembly - 自旋锁所需的最小 X86 组件是多少

Question

在汇编中实现自旋锁。在这里我发布一个我想出的解决方案。这是对的吗？你知道更短的吗？

锁定:

    mov ecx, 0
.loop:
    xchg [eax], ecx
    cmp ecx, 0
    je .loop

发布:

    lock dec dword [eax]

eax 初始化为-1(这意味着锁已释放)。这应该适用于许多线程(不一定是 2 个)。

Answer 1

最短的可能是:

acquire:
    lock bts [eax],0
    jc acquire

release:
    mov [eax],0

为了提高性能，最好使用“测试、测试和设置”方法，并使用暂停，如下所示:

acquire:
    lock bts [eax],0    ;Optimistic first attempt
    jnc l2              ;Success if acquired
l1:
    pause
    test [eax],1        
    jne l1              ;Don't attempt again unless there's a chance

    lock bts [eax],0    ;Attempt to acquire
    jc l1               ;Wait again if failed

l2:

release:
    mov [eax],0

为了进行调试，您可以添加额外的数据以更轻松地检测问题，如下所示:

acquire:
    lock bts [eax],31         ;Optimistic first attempt
    jnc l2                    ;Success if acquired

    mov ebx,[CPUnumber]
    lea ebx,[ebx+0x80000000]
    cmp [eax],ebx             ;Is the lock acquired by this CPU?
    je .bad                   ; yes, deadlock
    lock inc dword [eax+4]    ;Increase "lock contention counter"
l1:
    pause
    test [eax],0x80000000        
    jne l1                    ;Don't attempt again unless there's a chance

    lock bts [eax],31         ;Attempt to acquire
    jc l1                     ;Wait again if failed

l2: mov [eax],ebx             ;Store CPU number

release:
    mov ebx,[CPUnumber]
    lea ebx,[ebx+0x80000000]
    cmp [eax],ebx             ;Is lock acquired, and is CPU same?
    jne .bad                  ; no, either not acquired or wrong CPU
    mov [eax],0