問題描述
我在 Qt 的源代碼中看到了一些 x86 程序集:
I saw some x86 assembly in Qt's source:
q_atomic_increment:
movl 4(%esp), %ecx
lock
incl (%ecx)
mov $0,%eax
setne %al
ret
.align 4,0x90
.type q_atomic_increment,@function
.size q_atomic_increment,.-q_atomic_increment
通過谷歌搜索，我知道
lock指令會導致CPU鎖定總線，但我不知道CPU何時釋放總線?
From Googling, I knew
lockinstruction will cause CPU to lock the bus, but I don't know when CPU frees the bus?
關(guān)于上面的整個代碼，我不明白這段代碼是如何實現(xiàn)Add的?
About the whole above code, I don't understand how this code implements the Add?
推薦答案
LOCK本身不是一條指令:它是一個指令前綴，適用于后面的指令.該指令必須是對內(nèi)存(INC、XCHG、CMPXCHG等)執(zhí)行讀-修改-寫操作的指令——在此如果是incl (%ecx)指令，inc將long 字修改為ecx注冊.
LOCKis not an instruction itself: it is an instruction prefix, which applies to the following instruction. That instruction must be something that does a read-modify-write on memory (INC,XCHG,CMPXCHGetc.) --- in this case it is theincl (%ecx)instruction whichincrements thelong word at the address held in theecxregister.
LOCK 前綴確保 CPU 在操作期間擁有適當緩存行的獨占所有權(quán)，并提供某些額外的排序保證.這可以通過斷言總線鎖定來實現(xiàn)，但 CPU 將在可能的情況下避免這種情況.如果總線被鎖定，那么它只是在鎖定指令的持續(xù)時間內(nèi).
The LOCK prefix ensures that the CPU has exclusive ownership of the appropriate cache line for the duration of the operation, and provides certain additional ordering guarantees. This may be achieved by asserting a bus lock, but the CPU will avoid this where possible. If the bus is locked then it is only for the duration of the locked instruction.
這段代碼將要遞增的變量的地址從堆棧復制到ecx寄存器中，然后它以原子方式執(zhí)行lock incl (%ecx)將該變量加 1.如果變量的新值為 0，則接下來的兩條指令將 eax 寄存器(保存函數(shù)的返回值)設(shè)置為 0，否則設(shè)置為 1.該操作是一個增量，而不是一個添加(因此得名).
This code copies the address of the variable to be incremented off the stack into the ecx register, then it does lock incl (%ecx) to atomically increment that variable by 1. The next two instructions set the eax register (which holds the return value from the function) to 0 if the new value of the variable is 0, and 1 otherwise. The operation is an increment, not an add (hence the name).
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