What does the "lock" instruction mean in x86 assembly?

Solution 1:

1. LOCK is 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, CMPXCHG etc.) --- in this case it is the incl (%ecx) instruction which increments the long word at the address held in the ecx register.

   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.

2. 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).

Solution 2:

What you may be failing to understand is that the microcode required to increment a value requires that we read in the old value first.

The Lock keyword forces the multiple micro instructions that are actually occuring to appear to operate atomically.

If you had 2 threads each trying to increment the same variable, and they both read the same original value at the same time then they both increment to the same value, and they both write out the same value.

Instead of having the variable incremented twice, which is the typical expectation, you end up incrementing the variable once.

The lock keyword prevents this from happening.