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HomeBackend DevelopmentC++What are the intricacies of the \'lock\' prefix in x86 assembly, and how does it impact bus usage and memory operations?

 What are the intricacies of the

Understanding the Role of "lock" in x86 Assembly

The "lock" instruction is an intriguing element in x86 assembly, renowned for its ability to regulate bus usage by the CPU. This article delves into the complexities of this enigmatic command, answering two critical questions: its timing for releasing the bus and its implementation of addition.

When Does the CPU Release the Bus after a "lock" Instruction?

Contrary to popular belief, "lock" does not act as an independent instruction but serves as a prefix to the subsequent instruction. Typically, this instruction is one that executes a read-modify-write operation on memory, such as "incl" or "cmpxchg".

Upon encountering a "lock" prefix, the CPU assumes undivided control over the appropriate cache line for the duration of the operation. This exclusivity is often facilitated through the assertion of a bus lock, although the processor seeks to bypass this method whenever feasible. It's important to note that this bus lock remains in force exclusively for the duration of the locked instruction.

How does the "lock" Code Implement Addition?

In the provided code snippet:

  • "movl 4(%esp), �x" extracts the address to be incremented from the stack and stores it in register ecx.
  • "lock incl (�x)" subsequently increments the long word at the address held in ecx by one, ensuring atomicity.
  • The following instructions set the eax register's return value to 0 if the incremented variable becomes 0 or 1 otherwise.

It's crucial to recognize that this code performs an increment, not an addition.

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