关于Java内存模型的疑问：后续请求线程能否读取到volatile变量更新值？

Great question! You’re right to dig into the nuances of the Java Memory Model (JMM) here—this is a super common point of confusion, especially around how volatile and happens-before interact with real-world execution order. Let’s break this down to clear up your misunderstanding.

First, anchor to the actual execution constraints in your scenario #

Your client doesn’t just fire off two requests randomly: it waits for the first request to complete entirely before sending the second. That’s a critical detail. What this means in practice is:

• The thread handling handleFirstRequest() (let’s call it Thread A) has fully finished executing every line of that method—including the someFlag = 1 volatile write—and the response has been sent back to the client.
• Only after all that is done does the thread handling handleSecondRequest() (Thread B) start running its logic, including reading someFlag.

This "wait-for-completion" behavior creates an implicit happens-before relationship between every action Thread A takes and every action Thread B takes. In JMM terms, if Thread A’s work finishes before Thread B starts, all of Thread A’s actions happen-before all of Thread B’s actions.

How volatile rules combine with this happens-before relationship #

You already know volatile has core visibility rules:

• A volatile write synchronizes-with any subsequent volatile read of the same variable.
• If action X happens-before action Y, then all the results of X are visible to Y.

In your scenario:

• Thread A’s someFlag = 1 (a volatile write) is part of Thread A’s final actions, which happen-before Thread B’s volatile read of someFlag.
• This chain of happens-before guarantees that Thread B’s read must see the value written by Thread A—no exceptions.

Correcting your misconception about synchronization order #

You noted that the JMM doesn’t tie synchronization order to physical time, and that’s true—but it does enforce causality. If an operation physically happens after another (and there’s a clear causal link, like your client waiting for the first request), the synchronization order can’t flip those two operations.

The total order of synchronization operations exists to ensure consistency across threads, but it will never violate basic cause and effect. JMM’s design prioritizes making predictable, intuitive behavior possible for developers—so it can’t allow a read that happens after a write (in real execution time) to be ordered before that write in the synchronization sequence.

A quick look at underlying memory mechanics to reinforce this #

When Thread A executes the volatile write someFlag = 1:

1. It flushes the updated value of someFlag from its local cache to main memory.
2. It prevents any instruction reordering that would move this write before earlier operations in the method.

Your client’s wait ensures this flush is fully complete before Thread B starts. When Thread B does the volatile read:

1. It bypasses its local cache and loads someFlag directly from main memory.
2. It prevents reordering that would move this read after later operations in the method.

So even at the hardware level, Thread B is guaranteed to get the latest value of someFlag.

内容的提问来源于stack exchange，提问作者Alex W