英文:
Ensure all threads updating a single entry in a map in Java
问题
我有一个被多个线程同时调用的方法。在这个方法中，我试图处理下面片段所解释的情况：
Map<Object, Long> syncMap = Collections.synchronizedMap(normalHashMap);
Runnable mapOperations = () -> {
synchronized (syncMap) {
if (MapUtils.isNotEmpty(syncMap)) {
Object currentCount = syncMap.get(entryKey);
// 每个线程应该执行这部分代码一次。每个线程都会带一个值，这些值将会相加，并存储在这个“key”的条目中。
syncMap.put(key, value + valueFromThisThread);
}
}
};
mapOperations.run();
每个线程应该执行上述代码块一次。每个线程会带一个长整型值，这些值会相加并存储在这个“key”对应的地图条目中。
有人可以确认这是否可行？
英文:
I have a method that is called by multiple threads at the same time. Within it, I am trying to handle a scenario as explained by the snippet below:
Map<Object,Long> syncMap = Collections.synchronizedMap(normalHashMap);
Runnable mapOperations = () -> {
synchronized (syncMap) {
if (MapUtils.isNotEmpty(syncMap))
{
Object currentCount = syncMap.get(entryKey);
// Every thread should execute this once. Every thread brings a value and that gets added up and stored for this "key"
syncMap.put(key, value + valueFromThisThread);
}
}
};
mapOperations.run();
Every thread should execute the block above once. Every thread brings a long value and that gets added up and stored for this one "key" entry in the map.
Can someone please confirm if this would work?
答案1
得分: 0
你的代码应该能够正常工作。另外，如果你想确保不将该字段缓存在 CPU 缓存中，你可以将其定义为 volatile，例如：
volatile Map<Object, Long> syncMap = Collections.synchronizedMap(normalHashMap);
此外，如果你使用 SynchronizedMap，就不需要 synchronized { } 块。
英文:
Your code should work fine. Additionally, if you want to make sure you don't cache the field in the CPU cache, you might want to define it as a volatile, like:
volatile Map<Object,Long> syncMap = Collections.synchronizedMap(normalHashMap);
Also, you don't need a synchronized { } block if you use the SynchronizedMap.
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