Conditionally remove elements from a ConcurrentHashMap (thread-safe)












0















I have a ConcurrentHashMap<String, Instant> which is accessed concurrently. I'd say that



map.entrySet().removeIf(e -> e.getValue().isBefore(Instant.now()));


would be a thread-safe implementation as it would never remove a String which got a new Instant by another thread right after the predicate evaluation and before the actual removal. Even if multiple threads put strings with instants into this map, never an entry with an instant after Instant.now() is removed. That's because





  1. entrySet() is a view on the original map


  2. removeIf() uses the iterator from the entrySet() supporting Iterator.remove()


  3. removeIf() removes the String/Instant entry by calling remove(Object, Object)


I'm just not entirely sure about (3). Who can help me out?






java multithreading







share|improve this question























  • Note: Instant.now() can return a different value every time it is called. This means it doesn't remove up to the same point for all entries.

    – Peter Lawrey
    Nov 21 '18 at 19:06
















0















I have a ConcurrentHashMap<String, Instant> which is accessed concurrently. I'd say that



map.entrySet().removeIf(e -> e.getValue().isBefore(Instant.now()));


would be a thread-safe implementation as it would never remove a String which got a new Instant by another thread right after the predicate evaluation and before the actual removal. Even if multiple threads put strings with instants into this map, never an entry with an instant after Instant.now() is removed. That's because





  1. entrySet() is a view on the original map


  2. removeIf() uses the iterator from the entrySet() supporting Iterator.remove()


  3. removeIf() removes the String/Instant entry by calling remove(Object, Object)


I'm just not entirely sure about (3). Who can help me out?






java multithreading







share|improve this question























  • Note: Instant.now() can return a different value every time it is called. This means it doesn't remove up to the same point for all entries.

    – Peter Lawrey
    Nov 21 '18 at 19:06














0












0








0








I have a ConcurrentHashMap<String, Instant> which is accessed concurrently. I'd say that



map.entrySet().removeIf(e -> e.getValue().isBefore(Instant.now()));


would be a thread-safe implementation as it would never remove a String which got a new Instant by another thread right after the predicate evaluation and before the actual removal. Even if multiple threads put strings with instants into this map, never an entry with an instant after Instant.now() is removed. That's because





  1. entrySet() is a view on the original map


  2. removeIf() uses the iterator from the entrySet() supporting Iterator.remove()


  3. removeIf() removes the String/Instant entry by calling remove(Object, Object)


I'm just not entirely sure about (3). Who can help me out?






java multithreading







share|improve this question














I have a ConcurrentHashMap<String, Instant> which is accessed concurrently. I'd say that



map.entrySet().removeIf(e -> e.getValue().isBefore(Instant.now()));


would be a thread-safe implementation as it would never remove a String which got a new Instant by another thread right after the predicate evaluation and before the actual removal. Even if multiple threads put strings with instants into this map, never an entry with an instant after Instant.now() is removed. That's because





  1. entrySet() is a view on the original map


  2. removeIf() uses the iterator from the entrySet() supporting Iterator.remove()


  3. removeIf() removes the String/Instant entry by calling remove(Object, Object)


I'm just not entirely sure about (3). Who can help me out?






java multithreading




java multithreading






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 21 '18 at 17:17









steffensteffen

9,37022355




9,37022355













  • Note: Instant.now() can return a different value every time it is called. This means it doesn't remove up to the same point for all entries.

    – Peter Lawrey
    Nov 21 '18 at 19:06



















  • Note: Instant.now() can return a different value every time it is called. This means it doesn't remove up to the same point for all entries.

    – Peter Lawrey
    Nov 21 '18 at 19:06

















Note: Instant.now() can return a different value every time it is called. This means it doesn't remove up to the same point for all entries.

– Peter Lawrey
Nov 21 '18 at 19:06





Note: Instant.now() can return a different value every time it is called. This means it doesn't remove up to the same point for all entries.

– Peter Lawrey
Nov 21 '18 at 19:06












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