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The object has 6,7 fields. cache.get() is a map read. Which one is faster?

obj temp = cache.get(key);                              // temporary object creation
Int id = temp != null ? temp.getId() : null;

or

Int id = cache.get(key) != null ? cache.get(key).getId() : null;       // cache read twice

For Java with low latency, which of the above is better?

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    Your first one isn't creating a temporary object. It's assigning a value to a temporary variable. Commented Feb 1, 2023 at 10:19
  • 1
    The only thing potentially eating any perfomance here is the cache.get - why would you call it twice? Commented Feb 1, 2023 at 10:29

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In the first snippet, you are not creating a temporary reference - you are assigning an existing object to a local variable that will be go out of scope soon.

The first snippet is preferable, although if cache is a simple HashMap it would probably be unnoticeable. It becomes more important if cache is something fancier, e.g., some sort of synchronized map that needs to handle concurrency.

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2 Comments

would there be any cost is assignment, any shallow copy ?
In Java, assigning an object doesn't perform any copying of the data, shallow or otherwise - it's just another reference to the same object. You can think of it as assigning a pointer - the only thing you're copying is the address in memory.

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