Why does time for initialize array different

Question

Why

        long t = System.currentTimeMillis();
        int size = 3333333;
        int[][][] arr = new int[size][6][2];
//        int[][][] arr= new int[2][6][size];
        pr(System.currentTimeMillis() - t );

prints 5000 ms but

        long t = System.currentTimeMillis();
        int size = 3333333;
//        int[][][] arr = new int[size][6][2];
        int[][][] arr= new int[2][6][size];
        pr(System.currentTimeMillis() - t );

prints 44 ms

Second solution 115 time faster

How do you run this snippet? could you post the whole method? How many times have you run them? How many initial cycles did you run? Benchmarking Java operations is not a trivial thing... — ppeterka
– ppeterka, Commented Feb 1, 2013 at 9:25
See: stackoverflow.com/questions/504103/… - The way you are doing this is unlikely to produce any meaningful results. — Brian Roach
– Brian Roach, Commented Feb 1, 2013 at 9:35

dilix · Accepted Answer · 2013-02-01 08:23:53Z

7

It's simplier to test int[][]

int[][] arr = new int[size][2];

In this case you have to allocate size pieces of memory with size of 16 bytes.

And in this case

int[][] arr = new int[2][size];

you have only to allocate 2 pieces of memory of size*8 bytes.

And allocating is expensive operation.

answered Feb 1, 2013 at 8:23

dilix

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3 Comments

ppeterka Over a year ago

115 times faster? I think this is an issue related to JVM initialization, and initial optimization process...

dilix Over a year ago

With or without optimization in 1st sample you have to store many object (array) and allocate them all in heap and in second sample there is only 2 objects

ppeterka Over a year ago

I must admit it, you're right. I created a small, proper-ish benchmark around it, and the result is (roughly) the same. I think it's time for me to do some research around it...

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