CODEWITHSUNDEEP
javaperformancegarbage-collectionautoboxing
jellyfication
jellyfication

Reputation: 1605

Autoboxing performance

Why groups j=k*l and m=n*o have different performance, while first 3 groups have the same ?

int a = 42;
int b = 42;
int c = 42;

Integer d = 42;
int e = 42;
int f = 42;

int g = 42;
Integer h = 42;
int i = 42;

int j = 42;
int k = 42;
Integer l = 42;

Integer m = 42;
Integer n = 42;
Integer o = 42;

for(int z = 0; z < 1000000000; z++){
    // c = a*b; // 630 ms
    // f = d*e; // 630 ms
    // i = g*h; // 630 ms
    // l = j*k; // 6000 ms
    // o = m*n; // 6400 ms
}

Upvotes: 6

Views: 2057

Answers (7)

Bernhard Barker
Bernhard Barker

Reputation: 55609

Creating an object has an overhead (or, from other comments, it appears that it might be that, when creating an object, the loop can't optimised away because nothing is being done inside it).

int = int * int

No objects involved.

int = Integer * int or int = int * Integer

The Integer's int member get multiplied with the int, and no objects need to be created.

Integer = int * int

An Integer object needs to be created from the result (slow).

Integer = Integer * Integer

The two Integer's int members get multiplied, but then an Integer object needs to be created from the result (slow).

Upvotes: 0

Jayamohan
Jayamohan

Reputation: 12924

Basically this is what happening inside JVM and thats causing performance issue.

c = a*b; // 630  ms -> No boxing and unboxing
f = d*e; // 630  ms -> Unboxing d
i = g*h; // 630  ms -> Unboxing h
l = j*k; // 6000 ms -> Creates a new Integer object L, and assign J * K to L
o = m*n; // 6400 ms -> Creates a new Integer object 0, unbox m and n for calculation and assign m * n to o

In other words

l = j*k; // 6000 ms is equivalent to below code

temp = j*k;
Integer l = new Integer(temp);

o = m*n; // 6400 ms is equivalent to below code

temp = j.intValue() * k.intValue();
Integer l = new Integer(temp);

Your last two statements will create around 1000000000 (loop count) unnecessary Integer object. Which will be obviously a performance slog.

Upvotes: 4

Peter Lawrey
Peter Lawrey

Reputation: 533492

You have a loop which doesn't do anything useful. As such the JIT can take some time to detect this and eliminate the loop. In the first four examples, the code is taking an average of a fraction of a clock cycle. This is a strong hint that the loop has been optimised away and you are actually measuring how long it takes to detect and replace the loop with nothing,

In the later examples, the loop is not optimised away because there is something about the loop which the JIT doesn't eliminate. I suspected it was the object allocation for the Integer but -verbosegc showed that no objects were being created.

On my machine the loop take about 1.9 ns or about 6 clock cycles which is pretty quick so I am not sure what is left which takes that much time..

Upvotes: 6

Donato Szilagyi
Donato Szilagyi

Reputation: 4369

Creating a new object in a cycle always has a cost. In the last two examples you created new Integer objects in the cycle what has high performance cost. Primitives are stored in stack, objects are in heap. Putting something into stack is cheaper than into heap.

Upvotes: 4

imxylz
imxylz

Reputation: 7937

  1. Integer autobox has a small cache
  2. Creating an object costs a litter time

check this: java.lang.Integer.IntegerCache

Upvotes: 1

CloudyMarble
CloudyMarble

Reputation: 37566

m = n*o  ==> Integer = Integer * Integer

while:

l = j*k ==> Integer = int * int // Integer is an object so autoboxing costs some performance here.

Upvotes: 0

JB Nizet
JB Nizet

Reputation: 691715

Probably because l = j * k needs to box an int to an Integer, whereas c = a * b doesn't need to. The VM could also optimize the loop, understanding that executing a single iteration would lead to the same result as executing 1 billion. Beware of simplistic micro-benchmarks.

Upvotes: 0

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