java short,integer,long performance

Question

I read that JVM stores internally short, integer and long as 4 bytes. I read it from an article from the year 2000, so I don\'t know how true it is now.

For the newe

Answer 1

Integer types are stored in many bytes, depending on the exact type :

• byte on 8 bits
• short on 16 bits, signed
• int on 32 bits, signed
• long on 64 bits, signed

See the spec here.

As for performance, it depends on what you're doing with them. For example, if you're assigning a literal value to a byte or short, they will be upscaled to int because literal values are considered as ints by default.

byte b = 10;  // upscaled to int, because "10" is an int

That's why you can't do :

byte b = 10;
b = b + 1;  // Error, right member converted to int, cannot be reassigned to byte without a cast.

So, if you plan to use bytes or shorts to perform some looping, you won't gain anything.

for (byte b=0; b<10; b++) 
{ ... } 

On the other hand, if you're using arrays of bytes or shorts to store some data, you will obviously benefit from their reduced size.

byte[] bytes = new byte[1000];
int[] ints = new int[1000];  // 4X the size

So, my answer is : it depends :)

Answer 2

There's basically no difference. One has to "confuse" the JITC a bit so that it doesn't recognize that the increment/decrement operations are self-cancelling and that the results aren't used. Do that and the three cases come out about equal. (Actually, short seems to be a tiny bit faster.)

public class ShortTest {

    public static void main(String[] args){
        // Do the inner method 5 times to see how it changes as the JITC attempts to
        // do further optimizations.
        for (int i = 0; i < 5; i++) {
            calculate(i);
        }
    }

    public static void calculate(int passNum){

        System.out.println("Pass " + passNum);
        // Broke into two (nested) loop counters so the total number of iterations could
        // be large enough to be seen on the clock.  (Though this isn't as important when
        // the JITC over-optimizations are prevented.)
        int M = 100000;
        int N = 100000;
        java.util.Random r = new java.util.Random();
        short x = (short) r.nextInt(1);
        short y1 = (short) (x + 1);
        int y2 = x + 1;
        long y3 = x + 1;

        long time1=System.currentTimeMillis();
        short s=x;
        for (int j = 0; j<M;j++) {
            for(int i = 0; i<N;i++) {
                s+=y1;
                s-=1;
                if (s > 100) {
                    System.out.println("Shouldn't be here");
                }
            }
        }
        long time2=System.currentTimeMillis();
        System.out.println("Time elapsed for shorts: "+(time2-time1) + " (" + time1 + "," + time2 + ")");


        long time3=System.currentTimeMillis();
        int in=x;
        for (int j = 0; j<M;j++) {
            for(int i = 0; i<N;i++) {
                in+=y2;
                in-=1;
                if (in > 100) {
                    System.out.println("Shouldn't be here");
                }
            }
        }
        long time4=System.currentTimeMillis();
        System.out.println("Time elapsed for ints: "+(time4-time3) + " (" + time3 + "," + time4 + ")");


        long time5=System.currentTimeMillis();
        long l=x;
        for (int j = 0; j<M;j++) {
            for(int i = 0; i<N;i++) {
                l+=y3;
                l-=1;
                if (l > 100) {
                    System.out.println("Shouldn't be here");
                }
            }
        }
        long time6=System.currentTimeMillis();
        System.out.println("Time elapsed for longs: "+(time6-time5) + " (" + time5 + "," + time6 + ")");


        System.out.println(s+in+l);
    }
}

Results:

C:\JavaTools>java ShortTest
Pass 0
Time elapsed for shorts: 59119 (1422405830404,1422405889523)
Time elapsed for ints: 45810 (1422405889524,1422405935334)
Time elapsed for longs: 47840 (1422405935335,1422405983175)
0
Pass 1
Time elapsed for shorts: 58258 (1422405983176,1422406041434)
Time elapsed for ints: 45607 (1422406041435,1422406087042)
Time elapsed for longs: 46635 (1422406087043,1422406133678)
0
Pass 2
Time elapsed for shorts: 31822 (1422406133679,1422406165501)
Time elapsed for ints: 39663 (1422406165502,1422406205165)
Time elapsed for longs: 37232 (1422406205165,1422406242397)
0
Pass 3
Time elapsed for shorts: 30392 (1422406242398,1422406272790)
Time elapsed for ints: 37949 (1422406272791,1422406310740)
Time elapsed for longs: 37634 (1422406310741,1422406348375)
0
Pass 4
Time elapsed for shorts: 31303 (1422406348376,1422406379679)
Time elapsed for ints: 36583 (1422406379680,1422406416263)
Time elapsed for longs: 38730 (1422406416264,1422406454994)
0

C:\JavaTools>java -version
java version "1.7.0_65"
Java(TM) SE Runtime Environment (build 1.7.0_65-b19)
Java HotSpot(TM) 64-Bit Server VM (build 24.65-b04, mixed mode)

Answer 3

long  64 –9,223,372,036,854,775,808 to 9 ,223,372,036,854,775,807 
int   32 –2,147,483,648 to 2,147,483,647 
short 16 –32,768 to 32,767 
byte   8 –128 to 127 

Use what you need, I would think shorts are rarely used due to the small range and it is in big-endian format.

Any performance gain would be minimal, but like I said if your application requires a range more then that of a short go with int. The long type may be too extremly large for you; but again it all depends on your application.

You should only use short if you have a concern over space (memory) otherwise use int (in most cases). If you are creating arrays and such try it out by declaring arrays of type int and short. Short will use 1/2 of the space as opposed to the int. But if you run the tests based on speed / performance you will see little to no difference (if you are dealing with Arrays), in addition, the only thing you save is space.

Also being that a commentor mentioned long because a long is 64 bits. You will not be able to store the size of a long in 4 bytes (notice the range of long).

Answer 4

Calculations with a short type are extremely expensive.

Take the following useless loop for example:

short t=0;
//int t=0;
//long t=0;
for(many many times...)
{
  t+=1;
  t-=1;
}

If it is a short, it will take literally 1000s of times longer than if it's an int or a long.

Checked on 64-bit JVMs versions 6/7 on Linux

Answer 5

It's an implementation detail, but it's still true that for performance reasons, most JVMs will use a full word (or more) for each variable, since CPUs access memory in word units. If the JVM stored the variables in sub-word units and locations, it would actually be slower.

This means that a 32bit JVM will use 4 bytes for short (and even boolean) while a 64bit JVM will use 8 bytes. However, the same is not true for array elements.

Answer 6

I agree with user2391480, calculations with shorts seem to be way more expensive. Here is an example, where on my machine (Java7 64bit, Intel i7-3770, Windows 7) operations with shorts are around ~50 times slower than integers and longs.

public class ShortTest {

public static void main(String[] args){
    calculate();
    calculate();
}

public static void calculate(){
    int N = 100000000;

    long time1=System.currentTimeMillis();
    short s=0;
    for(int i = 0; i<N;i++) {
        s+=1;
        s-=1;
    }
    long time2=System.currentTimeMillis();
    System.out.println("Time elapsed for shorts: "+(time2-time1));


    long time3=System.currentTimeMillis();
    int in=0;
    for(int i = 0; i<N;i++) {
        in+=1;
        in-=1;
    }
    long time4=System.currentTimeMillis();
    System.out.println("Time elapsed for ints: "+(time4-time3));


    long time5=System.currentTimeMillis();
    long l=0;
    for(int i = 0; i<N;i++) {
        l+=1;
        l-=1;
    }
    long time6=System.currentTimeMillis();
    System.out.println("Time elapsed for longs: "+(time6-time5));


    System.out.println(s+in+l);
}

}

Output:

Time elapsed for shorts: 113
Time elapsed for ints: 2
Time elapsed for longs: 2
0
Time elapsed for shorts: 119
Time elapsed for ints: 2
Time elapsed for longs: 2
0

Note: specifying "1" to be a short (in order to avoid casting every time, as suggested by user Robotnik as a source of delay) does not seem to help, e.g.

    short s=0;
    short one = (short)1;
    for(int i = 0; i<N;i++) {
      s+=one;
      s-=one;
    }

EDIT: modified as per request of user Hot Licks in the comment, in order to invoke the calculate() method more than once outside the main method.