Goodbye PT. Simpli Mobile, 1 bulan lagi

Setelah hampir 3 tahun di PT Simpli Mobile, 1 bulan lagi gw pindah ke PT Cherrypicks.
2.5 tahun di Simpli telah memberikan banyak pengalaman dan peningkatan skill terutama di Java Programming.

Thanks buat semua teman-teman di Simpli, terutama yang sudah lama bareng gw di Simpli. 2.5 tahun yang menyenangkan dan seru :)

Di Cherrypicks sendiri walaupun belum diketahui pasti environment dan culture nya, mudah-mudahkan asik dan seru.

Buat teman-teman di Simpli, thanks buat support dan menjadi rekan yang oke selama ini.
Wish me luck ya :)

Integer, Byte, Long Constant Pool in Java

Following the topic about constant pool.
Let’s look inside the source code JDK 1.6, :

public static Integer valueOf(int i) {
final int offset = 128;
if (i >= -128 && i <= 127) { // must cache
return IntegerCache.cache[i + offset];
return new Integer(i);

From the source we can see when the value is -128 to 127 it’ll using cache / constant-pool.
So when we run the following code:

Integer i = 20;
Integer i2 = 20;
Integer i3 = 170;
Integer i4 = 170;
System.out.println(“i==i2 : ” + (i==i2) ); // true
System.out.println(“i==i2 : ” + (i3==i4) ); // false

Only value -128 until 127 will be cache. And of course the operator new wil create new object : new Integer(10)!=10.

Java Constant Pool : String

For those who already experienced in Java Programming should already familiar with the concept of String constant pool.

String s1 = “jim”;
String s2 = “jim”;

System.out.println(s1==s2); // true.

Object s2 is the same object with s1. But if you create using new operator:

String s1 = “jim”;
String s2 = new String(“jim”);
System.out.println(s1==s2); //false.

It will allocate new Object instead getting from String constant pool.
Another sample:

String s1 = “jim”;
String s2 = “j”+”im”;
String s3 = “j”;
String im = “im”;
s3+= im;
System.out.println(s1==s2); //true.
System.out.println(s1==s3); //false.

Concatenation during runtime will create new String Object.
Sample if using final modifier:

final String s1 = “j”;
final String s2 = “im”;
String jim = “jim”;
System.out.println(s1+s2 == jim); // returns true: Constant expression.

String s3 = “j”;
String s4 = “im”;
System.out.println(s3 + s4 == jim); // returns false: Not a constant expression.

Concatenation of 2 final String is a Constant.

Although is not necessary for us programmer to know about the detail of JVM implementation about String Constant Pool. Some would say “I’m not stupid, I used equals to check equality of Strings. And I don’t really care how JVM in my OS implemented the damn Constant Pool”. But for me knowing this knowledge is something great, and somehow I think it will help me in the future :)

More information about this : JLS – Lexical Structure

  • Literal strings within the same class (§8) in the same package (§7) represent references to the same String object (§4.3.1).
  • Literal strings within different classes in the same package represent references to the same String object.
  • Literal strings within different classes in different packages likewise represent references to the same String object.
  • Strings computed by constant expressions (§15.28) are computed at compile time and then treated as if they were literals.
  • Strings computed by concatenation at run time are newly created and therefore distinct.