As string gets stored in string constant pool(when we initialize it with =) on the basis on content than in that case why s1 == s3 false

Question

As string gets stored in string constant pool(when we initialize it with =) on the basis on content than in that case why s1 == s3 false ;

public static void main(String[] args) {
    String s1 = "ABC:5";
    String s2 = "ABC:5";
    String s3 = "ABC:"+s1.length();
    String s4 = "check";
    System.out.println(s1 == s2);
    System.out.println(s1 == s3);
    System.out.println(s1.length());
    System.out.println(s1.hashCode());
    System.out.println(s2.hashCode());
    System.out.println(s3.hashCode());
    System.out.println(s4.hashCode());
}

and even hashcode for s1 , s2 and s3 are same ;

No, only when you initialize it with string literal or intern it. There must be a duplicate for this. — Oleg, Sep 16 '17 at 23:21
The string pool is an interesting concept, but you shouldn't worry too much about it in your day-to-day programming life. And certainly don't think you can get away with using `==` instead of `.equals()` based on what you know or _think_ you know about how it works (:-)) — Kevin Anderson, Sep 16 '17 at 23:28

score 1 · Answer 1 · answered Sep 17 '17 at 00:01

@Pritam Banerjee's answer is not fully correct.

// both declarations are using the same object in memory
String s1 = "ABC:5"; // s1 == s1
String s2 = "ABC:5"; // s2 == s1

// this one creates at the beginning StringBuilder
String s3 = "ABC:" + s1.length(); // s3 != s1
String s4 = new StringBuilder().append("ABC:").append(s1.length()).toString(); 
// s4 != s1 && s4 != s3 this is what happens in byte code for s3

// the last case
String s5 = "ABC:" + "5"; // s5 == s1, because compiler knows exact value before compiling

The byte code for it:

  public StringBuilderTest();
    Code:
       0: aload_0
       1: invokespecial #1                  // Method java/lang/Object."<init>":()V
       4: return

  public static void main(java.lang.String[]);
    Code:
       0: ldc           #2                  // String ABC:5
       2: astore_1
       3: ldc           #2                  // String ABC:5
       5: astore_2
       6: new           #3                  // class java/lang/StringBuilder
       9: dup
      10: invokespecial #4                  // Method java/lang/StringBuilder."<init>":()V
      13: ldc           #5                  // String ABC:
      15: invokevirtual #6                  // Method java/lang/StringBuilder.append:(Ljava/lang/String;)Ljava/lang/Stri
ngBuilder;
      18: aload_1
      19: invokevirtual #7                  // Method java/lang/String.length:()I
      22: invokevirtual #8                  // Method java/lang/StringBuilder.append:(I)Ljava/lang/StringBuilder;
      25: invokevirtual #9                  // Method java/lang/StringBuilder.toString:()Ljava/lang/String;
      28: astore_3
      29: new           #3                  // class java/lang/StringBuilder
      32: dup
      33: invokespecial #4                  // Method java/lang/StringBuilder."<init>":()V
      36: ldc           #5                  // String ABC:
      38: invokevirtual #6                  // Method java/lang/StringBuilder.append:(Ljava/lang/String;)Ljava/lang/Stri
ngBuilder;
      41: aload_1
      42: invokevirtual #7                  // Method java/lang/String.length:()I
      45: invokevirtual #8                  // Method java/lang/StringBuilder.append:(I)Ljava/lang/StringBuilder;
      48: invokevirtual #9                  // Method java/lang/StringBuilder.toString:()Ljava/lang/String;
      51: astore        4
      53: ldc           #2                  // String ABC:5
      55: astore        5
      57: return
}

You can read more about declaring Strings and its equality here

Thanks for the answer :) . Could you please let me know how you have generated this bytecode , using any converter ? and how we can understand that bytecode ? — vikas srivastava, Sep 18 '17 at 10:40

As string gets stored in string constant pool(when we initialize it with =) on the basis on content than in that case why s1 == s3 false

1 Answers1