Is hash code of java.lang.String really cached?

Question

String s1 = "String1";      
System.out.println(s1.hashCode()); // return an integer i1 


Field field = String.class.getDeclaredField("value");  
field.setAccessible(true);  
char[] value = (char[])field.get(s1);  
value[0] = 'J';  
value[1] = 'a';  
value[2] = 'v';  
value[3] = 'a';  
value[4] = '1'; 
System.out.println(s1.hashCode()); // return same value of integer i1

Here even after I changed the characters with the help of reflection, same hash code value is mainatained.

Is there anything I need to know here?

If it's the same value, then it is cached. What other proof do you need? Go to the source code. — Sotirios Delimanolis
– Sotirios Delimanolis, Commented Jan 8, 2014 at 16:02
Using Reflection breaks the API of String, which causes this unexpected behaviour — Mirco
– Mirco, Commented Jan 8, 2014 at 16:03
@kocko: never, since String is immutable (unless the computed hashCode is 0, which also means "not computed yet") — JB Nizet
– JB Nizet, Commented Jan 8, 2014 at 16:05

Konstantin Yovkov · Accepted Answer · 2014-01-08 16:07:14Z

14

A String is meant to be immutable. As such, there is no point having to recalculate the hashcode. It is cached internally in a field called hash of type int.

String#hashCode() is implemented as (Oracle JDK7)

public int hashCode() {
    int h = hash;
    if (h == 0 && value.length > 0) {
        char val[] = value;

        for (int i = 0; i < value.length; i++) {
            h = 31 * h + val[i];
        }
        hash = h;
    }
    return h;
}

where hash initially has a value of 0. It will only be calculated the first time the method is called.

As stated in the comments, using reflection breaks the immutability of the object.

edited Jan 8, 2014 at 16:07

Konstantin Yovkov

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answered Jan 8, 2014 at 16:05

Sotirios Delimanolis

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