java - How is String in switch statement more efficient than corresponding if-else statement?

Question

Java documentation says

The Java compiler generates generally more efficient bytecode from switch statements that use String objects than from chained if-then-else statements.

AFAIK even String in switch uses .equals() internally in a case sensitive manner. So what efficiency do they mean in this context. Faster compilation? Less bytecodes ? better performance?

Answer 1

Using a switch statement is faster than equals (but only noticeably when there are more than just a few strings) because it first uses the hashCode of the string that switch on to determine the subset of the strings that could possibly match. If more than one string in the case labels has the same hashCode, the JVM will perform sequential calls to equals and even if there is only one string in the case labels that a hashCode, the JVM needs to call equals to confirm that the string in the case label is really equal to the one in the switch expression.

The runtime performance of a switch on String objects is comparable to a lookup in a HashMap.

This piece of code:

public static void main(String[] args) {
    String s = "Bar";
    switch (s) {
    case "Foo":
        System.out.println("Foo match");
        break;
    case "Bar":
        System.out.println("Bar match");
        break;
    }
}

Is internally compiled to and executed like this piece of code:

(not literally, but if you decompile both pieces of code you see that the exact same sequence of actions occurs)

final static int FOO_HASHCODE = 70822; // "Foo".hashCode();
final static int BAR_HASHCODE = 66547; // "Bar".hashCode();

public static void main(String[] args) {
    String s = "Bar";
    switch (s.hashCode()) {
    case FOO_HASHCODE:
        if (s.equals("Foo"))
            System.out.println("Foo match");
        break;
    case BAR_HASHCODE:
        if (s.equals("Bar"))
            System.out.println("Bar match");
        break;
    }
}