String.equals implementation

Question

In JDK 8, String.equals implements as

public boolean equals(Object anObject) {
    if (this == anObject) {
        return true;
    }
    if (anObject instanceof String) {
        String anotherString = (String)anObject;
        int n = value.length;
        if (n == anotherString.value.length) {
            char v1[] = value;
            char v2[] = anotherString.value;
            int i = 0;
            while (n-- != 0) {
                if (v1[i] != v2[i])
                    return false;
                i++;
            }
            return true;
        }
    }
    return false;
}

Why does the iteration uses two operations - incrementing i and decrementing n instead of something like that:

while (i < n) {
    if (v1[i] != v2[i])
        return false;
    i++;
}

or

while (i-- != 0) {
    if (v1[i] != v2[i])
        return false;
}

with one increment or decrement operation?

I suppose, it is somehow related to JVM bytecode optimization but don't understand how.

Answer 1

I think this is dead code, left over from the days when Strings still shared backing arrays and had offset and count, so that you needed to tweak the indexes a little.

The String implementation was changed in JDK 7 Update 6 (many people were upset about that, especially that it happened in a minor release). Strings don't share backing array anymore (unless the Strings are themselves equal, in which case a new deduplicator tries to re-share them).

You can see a discussion of the old implementation over at "how does String.equals() work".