Any problem with if(s1==s2 || s1.equals(s2))

Question

When you do not know how the strings are created either as �values of constant expressions� or as �new String objects�, i think most effective way to compare two strings for their equality is by:
if(s1==s2 || s1.equals(s2)){
System.out.println("s1 and s2 are equal");
}else {
System.out.println("s1 == s2 are not equal");
First part of the �if condition� (s1==s2) takes care if they are values of constant expressions, and is faster which is good.
Second part of the �if condition� s1.equals(s2) takes care if either s1 or s2 or both of them are new String objects which a bit slower as they compared char by char.
By the way i know that s1.equals(s2) is good enough to find if their values are or not. But i want to take advantage reference comparison(==) as it is faster than char by char comparison (equals()) method.
And also fact that if conditions are evaluated from left to right and if first one is good it does not do the rest of the contions for or operation.
Any issues with the above code?
Any better ideas?

Answer 1

1. What if s1 is null and s2 isn't?
2. Most equals methods do that == check first. In fact here is the source code for String's equals:
public boolean equals(Object anObject) {
     if (this == anObject) {
         return true;
     if (anObject instanceof String) {
         String anotherString = (String)anObject;
         int n = count;
         if (n == anotherString.count) {
          char v1[] = value;
          char v2[] = anotherString.value;
          int i = offset;
          int j = anotherString.offset;
          while (n-- != 0) {
              if (v1[i++] != v2[j++])
               return false;
          return true;
     return false;
    }3. Don't sweat about these sorts of minor efficiency details.