String equality check by only reference

Question

As we know if we do a check like below the output will be equal.

String s1 = "stackoverflow";
String s2 = "stackoverflow";
if(s1==s2){
    System.out.println("equal");
}

So my question is if i am not using new operator in my application to create String and all are strings are literals so can i use only reference equality as given above? Thanks in advance.

N.B: i am writing a crawler so i need to check whether i have already visited the given url that i am currently holding. I am using murmur hash which gives me a long for every url but there are collision so i need to check for the content if the url string if there is a hash collision. Hence for performance i am thinking of just comparing the reference equality of two string urls. And i am using jsoup for html parsing.

If you're writing a recursive crawler, not all of your strings are literals. However, look at interning. — SLaks, Dec 27 '13 at 13:31
If every string is a literal and you don't do any operations (substring, concatenation with unknown variables, etc) on it, then you should be able to just use `==`, yes. — Jeroen Vannevel, Dec 27 '13 at 13:34
Please refer http://stackoverflow.com/questions/513832/how-do-i-compare-strings-in-java — user3119954, Dec 27 '13 at 13:42

Ian Roberts · Answer 1 · 2013-12-27T14:03:34.420

if i am not using new operator in my application to create String and all are strings are literals so can i use only reference equality as given above?

If you are 100% sure that all the strings you are dealing with are plain string literals or compile-time constant expressions then yes. The Java Language Specification §15.28 mandates that

Compile-time constant expressions of type String are always "interned" so as to share unique instances, using the method String.intern.

But if you get strings from anywhere else (e.g. reading them from a web page retrieved by your crawler, or building them using concatenation expressions that are not compile-time constants) then you must use .equals to compare them by value rather than by reference or .intern() them explicitly.

It's not always obvious whether an expression is a compile-time constant or not:

String s1 = "Stack";
String s2 = s1 + "Overflow"; // not a CTC

but

final String s1 = "Stack";
String s2 = s1 + "Overflow"; // _is_ a CTC, because s1 is a "constant variable"
                             // (final, with an initializer that is itself a CTC)

score 1 · Answer 2 · answered Dec 27 '13 at 13:37

1

No, you cannot. The VM does not guarantee described behavior, it is an optimization. To guarantee this behavior, you need to call String#intern().

This and only this will guarantee reference equality.

But do a performance test, String#equals() is probably faster :-)

answered Dec 27 '13 at 13:37

Jan

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Can you provide a source that elaborates on why compile-time known strings aren't guaranteed to be interned? – Jeroen Vannevel Dec 27 '13 at 13:39
I probably did not read the question properly. My bad. The other answers are more precise. – Jan Dec 27 '13 at 15:10

Suresh Atta · Answer 3 · 2013-12-27T13:49:22.523

1

So my question is if i am not using new operator in my application to create String and all are strings are literals so can i use only reference equality as given above?

Yes , for sure. Since they are resolved at compile time , so no issues.

But keep in mind the below scenario

       String s3= s2;
       String s4= s1+"";   //resolved at run time 
      System.out.println(s3==s4);  //false
      System.out.println(s3.equals(s4));//true

So until unless you are sure that the strings not going to be change later, you can safely use ==.

edited Dec 27 '13 at 13:49

answered Dec 27 '13 at 13:38

Suresh Atta

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`s1+""` this is also string literal. please check your answer. – Trying Dec 27 '13 at 13:40
@Trying that expression is only a compile-time constant (and thus guaranteed to be interned) if `s1` is `final` – Ian Roberts Dec 27 '13 at 13:42
@Trying Yes. True. But s4 released the value at run-time. – Suresh Atta Dec 27 '13 at 13:43
@IanRoberts thats what i am saying but the OP has mentioned it in wrong way. – Trying Dec 27 '13 at 13:43
@IanRoberts so s3==s4 return true not false but op has mentioned false. – Trying Dec 27 '13 at 13:44
@Trying Are you sure ? – Suresh Atta Dec 27 '13 at 13:44
@sᴜʀᴇsʜᴀᴛᴛᴀ check it. – Trying Dec 27 '13 at 13:45
`String s1 = "stackoverflow"; String s2 = "stackoverflow"+""; if(s1==s2){ System.out.println("equal"); } `it prints equal. – Trying Dec 27 '13 at 13:46
@Trying May I have a demo, that how it's true ??? Here I can prove that It is false. https://ideone.com/9MQTjy – Suresh Atta Dec 27 '13 at 13:48
@Trying - Dont check s1 against s2... create s3 and s4 like suresh has done and try.. it returns false – TheLostMind Dec 27 '13 at 13:48
3

@Trying your example is a different case entirely - the concatenation of two string literals is a compile-time constant expression but the concatenation of one literal and one (non-constant) variable is not. – Ian Roberts Dec 27 '13 at 13:49

String equality check by only reference

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