Program considers two equal doubles to be different

Question

I am comparing two Doubles:

37.4238777160645

and

37.4238777160645

But Java does not consider them to be equal. I am comparing them in the following manner

if(object1.getLatitude()!=object2.getLatitude()){
    fail("objects are not equal "+object1.getLatitude()+":"+object2.getLatitude());
}

resulting in the following fail:

junit.framework.AssertionFailedError: objects are not equal  37.4238777160645:37.4238777160645

I don't understand why - Please advise.

Can you try with `if(!object1.getLatitude().equals(object2.getLatitude())){` — Pshemo, Mar 20 '14 at 13:55
@AndroidAddict are you sure it is `double` and not `Double`? — Pshemo, Mar 20 '14 at 13:56
@Pshemo that would require that I implement the equals and use the same method I am performing with the same issue. — Roy Hinkley, Mar 20 '14 at 13:56
@AndroidAddict Difference is that for non primitive types `==` compares equality of references, and `equals` checks equality of state of object (its value). — Pshemo, Mar 20 '14 at 13:58
Will someone please find a duplicate and put this to rest? I am ashamed to admit that I couldn't find it. — Marko Topolnik, Mar 20 '14 at 14:01
@AndroidAddict In future, when people ask you questions about your code, please answer them diligently and correctly. Saying it returned `double` when actually it returned `Double` just caused a flurry of comments/questions and wasted everybody's time. — Bohemian, Mar 20 '14 at 14:04

score 3 · Answer 1 · answered Mar 20 '14 at 14:04

The issue has already been pointed out. But if you are using junit, it would be simpler to use the appropriate method:

assertEquals(object1.getLatitude(), object2.getLatitude());

or

assertEquals(object1.getLatitude(), object2.getLatitude(), 0.001d);

instead of using fail. That would also solve your issue.

score 2 · Answer 2 · answered Mar 20 '14 at 13:57

2

Objects should be compared with .equals and not ==. By == you're comparing the references, which are not the same since you return a different object each time.

Use Double#equals to compare the values.

answered Mar 20 '14 at 13:57

Maroun

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I am comparing the `values` of property in the object, not the objects themselves – Roy Hinkley Mar 20 '14 at 13:58
@TimB Are you sure `double` cannot handle `37.4238777160645`? I tested it and it *worked*. Please check again. – Maroun Mar 20 '14 at 13:59
2

@AndroidAddict You mentioned that `getLatitude` returns a `Double`, which is an *object* and not a primitive. So you're comparing *references*. – Maroun Mar 20 '14 at 14:01
@MarounMaroun yes, I sit corrected. – Tim B Mar 20 '14 at 14:06

Hari · Accepted Answer · 2014-03-20T14:09:36.867

1

You should use java.lang.Double.compare() :

Double.compare(object1.getLatitude(), object2.getLatitude())

so you wil have :

if(Double.compare(object1.getLatitude(), object2.getLatitude()) != 0){
   fail("objects are not equal "+object1.getLatitude()+":"+object2.getLatitude());
}

edited Mar 20 '14 at 14:09

answered Mar 20 '14 at 13:58

Hari

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2

Your example should be != 0 since if they are identical the compare method would return the difference between the two values. If they are identical, the return from compare is therefore 0 and != would be something not 0. – Roy Hinkley Mar 20 '14 at 14:05
Edited answer as you suggested, I should have used `!=` instead of `==`. – Hari Mar 20 '14 at 14:10

score 1 · Answer 4 · answered Mar 20 '14 at 13:58

When applied to objects, the == operator returns true only when both operands are the same object.

Your method returns Double objects, so each call will produce a new object, and comparing them using == will always be false.

Use .equals(), which compares values of the Doubles:

if (!object1.getLatitude().equals(object2.getLatitude()))

Alternatively, change you methods to return double instead of Double, and your current code will work.

score 0 · Answer 5 · answered Mar 20 '14 at 14:03

Floating point comparison in may programming languages should always take the form of

if(a-b op c)

where a and b are the numbers being compared and c being the threshold and op is either > or

<

.

This is because floating point representation in binary to not map directly to what is printed.

Program considers two equal doubles to be different

5 Answers5