Why doesn't my equality comparison using = (a single equals) work correctly in Java?

Question

I have a syntax error in the following line. However I can't understand what is the reason of this error.

if (address1.compareTo(address2) = 1)
        System.out.println(address1 + " is greater than " + address2);

What I want to achieve is printing proper message if and only if compareTo returns 1.

i guess you need to look out this: https://docs.oracle.com/javase/tutorial/java/nutsandbolts/op2.html — Imran, Sep 08 '16 at 07:51
[Different language, but exactly the same problem.](http://stackoverflow.com/questions/38268329/why-doesnt-my-equality-comparison-using-a-single-equals-work-correctly) — T.J. Crowder, Sep 08 '16 at 07:52
`if (address1.compareTo(address2) > 0)` fits better the Javadoc of Comparable::compareTo — Eran, Sep 08 '16 at 07:52
Why do you even believe that this is the correct way to compare primitive values? ***Every*** beginners book covers that topic. — Tom, Sep 09 '16 at 12:14

score 4 · Accepted Answer · edited Feb 17 '17 at 16:12

4

You should compare (==) instead of assigning (=). It can be very dangerous! To avoid such situations you can use Yoda notation so instead of comparing

address1.compareTo(address2) == 1

You can compare:

1 == address1.compareTo(address2)

In case of missing =, there will be comparation error.

In your case, it would be better to compare:

address1.compareTo(address2) > 0

edited Feb 17 '17 at 16:12

T.J. Crowder

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answered Sep 08 '16 at 07:55

xenteros

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There's never any need for Yoda notation in Java. If you're dealing with a boolean, you don't use `==` at all except in a **very** small set of use cases. If you're dealing with a non-boolean, the compiler will refuse to compile `if (foo = 42)` because you can't use an `int` (etc.) as a condition. – T.J. Crowder Feb 17 '17 at 16:14

Why doesn't my equality comparison using = (a single equals) work correctly in Java?

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