Cannot create Set from String array

Question

I've been staring at the screen the last 5 minutes and can't seem to figure out what I'm doing wrong:

class Example {

    private final Set<String> values;

    public Example(String... values) {
        values = new HashSet<String>(Arrays.asList(values));
    }
}

I'm surprised why the String[] cannot be converted to List<String> to initialize the HashSet<String> with it.

I'm getting the build error:

incompatible types: java.util.HashSet<java.lang.String> cannot be converted to java.lang.String[]

What's wrong with my assignment?

The values parameter inside of your constructor is shadowing the class field values. — ajp15243
– ajp15243, Commented Dec 19, 2014 at 20:44
As @ajp15243 said. You can fix it by assigning to this.values, or by changing the name of the constructor argument. — John Bollinger
– John Bollinger, Commented Dec 19, 2014 at 20:45
Correct answers were already given so I thought I will make a comment — Abhijeet Kushe
– Abhijeet Kushe, Commented Dec 19, 2014 at 20:52
@AbhijeetKushe You're making a good point which I would definitely upvote, but it's up to you — janos
– janos, Commented Dec 19, 2014 at 20:53

Vogel612 · Accepted Answer · 2014-12-19 20:45:32Z

9

You're missing a qualification to actually access the private field. Currently you're trying to reassign the parameter passed to the constructor. Instead you should use the following code:

public Example(String... values) {
     this.values = new HashSet<String>(Arrays.asList(values));
}

This can be shortened even further by using the "Diamond Operator", which is avaliable since Java 7:

public Example(String... values) {
     this.values = new HashSet<>(Arrays.asList(values));
}

answered Dec 19, 2014 at 20:45

Vogel612

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1 Comment

janos Over a year ago

Facepalm... Thanks Vogel!

Abhijeet Kushe · Accepted Answer · 2014-12-19 21:02:48Z

3

This is how you can do it

 if(values != null)
      this.values = new HashSet<>(Arrays.asList(values));
 else
      this.values = Collections.emptySet();

Add the if(values != null) check before the assignment.Whenever you use var args you are exposing a contract which will permit your clients to create an valid Example object without any arguments.If you want to avoid that from happening then just use String[] values directly and throw an exception incase if it is null

edited Dec 19, 2014 at 21:02

answered Dec 19, 2014 at 20:55

Abhijeet Kushe

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2 Comments

janos Over a year ago

You're right that I should validate my inputs, excellent point. Actually I think throwing IllegalArgumentException might be best for a null arg.

Abhijeet Kushe Over a year ago

Yeah IllegalArgumentException would be ideal if use Sting[] values as the parameter

Edward · Accepted Answer · 2014-12-19 21:03:38Z

3

Other answers have addressed the cause, but wouldn't it be better to simply rename the parameter to avoid the shadowing?

answered Dec 19, 2014 at 21:03

Edward

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1 Comment

janos Over a year ago

Good point. I wouldn't have been in this situation in the first place.

VJ THAKUR · Accepted Answer · 2014-12-19 20:58:55Z

0

Use this.values inside a constructor in place of value because the parameter of constructor is string array and you are trying to convert a List to array.

this.values = ......

answered Dec 19, 2014 at 20:58

VJ THAKUR

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2 Comments

Vogel612 Over a year ago

I don't understand how you ended at the correct result with such a wretched premise... Your logic is... abysmal

Simon Forsberg Over a year ago

@Vogel612 perhaps the formulation of this answer is a bit confusing, but the conclusion seems to be totally correct.

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