CODEWITHSUNDEEP
groovydefault-value
icfantv
icfantv

Reputation: 4643

Applying default groovy method parameter value when passing null

In Groovy, if I have:

def say(msg = 'Hello', name = 'world') {
  "$msg $name!"
}

And then call:

say() // Hello world!
say("hi") // Hi world!
say(null) // null world!

Why is the last one getting interpreted literally as null and not applying the default value? Doesn't this defeat the purpose of default method argument values? I do get that passing null is different from not passing anything w/r/t argument length.

My problem here is that if I now have a method that takes a collection as an argument:

def items(Set<String> items = []) {
  new HashSet<>(items)
}

This will throw a NullPointerException if I call items(null) but work fine if I just say items(). In order for this to work right, I have to change the line to be new HashSet<>(items ?: []) which, again, seems to defeat the entire purpose of having default method argument values.

What am I missing here?

Upvotes: 9

Views: 8667

Answers (1)

Will
Will

Reputation: 14529

In Groovy, default parameters generates overloaded methods. Thus, this:

def items(Set<String> items = []) {
  new HashSet<>(items)
}

Will generate these two methods (I used javap to get these values):

public java.lang.Object items(java.util.Set<java.lang.String>);
public java.lang.Object items();

So when you call items(null) you are, in fact, passing some value, and items(Set) method will be used.

You can also refer to this question about default parameters.

Upvotes: 12

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