Groovy - check if param is a function

Question

In JavaScript, this is how I check if a function parameter is a function:

function foo ( p ) {
    if ( typeof p === 'function' ) {
        p();
    }
    // ....
}

How can I do the same in Groovy?

Answer 1

Groovy makes closures a first-class citizen. Each closure extends abstract class groovy.lang.Closure<V> and in case of undefined argument type you can use instanceof to check if parameter that was passed to a method is a closure. Something like that:

def closure = {
    println "Hello!"
}

def foo(p) {
    if (p instanceof Closure) {
        p()
    }
}

foo(closure)

Running this script generates output:

Hello!

Using concrete parameter type

Groovy allows you (and it's worth doing actually) to define a type of a method parameter. Instead of checking if p was a closure, you can require that caller passes a closure. Consider following example:

def closure = {
    println "Hello!"
}

def foo2(Closure cl) {
    cl()
}

foo2(closure)
foo2("I'm not a closure")

First call will do what closure does (prints "Hello!"), but second call will throw an exception:

Hello!
Caught: groovy.lang.MissingMethodException: No signature of method: test.foo2() is applicable for argument types: (java.lang.String) values: [I'm not a closure]
Possible solutions: foo2(groovy.lang.Closure), foo(java.lang.Object), find(), find(groovy.lang.Closure), wait(), run()
groovy.lang.MissingMethodException: No signature of method: test.foo2() is applicable for argument types: (java.lang.String) values: [I'm not a closure]
Possible solutions: foo2(groovy.lang.Closure), foo(java.lang.Object), find(), find(groovy.lang.Closure), wait(), run()
    at test.run(test.groovy:18)

It's always a good practice to make your code type-safe, so you don't have to worry if a value passed as a parameter is a type you expect.