Even in a language like Groovy that is normally so clean and intuitive, there are traps for the unwary. I fell into one again today (in front of a room full of students), and I think it’s high time I documented it, at least so I’ll remember it for next time.

I’m teaching a Groovy / Grails class this week, and one of the problems I posed to the students was to write an `isPrime`

method that worked for integers less than, say, 1000. The easiest brute force way to solve the problem is to divide the given number by every integer from 2 up to the square root of the number, rounded up.

Here’s a first try:

[sourcecode language=”groovy”]

// NOTE: THIS DOESN’T WORK

boolean isPrime1(int x) {

if (x == 2) return true

int limit = Math.sqrt(x) + 1

(2..limit).each { n ->

if (x % n == 0) return false // Danger, Will Robinson!

}

return true

}

assert (2..20).findAll { isPrime1(it) } == (2..20) // Wait, what??

[/sourcecode]

In other words, the method returns `true`

for every number, whether it’s prime or not.

The problem is the `return`

statement inside the `each`

loop. The argument to the `each`

method is a closure, and returning from a closure returns only from the closure, not the whole method.

I should emphasize that, so maybe next time I’ll remember it:

A

`return`

inside a closure returnsonly from the closure, not from the method that called it.

It’s actually not that hard to understand, when I think it through. Calling the closure is like calling another method. When I use the `return`

keyword inside the closure, it’s like returning from that method, so it only returns from there and not from the `each`

method itself.

So, how do I fix this? One way is to define a boolean variable outside the closure and then set it inside, but that leads to a minor problem as well.

[sourcecode language=”groovy”]

boolean isPrime2(int x) {

if (x == 2) return true

boolean result = true // local var to set in closure

int limit = Math.sqrt(x) + 1

(2..limit).each { n ->

if (x % n == 0) {

result = false

// don’t you wish you could break here?

}

}

return result

}

assert (2..20).findAll { isPrime2(it) } ==

[2, 3, 5, 7, 11, 13, 17, 19] // works, but no break

[/sourcecode]

I defined a boolean variable called `result`

and initialized it to `true`

. Then, inside the closure, if a number is not prime, I set the variable to `false`

, and returned the variable at the end.

As the comment shows, however, I can’t break out of the loop. You can only use `break`

inside loops, and some reason (I don’t know why), `each`

loops don’t qualify. I’m forced to follow the loop to the end.

There is a way to solve that problem, too. I just need to replace the `each`

loop with Groovy’s for-in loop.

[sourcecode language=”groovy”]

boolean isPrime3(int x) {

if (x == 2) return true

boolean result = true

int limit = Math.sqrt(x) + 1

for (n in 2..limit) {

if (x % n == 0) {

result = false

break // yay!

}

}

return result

}

assert (2..20).findAll { isPrime3(it)} ==

[2, 3, 5, 7, 11, 13, 17, 19] // works

[/sourcecode]

Now it works. This is also one reason to sometimes favor the for-in loop over the each iterator, but usually they’re interchangeable.

As usual, I couldn’t leave well enough alone. A common idiom in the Groovy JDK is to use metaprogramming to replace a static method in one class with an instance method in another. For example, `java.lang.Math`

has the static `abs`

method, but the Groovy JDK adds `abs`

as an instance method to `java.lang.Number`

. It’s not hard to do the same thing here:

[sourcecode language=”groovy”]

Number.metaClass.isPrime = { ->

Integer x = delegate as Integer

if (x == 2) return true

boolean result = true

int limit = Math.sqrt(x) + 1

for (n in 2..limit) {

if (x % n == 0) {

result = false

break

}

}

return result

}

assert (2..20).findAll { it.isPrime() } == [2, 3, 5, 7, 11, 13, 17, 19][/sourcecode]

The `delegate`

of the closure is the number the `isPrime`

method was called on, so if I assign it to `x`

I can copy and paste my implementation from above.

The truly fun part was that I assigned this problem to the students, and then decided to live code the implementation in front of them. Of course I went for the `each`

implementation, falling right into the trap. Since I’ve seen this problem many times, I recognized it pretty quickly and therefore wrote the other two implementations after that. I’d like to believe that the students got some value out of seeing me mess it up and then fix it. If nothing else, they learned the value of test cases. Without those `assert`

statements, I probably wouldn’t have noticed the error in the first place.

Baruch Sadogursky (@jbaruch on Twitter) is doing a presentation on Groovy and Grails Puzzlers at Gr8conf in Minneapolis at the end of July, and he’s always interested in issues like this, so I made sure to send it along to him. If you know of any others, I’m sure he’d appreciate them as well.

## 9 replies on “The Closure Of No Return”

you can actually keep the open block, if you replace the usage of “each” with find. You have to reverse the booleans a bit, but the following works:

boolean isPrime1(int x) {

if (x == 2) return true

int limit = Math.sqrt(x) + 1

return (2..limit).find { x % it == 0 } == null

}

You could replace return with throw and wrap the function in try-catch.

A good way to challenge students is to define the general, lazy prime stream (https://gist.github.com/mperry/10878344):

@Grab(“org.functionaljava:functionaljava:3.1”)

import fj.F

import fj.data.Stream

import groovy.transform.Memoized

@Memoized

Stream primes() {

Stream.cons(2, { ->

Stream.range(3).filter { Integer i ->

primes().takeWhile({ Integer j -> j * j i % k != 0 }

}

})

}

println primes().takeWhile { it <= 20 }

Since you changed the each-loop to a for-loop, you no longer need the ‘break’, you can use ‘return’ as you did first. Since the for-loop isn’t a closure, a ‘return’ will return the method just as expected.

Ah, of course. That didn’t occur to me, and it probably should have. 🙂 Thanks for the info.

Have you tried using the any function rather than each? I am making the assumption it breaks out of checking each value when it gets a true value.

boolean isPrime3(int x) {

if (x == 2) return true

int limit = Math.sqrt(x) + 1

!(2..limit).any { n ->

x % n == 0

}

}

assert (2..20).findAll { isPrime3(it)} == [2, 3, 5, 7, 11, 13, 17, 19] // works

My go at making it even more Groovy:

boolean isPrime(int x) {

int limit = Math.sqrt(x) + 1

x == 2 || (2..limit).every { n ->

x % n

}

}

i like the ‘any’ solution .. here is a slight variation for 3..n primes and trying to keep it in a single line 🙂 http://vasya10.wordpress.com/2012/01/10/groovy-olc-1/

[…] blogged about that before, and even submitted a variation to Baruch Sadogursky as a Groovy Puzzler, but this is where I first […]