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I'm super new to programming and I'm wondering how to tackle one of the most basic problems out there-- the "FizzBuzz" thing. I'm doing this in Groovy.

I have a pretty specific idea of how I want the code to be constructed, but I can't for the life of me figure out how to test something against each value in an array.

What I mean is, say, for each value in list [1,2,3,4], how would I go about checking if each value is even? I know I can select each specific point in the array but that's not what I want-- I'd like to be able to say something like "if n%2=0, label this even."

That's a very basic example, but you probably get the idea. Any help would be greatly appreciated.

Tomasz Nurkiewicz
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Ben
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2 Answers2

6

Groovy allows you to tackle this problem with a functional approach. By applying a mapping transformation, you can generate a list of pairs containing the number and whether it's even or odd.

All groovy lists have a method called collect for mapping a closure over each element. The return value is a list containing the result the closure being called on each element. For example:

 [1, 2, 3, 4].collect {
      [it, it % 2 ? 'odd' : 'even']
 }
 ===> [[1:odd], [2:even], [3:odd], [4:even]]

This results in a list of pairs (actually 2 element lists). It's pretty common to want the result to be a map instead of a list, and groovy has a specialized version of collect, called collectEntries just for this. Here's an alternative that returns a map:

[1, 2, 3, 4].collectEntries {
    [it, it % 2 ? 'odd' : 'even']
}
===> {1=odd, 2=even, 3=odd, 4=even}
ataylor
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0

I would do something like:

def list = [1, 2, 3, 4]

def map = [:]

list.each {
    if (it % 2 == 0)
        map.putAt(it, "even")
    else
        map.putAt(it, "odd")
}

println map

this would print: [1:odd, 2:even, 3:odd, 4:even], you can do whatever you want within the 'if else' statements, but i think what you're asking is how to iterate through a collection one member at a time, the 'each' closure is the primary way to do this.

awfulHack
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