CODEWITHSUNDEEP
testinggroovybddeasyb
matcauthon
matcauthon

Reputation: 2261

EasyB testing multiple input/ouput values

How do I test a functionality with multiple inputs/expected outputs?

Here is a really simple example:

scenario "Can add two numbers", {
    given "Two numbers", {
        num1 = 2
        num2 = 3
    }

    when "I trigger add.", {
        result = add(num1,num2)
    }

    then "The result should be correct.", {
        result.shouldBe 5
    }
}

I want to test this with multiple values, say add(4,8).shouldBe 12, ....

Whats the best practice to do this? In other BDD frameworks I have seen table like structures to implement this, but cannot find something like that in EasyB. Should I create multiple scenarios to cover this (appending (1), (2) to the scenario name), or should I put the inputs and expected outputs into an array, and check this for equality? If I use the latter approach, how do I get meaningfull failures?

Upvotes: 1

Views: 343

Answers (1)

KarlM
KarlM

Reputation: 1662

Use the where/example clause http://code.google.com/p/easyb/wiki/ChangesInEasyb098

package org.easyb.where

/*
Example tests a map at the story level
 */

numberArray = [12, 8, 20, 199]

where "we are using sample data at a global level", [number:numberArray]

before "Before we start running the examples", {
  given "an initial value for counters", {
    println "initial"
    whenCount = 0
    thenCount = 0
    numberTotal = 0
  }
}

scenario "Number is #number and multiplier is #multiplier and total is #{number * multiplier}", {
  when "we multiply #number by #multiplier", {
    whenCount ++
    num = number * multiplier
  }
  then "our calculation (#num) should equal #{number * multiplier}", {
    num.shouldBeGreaterThan 0
    numberTotal += num
    thenCount ++
  }
  where "Multipliers should be", {
    multiplier = [1,2,3]
  }
}


after "should be true after running example data", {
  then "we should have set totals", {
    whenCount.shouldBe 12
    thenCount.shouldBe 12
    num = 0
    numberArray.each { n ->
      num = num + (n + (2*n) + (3*n))
    }

    num.shouldBe numberTotal
  }
}

Upvotes: 1

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