One part of the story is that you should not repeat the algorithm in the test. The only thing you’re testing is that you can write the same code in two places. Every error in your reasoning is repeated. It’s natural to start with a known return value and create input values that lead to this result.
I’ll demonstrate the principle with a functionality I recently used: k-permutation with repetition. (k defines the size of the output values. Permutation indicates that we care about order. Repetition says that output values can be created by reusing input values.) For example the k-permutation with repetition for the input set {A B} and k = 3 is {AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB}. We know that the size of the permutation is n^k. (n is the size of the input set.)
@Test public void kPermutationWithRepetitions() { int maxSize = 6; for(String expected : toIterable(strings(1, maxSize))) { Set<Character> allowed = newHashSet(Chars.asList(expected.toCharArray())); Set<Character> additional = anySet(characters(), 0, maxSize - allowed.size()); allowed.addAll(additional); int k = expected.length(); Iterable<String> actual = kPermutationWithRepetition(allowed, STRING_ADD, k); assertEquals( Math.pow(allowed.size(), k), Iterables.size(actual), 0); assertTrue(Iterables.contains(actual, expected)); } }
The test uses Quickcheck and Guava. It starts by generating one expected result value. The interesting input value is the set of allowed characters. This set contains all characters of the result string plus some additional characters. We have to take care that the input set is small enough, otherwise the result size is too large to run the test in practice. With this in place we can call kPermutationWithRepetition that generates the actual result. The test checks that the expected value is in the result and that the size of the permutation is correct.
This example demonstrates that it can be much easier to start with an expected return value. Depending on the situation it’s appropriate to create the entire result or just aspects of it. If you can generate at least one result value, the probability of the generated values is evenly distributed over all results and you know the total result size, your test is sufficient.
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