Programmers Stack Exchange is a question and answer site for professional programmers interested in conceptual questions about software development. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have received comments from the supervisor reviewing my thesis. He asked two questions I cannot answer right now:

  1. If ISO 12207 says under "Integration verification" that it "checks that components are correctly and completely integrated into a system", how this can be verified without testing, if all testing is validation? How without testing can I know that system is integrated correctly and fully?
  2. If unit testing is validation, how does it match the ISO definiton of validation "that requirements for intended use were fulfilled" if its so low level?
share|improve this question
up vote 1 down vote accepted
  1. There's a reason ISO standards have the reputation of being impractical beaurocratic nonsense. Theoretically, it's possible to check that the artifacts that represent the components are all present, complete and configured the way they should be, by inspecting directories, file contents, perhaps comparing checksums, all without actually running the system. In practice, that's a lot of work and does very little to prove that the system actually works. The closest thing that is done in practice is to check that the automated build script has not produced an error. Perhaps the solution to the puzzle is that ISO 12207 was written to cover integration processes that are not automated (in which case you better have a reason why it cannot be automated, otherwise automate that damn build already instead of wasting time on ISO standards).
  2. Requirements can be very low level, e.g. business rules.
share|improve this answer
I agree in general. In an application with dependency injection or plug-in architecture, tests like this become less trivial and more useful. – neontapir Nov 13 '12 at 20:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.