7 ms·

Yes. This is the only way you can justify that you know better than the compiler.

winstonewert 9y ago
So, if you wrote the following code: double sum(double[] numbers) { double total = 0.0; for (int index = 0; index < numbers.length; index++) { total += numbers[index]; } return total } You would write a formal proof on paper that the index won't go out of bounds?
- catnaroek 9y ago
  Theorem: Whenever `numbers[index]` is used, `0 <= index < numbers.length`. Proof: Use the loop invariant `0 <= index <= numbers.length`. The invariant holds before the loop begins because `index == 0` (established by you) and `numbers.length >= 0` (established by the implementor of the array type) imply it. Inductively assume that prior iterations of the loop preserved the invariant and didn't perform any invalid indexing operations. Now, every time the loop's condition is evaluated, there are two possibilities: either `0 <= index == numbers.length` or `0 <= index < numbers.length`. In the former case, the loop terminates and no further indexing operations are used, establishing the claim. In the latter case, `numbers[index]` is a valid indexing operation, and the subsequent `index++` operation does not destroy the invariant, because, given the precondition `0 <= index < numbers.length`, the operation `index++` establishes the postcondition `0 < index <= numbers.length`, which implies the invariant. --- Note, however, that in this case it would have been more efficient to use an iterator, because the iterator's implementation already packages this proof in a reusable way, so that you don't need to reprove it every time you write a loop similar to this.
  - winstonewert 9y ago
    Well, I've never seen anyone write a proof for a function so trivial. Instead, most people will simply deem the proof to be sufficiently obvious they don't need to write one out. I've only ever seen someone offer something in terms of a proof for more complex situation where its decidely non-intuitive that array indexes will not go out of bounds.
    - catnaroek 9y ago
      You can only deem it obvious if you can actually prove it. --- @winstonewert > True, but the key word is "can". I could write a proof that my dynamically typed programs are correct If you could actually write the proof, then you would not want the dynamic checks, as all they offer is protection against errors that you can prove you have not made. > or that my functions do not index out of bounds, but I don't. Are you actually sure you can? --- @AnimalMuppet > By a certain point, you've seen enough of them that you don't have to prove it, you can just see it. If all your loops are so similar to each other that you can “just see” that an iteration pattern is correct, you should consider writing an iterator library, so that others can benefit from your hard-earned wisdom without going through the pain themselves. Or else, if you are implying that you could be given an arbitrary loop and “just see” that it is correct, I am afraid you are wrong. > But of course, everybody thinks they're at that point well before they actually are... I have never even entertained the possibility.
      - AnimalMuppet 9y ago
        I disagree. I've been programming professionally for 33 years. I may have written one wrongly-indexed "for" loop in the last 10 years. By a certain point, you've seen enough of them that you don't have to prove it, you can just see it. But of course, everybody thinks they're at that point well before they actually are...
      - winstonewert 9y ago
        True, but the key word is "can". I could write a proof that my dynamically typed programs are correct or that my functions do not index out of bounds, but I don't. Nor has any programmer I've ever worked with. Nor has any open source project I've ever seen provide such proofs. People do not write these imaginary proofs, because it would be a waste of time.
      - winstonewert 9y ago
        > If you could actually write the proof, then you would not want the dynamic checks, as all they offer is protection against errors that you can prove you have not made. Even if I write a formal proof, I'm not guaranteed that my program won't index out of bounds. Formals proofs have errors in them all the time. I want dynamic checks to catch me in those cases. Further to the point, let me reiterate: programmers do not (in almost all cases) write these proofs you are talking about.

So, if you wrote the following code: double sum(double[] numbers) { double total = 0.0; for (int index = 0; index < numbers.length; index++) { total += numbers[index]; } return total } You would write a formal proof on paper that the index won't go out of bounds?

catnaroek 9y ago
Theorem: Whenever `numbers[index]` is used, `0 <= index < numbers.length`. Proof: Use the loop invariant `0 <= index <= numbers.length`. The invariant holds before the loop begins because `index == 0` (established by you) and `numbers.length >= 0` (established by the implementor of the array type) imply it. Inductively assume that prior iterations of the loop preserved the invariant and didn't perform any invalid indexing operations. Now, every time the loop's condition is evaluated, there are two possibilities: either `0 <= index == numbers.length` or `0 <= index < numbers.length`. In the former case, the loop terminates and no further indexing operations are used, establishing the claim. In the latter case, `numbers[index]` is a valid indexing operation, and the subsequent `index++` operation does not destroy the invariant, because, given the precondition `0 <= index < numbers.length`, the operation `index++` establishes the postcondition `0 < index <= numbers.length`, which implies the invariant. --- Note, however, that in this case it would have been more efficient to use an iterator, because the iterator's implementation already packages this proof in a reusable way, so that you don't need to reprove it every time you write a loop similar to this.
- winstonewert 9y ago
  Well, I've never seen anyone write a proof for a function so trivial. Instead, most people will simply deem the proof to be sufficiently obvious they don't need to write one out. I've only ever seen someone offer something in terms of a proof for more complex situation where its decidely non-intuitive that array indexes will not go out of bounds.
  - catnaroek 9y ago
    You can only deem it obvious if you can actually prove it. --- @winstonewert > True, but the key word is "can". I could write a proof that my dynamically typed programs are correct If you could actually write the proof, then you would not want the dynamic checks, as all they offer is protection against errors that you can prove you have not made. > or that my functions do not index out of bounds, but I don't. Are you actually sure you can? --- @AnimalMuppet > By a certain point, you've seen enough of them that you don't have to prove it, you can just see it. If all your loops are so similar to each other that you can “just see” that an iteration pattern is correct, you should consider writing an iterator library, so that others can benefit from your hard-earned wisdom without going through the pain themselves. Or else, if you are implying that you could be given an arbitrary loop and “just see” that it is correct, I am afraid you are wrong. > But of course, everybody thinks they're at that point well before they actually are... I have never even entertained the possibility.
    - AnimalMuppet 9y ago
      I disagree. I've been programming professionally for 33 years. I may have written one wrongly-indexed "for" loop in the last 10 years. By a certain point, you've seen enough of them that you don't have to prove it, you can just see it. But of course, everybody thinks they're at that point well before they actually are...
    - winstonewert 9y ago
      True, but the key word is "can". I could write a proof that my dynamically typed programs are correct or that my functions do not index out of bounds, but I don't. Nor has any programmer I've ever worked with. Nor has any open source project I've ever seen provide such proofs. People do not write these imaginary proofs, because it would be a waste of time.
    - winstonewert 9y ago
      > If you could actually write the proof, then you would not want the dynamic checks, as all they offer is protection against errors that you can prove you have not made. Even if I write a formal proof, I'm not guaranteed that my program won't index out of bounds. Formals proofs have errors in them all the time. I want dynamic checks to catch me in those cases. Further to the point, let me reiterate: programmers do not (in almost all cases) write these proofs you are talking about.