If you were pair programming, your pair could always create a new failing test with the current implementation.
But I'm not pair programming. And you can't always create a new failing test because int is a finite type. There are only about 4 billion cases to handle.
Which might take a while to type up manually, but that's why we have meta-programming: Code that generates code. (In C++ you could even use templates, but you might run into compiler recursion limits.)
More to the point, the risk with TDD is that all development is driven by failing test cases, so a naive approach will end up "overfitting", producing exactly the code required to make a particular set of tests pass and nothing more. "It can't pass all test cases"? It doesn't have to. For TDD, it only needs to pass the tests that have actually been written. You can't test all combinations of all inputs.
(Also, if you changed this function to use modulus, it would handle more cases than before, which is a change in behavior. You're not supposed to do that when refactoring; refactoring should preserve semantics.)
Both of those declarations look weird to me. In Haskell it would be:
a :: Stringbob :: (String, Int, Double) -> [String]bob (a, b, c) = ...
... except that makes bob a function taking a tuple and it's much more idiomatic to curry it instead:
bob :: String -> Int -> Double -> [String]bob a b c = ...-- syntactic sugar for:-- bob = \a -> \b -> \c -> ...
The [T] syntax also has a prefix form [] T, so [String] could also be written [] String.
OCaml makes the opposite choice. In OCaml, a list of strings would be written string list, and a set of lists of strings would be string list set, a list of lists of integers int list list, etc.
Because let x: y is syntactically unambiguous, but you need to know that y names a type in order to correctly parse y x. (Or at least that's the case in C where a(b) may be a variable declaration or a function call depending on what typedefs are in scope.)
The same is true of std::endl. std::endl is simply defined as << '\n' << std::flush; nothing more, nothing less. In all cases where endl gives you a "properly translated" newline, so does \n.
@tetris11 @Aedis More like:
(If all the line breaks are gone from this code snippet, blame Lemmy. It looks fine here.)