3 ms·

Even in Haskell you can fairly easily get non-reachable code: if a == a then ... else -- unreachable Combine this with the unfortunate habit (

Even in Haskell you can fairly easily get non-reachable code: if a == a then ... else -- unreachable Combine this with the unfortunate habit (which I'm also guilty of) of allowing variables to be called a' and you'll an easy-to-miss bug.

WolfeReader 13y ago
Given that types may define their own instances of Eq, the else clause might be reachable.
- mercurial 13y ago
  And the content of the first block may be unreachable all the time, depending on the implementation of Eq. But you probably have a number of issues to worry about if your implementation of Eq doesn't work for the identity case.
  - loqi 13y ago
    let nan = 0.0/0.0 in nan == nan