4 ms·

It was meant to be pithy jibberish, yes; I tried to pick three characters in an operator that I was semi-confident that I hadn't seen while writing Haskell, and

• nybble41 6y ago
  It can't mutate `c` because it doesn't know what `c` is. There might not even be a value to mutate; `c` could be Void, or a phantom type parameter. But given a definition for `M c` like: data M c = M (A -> c) there would only be one reasonable implementation: a -*/ (M f) = f a If `M c` does not contain a function from `A` to `c` then the only possible result is non-termination. You weren't given a value of type `c`, and you have no way of producing one from the inputs, so you can't return such a value, and the only alternative is not returning at all. Of course the real problem here is that (at least without context) this operator name doesn't mean much. You could make the same complaint about a perfectly ordinary function named `ixfni`. The only real difference in Haskell is that symbolic operators is written infix by default, while named functions are written prefix, but it's very easy to flip that around: (-*/) a b a `ixfni` b Note that you can even define fixity and precedence for named functions in their infix forms. Unlike most other languages with user-defined operators there is remarkably little pressure to rely on symbols in situations where they would be unclear. Symbolic operators should only be defined in situations where the symbols are reasonably well-known (within some context) or "intuitively" related to existing operators.