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No. Haskell made me decide that operators should not be definable by users. It is truly a miserable affair to read terse point free code riddled with operators
by knome 3y ago
No. Haskell made me decide that operators should not be definable by users. It is truly a miserable affair to read terse point free code riddled with operators from a half dozen libraries ( or one overly clever library ).
Cleverness is a register that is easy to overflow, and too many don't have to good taste to avoid doing so.
Give me names every time.
- hollerith 3y agoDo you consider them as bad as user-definable Lisp or Scheme macros?
- j16sdiz 3y agoIt depends on what you tries to do. Doing linear algebra without any operator overloading is just unreadable
- layer8 3y agoI think GP meant users should not be able to define new operators (as is possible and fairly common in Haskell), but may still be allowed to overload the existing operators.
- wwalexander 3y agoThat’s a good point. I suppose I was imagining something like internal-only operators that could be defined within the scope of a single module but not exported; though I suppose with any sufficiently large module with enough orthogonal pieces you’d still run into arcane and overly-terse operators.