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Well I think you're conflating the simple operator syntax with broader issues of static typing. There's no reason you couldn't have Haskell-style operators in a

by chancho 18y ago

Well I think you're conflating the simple operator syntax with broader issues of static typing. There's no reason you couldn't have Haskell-style operators in a Python-like language, where the definition of each operator is given by a method of its first argument. The 'no-overloading' rule of Haskell is kind of annoying. You can import different definitions of the same function, its just that you then have to disambiguate with a prefix. Like 'List.foldl (+) 0 xs' versus 'Map.foldl (+) 0 xs'. This is made more annoying by the fact that core functions such as foldl aren't typeclass methods. AFAICT the only reason you are forced to disambiguate is because otherwise the compiler would need to use Prolog-style backtracking to typecheck your code on account of type inference. It's perfectly possible, its just that the designers didn't want to force this huge requirement on implementors. Plus a lot of people such as yourself feel that having multiple functions with the same name is a bad thing. I personally don't, as long as they have different types and the compiler is able to disambiguate which one you are referring to based on that type.

dkarl 18y ago
There's no reason you couldn't have Haskell-style operators in a Python-like language, where the definition of each operator is given by a method of its first argument. That's how Scala does it. Scala also lets you define new operators and use infix notation for methods, which gives programmers no excuse for violating conventional operator semantics. (In C++, the limited choice of operators tempts people to commit such atrocities as using + and * to denote non-commutative operations.) Operators like + and * seem to cry out for generic treatment a la Common Lisp, but I don't know how well that technique scales to large codebases. Java's single-dispatch model is limited, but it seems to scale acceptably.
jmillikin 18y ago
Python, by necessity, is more Haskell-ish. I was primarily considering Haskell as opposed to C++, where multiple functions share a name and differ by parameter types. This sort of situation is a true nightmare to debug, and I believe is responsible for some of the backlash against user-defined operators in general.
- chancho 18y ago
  I don't see how Haskell typeclass methods are any different from C++ overloaded functions in this regard. In both languages you refer to a concrete function by its signature rather than its name. Eg in Haskell you'd have "(+) :: Int -> Int -> Int" or "(+) :: Float -> Float -> Float". You have this additional way to refer to the abstract version, "Num a => (+) :: a -> a -> a" but I don't see how that helps in debugging, other than to simultaneously set breakpoints on all the concrete functions which constitute the abstract method. Also C++0x concepts bring these two frameworks closer together. On the larger issue of free-for-all overloading I mostly agree with you: it causes unnecessary confusion. I'm comfortable with it and like it in my own personal code but I could see how you would want to avoid it in larger projects.
  - SamReidHughes_ 18y ago
    C++ requires that all overloaded versions exist at compile time; Haskell allows them to exist at runtime. For example, in Haskell you can make a datatype like data Foo a = Blah a | Bar (Foo [a]) and you'll have no trouble making a function "showFoo :: Show a => Foo a -> String" using straightforward programming. If you try to do that in C++ you'll get infinite template recursion. Also, C++ templates don't restrict you to match the right type signatures, the way Haskell typeclasses do. But other than that, they are pretty much the same kind of thing and are used in the same way.
    - chancho 18y ago
      You've mixed together a lot of stuff here. your Foo data type is recursive. I can achieve something similar in C++ without infinite template recursion, but I have to use pointers to do it. (Your Haskell has pointers too, they're just hidden from you.) Also there is nothing 'runtime' about the example you've got here. I can do the same thing with a tagged-union type in C/C++ and a switch statement, which is what your Haskell compiles down to. Maybe you were thinking of existential types? http://en.wikibooks.org/wiki/Haskell/Existentially_quantified_types http://en.wikibooks.org/wiki/Haskell/Existentially_quantifie... . That's something C++ can't do. --------- Edit: nm I see what you did there with the Foo a = ... | Blah (Foo [a]). That would probably break C++.

by chancho 18y ago

dkarl 18y ago

There's no reason you couldn't have Haskell-style operators in a Python-like language, where the definition of each operator is given by a method of its first argument. That's how Scala does it. Scala also lets you define new operators and use infix notation for methods, which gives programmers no excuse for violating conventional operator semantics. (In C++, the limited choice of operators tempts people to commit such atrocities as using + and * to denote non-commutative operations.) Operators like + and * seem to cry out for generic treatment a la Common Lisp, but I don't know how well that technique scales to large codebases. Java's single-dispatch model is limited, but it seems to scale acceptably.

jmillikin 18y ago

Python, by necessity, is more Haskell-ish. I was primarily considering Haskell as opposed to C++, where multiple functions share a name and differ by parameter types. This sort of situation is a true nightmare to debug, and I believe is responsible for some of the backlash against user-defined operators in general.

chancho 18y ago
I don't see how Haskell typeclass methods are any different from C++ overloaded functions in this regard. In both languages you refer to a concrete function by its signature rather than its name. Eg in Haskell you'd have "(+) :: Int -> Int -> Int" or "(+) :: Float -> Float -> Float". You have this additional way to refer to the abstract version, "Num a => (+) :: a -> a -> a" but I don't see how that helps in debugging, other than to simultaneously set breakpoints on all the concrete functions which constitute the abstract method. Also C++0x concepts bring these two frameworks closer together. On the larger issue of free-for-all overloading I mostly agree with you: it causes unnecessary confusion. I'm comfortable with it and like it in my own personal code but I could see how you would want to avoid it in larger projects.
- SamReidHughes_ 18y ago
  C++ requires that all overloaded versions exist at compile time; Haskell allows them to exist at runtime. For example, in Haskell you can make a datatype like data Foo a = Blah a | Bar (Foo [a]) and you'll have no trouble making a function "showFoo :: Show a => Foo a -> String" using straightforward programming. If you try to do that in C++ you'll get infinite template recursion. Also, C++ templates don't restrict you to match the right type signatures, the way Haskell typeclasses do. But other than that, they are pretty much the same kind of thing and are used in the same way.
  - chancho 18y ago
    You've mixed together a lot of stuff here. your Foo data type is recursive. I can achieve something similar in C++ without infinite template recursion, but I have to use pointers to do it. (Your Haskell has pointers too, they're just hidden from you.) Also there is nothing 'runtime' about the example you've got here. I can do the same thing with a tagged-union type in C/C++ and a switch statement, which is what your Haskell compiles down to. Maybe you were thinking of existential types? http://en.wikibooks.org/wiki/Haskell/Existentially_quantified_types http://en.wikibooks.org/wiki/Haskell/Existentially_quantifie... . That's something C++ can't do. --------- Edit: nm I see what you did there with the Foo a = ... | Blah (Foo [a]). That would probably break C++.