Haskell相当于C ++类型的对

Question

I'm think of converting following C++ code into Haskell

#include <utility>

template<typename Pair>
struct inv_pair
{
    typedef std::pair<
        typename Pair::second_type, 
        typename Pair::first_type
    > type;
};

The inv_pair basically invert pair's first_type and second_type. It can be used as follows

typedef std::pair<int, std::string> pair_t1;    
typedef inv_pair<pair_t1>::type inv_par_t1; 
// of type std::pair<std::string, int>

Haskell

data Pair' = Pair' Int String
-- then?

Maybe it's not an useful pattern. Still curious and willing to learn.

Answer 1

In Data.Tuple there's already a function called swap which does what you need. It's type is:

swap :: (a, b) -> (b, a)

As an example the following:

import Data.Tuple

tuple :: (Int, String)
tuple = (1, "OK")

main = putStr $ (fst . swap) tuple

Live demo

will print OK .

---

On a side note, the data constructor for pairs is (,) which can be called (thanks to syntactic sugar) as:

(a, b)

instead of:

(,) a b

So you could also flip the data constructor for pairs. For example:

flip (,)

will produce a data constructor with reversed arguments. So that:

reversedTuple :: b -> a -> (a, b)
reversedTuple = flip (,)

main = putStr $ fst $ reversedTuple "First" "Second"

Live demo

will print Second .

Answer 2

Haskell has built-in tuples. For example (3, "foo") has type (Int, String) .

Creating the "inverted" pair is easy. For example the following function swaps the two entries of a pair:

swap (x, y) = (y, x)

and its type can be written down as

swap :: (a, b) -> (b, a)

Answer 3

A possible Haskell translation of the following C++ type hackery

template<typename Pair>
struct inv_pair
{
    typedef std::pair<
        typename Pair::second_type, 
        typename Pair::first_type
    > type;
};

typedef std::pair<int, std::string> pair_t1;
typedef inv_pair<pair_t1>::type inv_pair_t1; 

could be the following

{-# LANGUAGE TypeFamilies #-}
type family Swap t 
type instance Swap (a,b) = (b,a)

type Pair_t1     = (Int, String)
type Inv_Pair_t1 = Swap Pair_t1

I am not so sure this would be useful in practice, though.