How to curry functions in Haskell
0
I have a function multThree for multiplying 3 numbers which works with currying. However, when I tried extending this to multiplying four numbers using the same structure it doesn't work. Why is this and how could it be fixed?
multThree :: Num a => a -> (a -> (a -> a))
multThree x = (*) . (*) x
multFour :: Num a => a -> (a -> (a -> (a -> a)))
multFour x = (*) . (*) . (*) x
Error given:
• Occurs check: cannot construct the infinite type: a ~ a -> a
Expected type: a -> a -> a -> a
Actual type: a -> (a -> a) -> a -> a
• In the expression: (*) . (*) . (*) x
In an equation for ‘multFour’: multFour x = (*) . (*) . (*) x
• Relevant bindings include
x :: a (bound at test2.hs:19:10)
multFour :: a -> a -> a -> a -> a
haskell currying
asked Jan 2 at 12:08
KhamKham
85
add a comment |
0
I have a function multThree for multiplying 3 numbers which works with currying. However, when I tried extending this to multiplying four numbers using the same structure it doesn't work. Why is this and how could it be fixed?
multThree :: Num a => a -> (a -> (a -> a))
multThree x = (*) . (*) x
multFour :: Num a => a -> (a -> (a -> (a -> a)))
multFour x = (*) . (*) . (*) x
Error given:
• Occurs check: cannot construct the infinite type: a ~ a -> a
Expected type: a -> a -> a -> a
Actual type: a -> (a -> a) -> a -> a
• In the expression: (*) . (*) . (*) x
In an equation for ‘multFour’: multFour x = (*) . (*) . (*) x
• Relevant bindings include
x :: a (bound at test2.hs:19:10)
multFour :: a -> a -> a -> a -> a
haskell currying
asked Jan 2 at 12:08
KhamKham
85
4
This isn't about currying (multThreeis already curried, as its type isa -> a -> a -> a, not(a, a, a) -> a); it's about composing higher-order functions.
– chepner
Jan 2 at 12:56
The hideous point-free form ofmultThreewould bemultThree = ((*) .) . (*), which (to be honest), is barely worth understanding instead of simply writing the non-point-free version. (multFour = ((((*) .) . (*)) .) . (*)is even worse.)
– chepner
Jan 2 at 18:44
If you really want a point-free version, you could usemultThree = fmap (*) . (*)andmultFour = fmap (fmap (*)) . fmap (*) . (*)=fmap (fmap (*) . (*)) . (*)=fmap multThree . (*). Thefmapis the same as(.)but in my opinion makes it a little clearer where you’re “skipping over” arguments. Another way to build a point-free solution is e.g.multThree = curry ((*) . uncurry (*)): you’re pairing up two arguments in a tuple, multiplying them together, then multiplying the result by the other argument.
– Jon Purdy
Jan 5 at 0:00
add a comment |
0
0
0
I have a function multThree for multiplying 3 numbers which works with currying. However, when I tried extending this to multiplying four numbers using the same structure it doesn't work. Why is this and how could it be fixed?
multThree :: Num a => a -> (a -> (a -> a))
multThree x = (*) . (*) x
multFour :: Num a => a -> (a -> (a -> (a -> a)))
multFour x = (*) . (*) . (*) x
Error given:
• Occurs check: cannot construct the infinite type: a ~ a -> a
Expected type: a -> a -> a -> a
Actual type: a -> (a -> a) -> a -> a
• In the expression: (*) . (*) . (*) x
In an equation for ‘multFour’: multFour x = (*) . (*) . (*) x
• Relevant bindings include
x :: a (bound at test2.hs:19:10)
multFour :: a -> a -> a -> a -> a
haskell currying
asked Jan 2 at 12:08
KhamKham
85
I have a function multThree for multiplying 3 numbers which works with currying. However, when I tried extending this to multiplying four numbers using the same structure it doesn't work. Why is this and how could it be fixed?
multThree :: Num a => a -> (a -> (a -> a))
multThree x = (*) . (*) x
multFour :: Num a => a -> (a -> (a -> (a -> a)))
multFour x = (*) . (*) . (*) x
Error given:
• Occurs check: cannot construct the infinite type: a ~ a -> a
Expected type: a -> a -> a -> a
Actual type: a -> (a -> a) -> a -> a
• In the expression: (*) . (*) . (*) x
In an equation for ‘multFour’: multFour x = (*) . (*) . (*) x
• Relevant bindings include
x :: a (bound at test2.hs:19:10)
multFour :: a -> a -> a -> a -> a
haskell currying
haskell currying
asked Jan 2 at 12:08
KhamKham
85
asked Jan 2 at 12:08
KhamKham
85
asked Jan 2 at 12:08
KhamKham
85
asked Jan 2 at 12:08
KhamKham
85
asked Jan 2 at 12:08
KhamKham
85
85
4
This isn't about currying (multThreeis already curried, as its type isa -> a -> a -> a, not(a, a, a) -> a); it's about composing higher-order functions.
– chepner
Jan 2 at 12:56
The hideous point-free form ofmultThreewould bemultThree = ((*) .) . (*), which (to be honest), is barely worth understanding instead of simply writing the non-point-free version. (multFour = ((((*) .) . (*)) .) . (*)is even worse.)
– chepner
Jan 2 at 18:44
If you really want a point-free version, you could usemultThree = fmap (*) . (*)andmultFour = fmap (fmap (*)) . fmap (*) . (*)=fmap (fmap (*) . (*)) . (*)=fmap multThree . (*). Thefmapis the same as(.)but in my opinion makes it a little clearer where you’re “skipping over” arguments. Another way to build a point-free solution is e.g.multThree = curry ((*) . uncurry (*)): you’re pairing up two arguments in a tuple, multiplying them together, then multiplying the result by the other argument.
– Jon Purdy
Jan 5 at 0:00
add a comment |
4
This isn't about currying (multThreeis already curried, as its type isa -> a -> a -> a, not(a, a, a) -> a); it's about composing higher-order functions.
– chepner
Jan 2 at 12:56
The hideous point-free form ofmultThreewould bemultThree = ((*) .) . (*), which (to be honest), is barely worth understanding instead of simply writing the non-point-free version. (multFour = ((((*) .) . (*)) .) . (*)is even worse.)
– chepner
Jan 2 at 18:44
If you really want a point-free version, you could usemultThree = fmap (*) . (*)andmultFour = fmap (fmap (*)) . fmap (*) . (*)=fmap (fmap (*) . (*)) . (*)=fmap multThree . (*). Thefmapis the same as(.)but in my opinion makes it a little clearer where you’re “skipping over” arguments. Another way to build a point-free solution is e.g.multThree = curry ((*) . uncurry (*)): you’re pairing up two arguments in a tuple, multiplying them together, then multiplying the result by the other argument.
– Jon Purdy
Jan 5 at 0:00
4
4
This isn't about currying (
multThree is already curried, as its type is a -> a -> a -> a, not (a, a, a) -> a); it's about composing higher-order functions.– chepner
Jan 2 at 12:56
This isn't about currying (
multThree is already curried, as its type is a -> a -> a -> a, not (a, a, a) -> a); it's about composing higher-order functions.– chepner
Jan 2 at 12:56
The hideous point-free form of
multThree would be multThree = ((*) .) . (*), which (to be honest), is barely worth understanding instead of simply writing the non-point-free version. (multFour = ((((*) .) . (*)) .) . (*) is even worse.)– chepner
Jan 2 at 18:44
The hideous point-free form of
multThree would be multThree = ((*) .) . (*), which (to be honest), is barely worth understanding instead of simply writing the non-point-free version. (multFour = ((((*) .) . (*)) .) . (*) is even worse.)– chepner
Jan 2 at 18:44
If you really want a point-free version, you could use
multThree = fmap (*) . (*) and multFour = fmap (fmap (*)) . fmap (*) . (*) = fmap (fmap (*) . (*)) . (*) = fmap multThree . (*). The fmap is the same as (.) but in my opinion makes it a little clearer where you’re “skipping over” arguments. Another way to build a point-free solution is e.g. multThree = curry ((*) . uncurry (*)): you’re pairing up two arguments in a tuple, multiplying them together, then multiplying the result by the other argument.– Jon Purdy
Jan 5 at 0:00
If you really want a point-free version, you could use
multThree = fmap (*) . (*) and multFour = fmap (fmap (*)) . fmap (*) . (*) = fmap (fmap (*) . (*)) . (*) = fmap multThree . (*). The fmap is the same as (.) but in my opinion makes it a little clearer where you’re “skipping over” arguments. Another way to build a point-free solution is e.g. multThree = curry ((*) . uncurry (*)): you’re pairing up two arguments in a tuple, multiplying them together, then multiplying the result by the other argument.– Jon Purdy
Jan 5 at 0:00
add a comment |
1 Answer
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4
Let’s write it out without (.):
multFour x = (*) . (*) . (*) x
= (*) . (y -> (y*)) . (x*)
= (w -> (w*)) . (z -> ((x*z)*))
= (w -> (w*)) . (z v -> x*z*v)
= z -> u -> (v -> x*z*v) * u
And so we see that we are trying to multiply a function by a number.
The key error is this:
multFour x = (*) . multThree x
And the types are:
(*) :: Num a => a -> (a -> a)
multThree x :: Num b => b -> (b -> b)
x :: b
(.) :: (y -> z) -> (x -> y) -> (x -> z)
So the types unify as:
a = y
z = (a -> a)
b = x
y = b -> b
multFour :: Num b => b -> x -> z
multFour :: (Num b, Num (b -> b)) => b -> b -> (b -> b) -> (b -> b)
Which is not the type you want it to be.
To fix your code, I recommend:
multFour a b c d = a * b * c * d
This is much more readable.
answered Jan 2 at 12:31
Dan RobertsonDan Robertson
3,313716
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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oldest
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oldest
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4
Let’s write it out without (.):
multFour x = (*) . (*) . (*) x
= (*) . (y -> (y*)) . (x*)
= (w -> (w*)) . (z -> ((x*z)*))
= (w -> (w*)) . (z v -> x*z*v)
= z -> u -> (v -> x*z*v) * u
And so we see that we are trying to multiply a function by a number.
The key error is this:
multFour x = (*) . multThree x
And the types are:
(*) :: Num a => a -> (a -> a)
multThree x :: Num b => b -> (b -> b)
x :: b
(.) :: (y -> z) -> (x -> y) -> (x -> z)
So the types unify as:
a = y
z = (a -> a)
b = x
y = b -> b
multFour :: Num b => b -> x -> z
multFour :: (Num b, Num (b -> b)) => b -> b -> (b -> b) -> (b -> b)
Which is not the type you want it to be.
To fix your code, I recommend:
multFour a b c d = a * b * c * d
This is much more readable.
answered Jan 2 at 12:31
Dan RobertsonDan Robertson
3,313716
add a comment |
4
Let’s write it out without (.):
multFour x = (*) . (*) . (*) x
= (*) . (y -> (y*)) . (x*)
= (w -> (w*)) . (z -> ((x*z)*))
= (w -> (w*)) . (z v -> x*z*v)
= z -> u -> (v -> x*z*v) * u
And so we see that we are trying to multiply a function by a number.
The key error is this:
multFour x = (*) . multThree x
And the types are:
(*) :: Num a => a -> (a -> a)
multThree x :: Num b => b -> (b -> b)
x :: b
(.) :: (y -> z) -> (x -> y) -> (x -> z)
So the types unify as:
a = y
z = (a -> a)
b = x
y = b -> b
multFour :: Num b => b -> x -> z
multFour :: (Num b, Num (b -> b)) => b -> b -> (b -> b) -> (b -> b)
Which is not the type you want it to be.
To fix your code, I recommend:
multFour a b c d = a * b * c * d
This is much more readable.
answered Jan 2 at 12:31
Dan RobertsonDan Robertson
3,313716
add a comment |
4
4
4
Let’s write it out without (.):
multFour x = (*) . (*) . (*) x
= (*) . (y -> (y*)) . (x*)
= (w -> (w*)) . (z -> ((x*z)*))
= (w -> (w*)) . (z v -> x*z*v)
= z -> u -> (v -> x*z*v) * u
And so we see that we are trying to multiply a function by a number.
The key error is this:
multFour x = (*) . multThree x
And the types are:
(*) :: Num a => a -> (a -> a)
multThree x :: Num b => b -> (b -> b)
x :: b
(.) :: (y -> z) -> (x -> y) -> (x -> z)
So the types unify as:
a = y
z = (a -> a)
b = x
y = b -> b
multFour :: Num b => b -> x -> z
multFour :: (Num b, Num (b -> b)) => b -> b -> (b -> b) -> (b -> b)
Which is not the type you want it to be.
To fix your code, I recommend:
multFour a b c d = a * b * c * d
This is much more readable.
answered Jan 2 at 12:31
Dan RobertsonDan Robertson
3,313716
Let’s write it out without (.):
multFour x = (*) . (*) . (*) x
= (*) . (y -> (y*)) . (x*)
= (w -> (w*)) . (z -> ((x*z)*))
= (w -> (w*)) . (z v -> x*z*v)
= z -> u -> (v -> x*z*v) * u
And so we see that we are trying to multiply a function by a number.
The key error is this:
multFour x = (*) . multThree x
And the types are:
(*) :: Num a => a -> (a -> a)
multThree x :: Num b => b -> (b -> b)
x :: b
(.) :: (y -> z) -> (x -> y) -> (x -> z)
So the types unify as:
a = y
z = (a -> a)
b = x
y = b -> b
multFour :: Num b => b -> x -> z
multFour :: (Num b, Num (b -> b)) => b -> b -> (b -> b) -> (b -> b)
Which is not the type you want it to be.
To fix your code, I recommend:
multFour a b c d = a * b * c * d
This is much more readable.
answered Jan 2 at 12:31
Dan RobertsonDan Robertson
3,313716
answered Jan 2 at 12:31
Dan RobertsonDan Robertson
3,313716
answered Jan 2 at 12:31
Dan RobertsonDan Robertson
3,313716
answered Jan 2 at 12:31
Dan RobertsonDan Robertson
3,313716
3,313716
add a comment |
add a comment |
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4
This isn't about currying (
multThreeis already curried, as its type isa -> a -> a -> a, not(a, a, a) -> a); it's about composing higher-order functions.– chepner
Jan 2 at 12:56
The hideous point-free form of
multThreewould bemultThree = ((*) .) . (*), which (to be honest), is barely worth understanding instead of simply writing the non-point-free version. (multFour = ((((*) .) . (*)) .) . (*)is even worse.)– chepner
Jan 2 at 18:44
If you really want a point-free version, you could use
multThree = fmap (*) . (*)andmultFour = fmap (fmap (*)) . fmap (*) . (*)=fmap (fmap (*) . (*)) . (*)=fmap multThree . (*). Thefmapis the same as(.)but in my opinion makes it a little clearer where you’re “skipping over” arguments. Another way to build a point-free solution is e.g.multThree = curry ((*) . uncurry (*)): you’re pairing up two arguments in a tuple, multiplying them together, then multiplying the result by the other argument.– Jon Purdy
Jan 5 at 0:00