Haskell Type Errors with division
I have the below simple program to find the average of a list. I know my error is related to the inference of types but I cannot correct it. Can I get a correction and a simple explanation for this?
average :: Float
average= uncurry (/) . sumlen
sumlen ::[Int]-> (Int,Int)
sumlen = foldl f (0,0)
where f (s,n) x = (s+x,n+1)
The error is:
• Couldn't match expected type ‘Float’
with actual type ‘[Int] -> Int’
• Probable cause: ‘(.)’ is applied to too few arguments
In the expression: uncurry (/) . sumlen
In an equation for ‘average’: average = uncurry (/) . sumlen
haskell
asked Nov 26 '18 at 10:43
GakuoGakuo
394213
add a comment |
I have the below simple program to find the average of a list. I know my error is related to the inference of types but I cannot correct it. Can I get a correction and a simple explanation for this?
average :: Float
average= uncurry (/) . sumlen
sumlen ::[Int]-> (Int,Int)
sumlen = foldl f (0,0)
where f (s,n) x = (s+x,n+1)
The error is:
• Couldn't match expected type ‘Float’
with actual type ‘[Int] -> Int’
• Probable cause: ‘(.)’ is applied to too few arguments
In the expression: uncurry (/) . sumlen
In an equation for ‘average’: average = uncurry (/) . sumlen
haskell
asked Nov 26 '18 at 10:43
GakuoGakuo
394213
The average of which list are you trying to find? Do you maybe wantaverageto be a function, not aFloat?
– sepp2k
Nov 26 '18 at 10:47
1
firstly:averageis not aFloat. It is a function[Int] -> Float. Also, you should explictly convert yourInts intoFloatwhen applying(/)
– Luis Morillo
Nov 26 '18 at 10:49
Possible duplicate of Haskell types frustrating a simple 'average' function
– Redu
Nov 26 '18 at 14:31
add a comment |
I have the below simple program to find the average of a list. I know my error is related to the inference of types but I cannot correct it. Can I get a correction and a simple explanation for this?
average :: Float
average= uncurry (/) . sumlen
sumlen ::[Int]-> (Int,Int)
sumlen = foldl f (0,0)
where f (s,n) x = (s+x,n+1)
The error is:
• Couldn't match expected type ‘Float’
with actual type ‘[Int] -> Int’
• Probable cause: ‘(.)’ is applied to too few arguments
In the expression: uncurry (/) . sumlen
In an equation for ‘average’: average = uncurry (/) . sumlen
haskell
asked Nov 26 '18 at 10:43
GakuoGakuo
394213
I have the below simple program to find the average of a list. I know my error is related to the inference of types but I cannot correct it. Can I get a correction and a simple explanation for this?
average :: Float
average= uncurry (/) . sumlen
sumlen ::[Int]-> (Int,Int)
sumlen = foldl f (0,0)
where f (s,n) x = (s+x,n+1)
The error is:
• Couldn't match expected type ‘Float’
with actual type ‘[Int] -> Int’
• Probable cause: ‘(.)’ is applied to too few arguments
In the expression: uncurry (/) . sumlen
In an equation for ‘average’: average = uncurry (/) . sumlen
haskell
haskell
asked Nov 26 '18 at 10:43
GakuoGakuo
394213
asked Nov 26 '18 at 10:43
GakuoGakuo
394213
asked Nov 26 '18 at 10:43
GakuoGakuo
394213
asked Nov 26 '18 at 10:43
GakuoGakuo
394213
asked Nov 26 '18 at 10:43
GakuoGakuo
394213
394213
The average of which list are you trying to find? Do you maybe wantaverageto be a function, not aFloat?
– sepp2k
Nov 26 '18 at 10:47
1
firstly:averageis not aFloat. It is a function[Int] -> Float. Also, you should explictly convert yourInts intoFloatwhen applying(/)
– Luis Morillo
Nov 26 '18 at 10:49
Possible duplicate of Haskell types frustrating a simple 'average' function
– Redu
Nov 26 '18 at 14:31
add a comment |
The average of which list are you trying to find? Do you maybe wantaverageto be a function, not aFloat?
– sepp2k
Nov 26 '18 at 10:47
1
firstly:averageis not aFloat. It is a function[Int] -> Float. Also, you should explictly convert yourInts intoFloatwhen applying(/)
– Luis Morillo
Nov 26 '18 at 10:49
Possible duplicate of Haskell types frustrating a simple 'average' function
– Redu
Nov 26 '18 at 14:31
The average of which list are you trying to find? Do you maybe want
average to be a function, not a Float?– sepp2k
Nov 26 '18 at 10:47
The average of which list are you trying to find? Do you maybe want
average to be a function, not a Float?– sepp2k
Nov 26 '18 at 10:47
1
1
firstly:
average is not a Float. It is a function [Int] -> Float. Also, you should explictly convert your Ints into Float when applying (/)– Luis Morillo
Nov 26 '18 at 10:49
firstly:
average is not a Float. It is a function [Int] -> Float. Also, you should explictly convert your Ints into Float when applying (/)– Luis Morillo
Nov 26 '18 at 10:49
Possible duplicate of Haskell types frustrating a simple 'average' function
– Redu
Nov 26 '18 at 14:31
Possible duplicate of Haskell types frustrating a simple 'average' function
– Redu
Nov 26 '18 at 14:31
add a comment |
1 Answer
1
active
oldest
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Summary: Haskell is a strongly typed language.
You force the type of sumlen to be [Int] -> (Int,Int).
The type of (/) is Fractional a => a -> a -> a.
Let us typecheck your function. First we will try to infer its type:
average = (uncurry (/)) . sumlen
uncurry :: (a -> b -> c) -> (a,b) -> c
(/) :: Fractional d => d -> d -> d
(.) :: (f -> g) -> (e -> f) -> e -> g
sumlen :: [Int] -> (Int,Int)
So:
e = [Int]
f = (Int,Int) = (a,b) = (d,d)
g = d = Int
average :: Fractional Int => [Int] -> Int
Now you tell Haskell that average :: Float and Haskell tells you that a float is not a function. This is your error.
If you remove the annotation then you will be told that there is no instance Fractional Int. This is because you can’t represent or reasonably approximate 1 / 2 as an Int.
How to fix this?
Step 0 is to not write down a type signature that is wrong (ie average :: Float)
One thing you could do is have a more general sumlen. Try Num a => [a] -> (a,a) instead. You could even have (Num a, Num b) => [a] -> (a,b). Then you can have the sensible type average :: Fractional a => [a] -> a
Another thing you could do is convert from integers:
average xs = fromIntegral s / fromIntegral l where
(s,l) = sumlen xs
answered Nov 26 '18 at 11:24
Dan RobertsonDan Robertson
3,443716
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Summary: Haskell is a strongly typed language.
You force the type of sumlen to be [Int] -> (Int,Int).
The type of (/) is Fractional a => a -> a -> a.
Let us typecheck your function. First we will try to infer its type:
average = (uncurry (/)) . sumlen
uncurry :: (a -> b -> c) -> (a,b) -> c
(/) :: Fractional d => d -> d -> d
(.) :: (f -> g) -> (e -> f) -> e -> g
sumlen :: [Int] -> (Int,Int)
So:
e = [Int]
f = (Int,Int) = (a,b) = (d,d)
g = d = Int
average :: Fractional Int => [Int] -> Int
Now you tell Haskell that average :: Float and Haskell tells you that a float is not a function. This is your error.
If you remove the annotation then you will be told that there is no instance Fractional Int. This is because you can’t represent or reasonably approximate 1 / 2 as an Int.
How to fix this?
Step 0 is to not write down a type signature that is wrong (ie average :: Float)
One thing you could do is have a more general sumlen. Try Num a => [a] -> (a,a) instead. You could even have (Num a, Num b) => [a] -> (a,b). Then you can have the sensible type average :: Fractional a => [a] -> a
Another thing you could do is convert from integers:
average xs = fromIntegral s / fromIntegral l where
(s,l) = sumlen xs
answered Nov 26 '18 at 11:24
Dan RobertsonDan Robertson
3,443716
add a comment |
Summary: Haskell is a strongly typed language.
You force the type of sumlen to be [Int] -> (Int,Int).
The type of (/) is Fractional a => a -> a -> a.
Let us typecheck your function. First we will try to infer its type:
average = (uncurry (/)) . sumlen
uncurry :: (a -> b -> c) -> (a,b) -> c
(/) :: Fractional d => d -> d -> d
(.) :: (f -> g) -> (e -> f) -> e -> g
sumlen :: [Int] -> (Int,Int)
So:
e = [Int]
f = (Int,Int) = (a,b) = (d,d)
g = d = Int
average :: Fractional Int => [Int] -> Int
Now you tell Haskell that average :: Float and Haskell tells you that a float is not a function. This is your error.
If you remove the annotation then you will be told that there is no instance Fractional Int. This is because you can’t represent or reasonably approximate 1 / 2 as an Int.
How to fix this?
Step 0 is to not write down a type signature that is wrong (ie average :: Float)
One thing you could do is have a more general sumlen. Try Num a => [a] -> (a,a) instead. You could even have (Num a, Num b) => [a] -> (a,b). Then you can have the sensible type average :: Fractional a => [a] -> a
Another thing you could do is convert from integers:
average xs = fromIntegral s / fromIntegral l where
(s,l) = sumlen xs
answered Nov 26 '18 at 11:24
Dan RobertsonDan Robertson
3,443716
add a comment |
Summary: Haskell is a strongly typed language.
You force the type of sumlen to be [Int] -> (Int,Int).
The type of (/) is Fractional a => a -> a -> a.
Let us typecheck your function. First we will try to infer its type:
average = (uncurry (/)) . sumlen
uncurry :: (a -> b -> c) -> (a,b) -> c
(/) :: Fractional d => d -> d -> d
(.) :: (f -> g) -> (e -> f) -> e -> g
sumlen :: [Int] -> (Int,Int)
So:
e = [Int]
f = (Int,Int) = (a,b) = (d,d)
g = d = Int
average :: Fractional Int => [Int] -> Int
Now you tell Haskell that average :: Float and Haskell tells you that a float is not a function. This is your error.
If you remove the annotation then you will be told that there is no instance Fractional Int. This is because you can’t represent or reasonably approximate 1 / 2 as an Int.
How to fix this?
Step 0 is to not write down a type signature that is wrong (ie average :: Float)
One thing you could do is have a more general sumlen. Try Num a => [a] -> (a,a) instead. You could even have (Num a, Num b) => [a] -> (a,b). Then you can have the sensible type average :: Fractional a => [a] -> a
Another thing you could do is convert from integers:
average xs = fromIntegral s / fromIntegral l where
(s,l) = sumlen xs
answered Nov 26 '18 at 11:24
Dan RobertsonDan Robertson
3,443716
Summary: Haskell is a strongly typed language.
You force the type of sumlen to be [Int] -> (Int,Int).
The type of (/) is Fractional a => a -> a -> a.
Let us typecheck your function. First we will try to infer its type:
average = (uncurry (/)) . sumlen
uncurry :: (a -> b -> c) -> (a,b) -> c
(/) :: Fractional d => d -> d -> d
(.) :: (f -> g) -> (e -> f) -> e -> g
sumlen :: [Int] -> (Int,Int)
So:
e = [Int]
f = (Int,Int) = (a,b) = (d,d)
g = d = Int
average :: Fractional Int => [Int] -> Int
Now you tell Haskell that average :: Float and Haskell tells you that a float is not a function. This is your error.
If you remove the annotation then you will be told that there is no instance Fractional Int. This is because you can’t represent or reasonably approximate 1 / 2 as an Int.
How to fix this?
Step 0 is to not write down a type signature that is wrong (ie average :: Float)
One thing you could do is have a more general sumlen. Try Num a => [a] -> (a,a) instead. You could even have (Num a, Num b) => [a] -> (a,b). Then you can have the sensible type average :: Fractional a => [a] -> a
Another thing you could do is convert from integers:
average xs = fromIntegral s / fromIntegral l where
(s,l) = sumlen xs
answered Nov 26 '18 at 11:24
Dan RobertsonDan Robertson
3,443716
edited Dec 15 '18 at 14:07
answered Nov 26 '18 at 11:24
Dan RobertsonDan Robertson
3,443716
answered Nov 26 '18 at 11:24
Dan RobertsonDan Robertson
3,443716
answered Nov 26 '18 at 11:24
Dan RobertsonDan Robertson
3,443716
3,443716
add a comment |
add a comment |
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The average of which list are you trying to find? Do you maybe want
averageto be a function, not aFloat?– sepp2k
Nov 26 '18 at 10:47
1
firstly:
averageis not aFloat. It is a function[Int] -> Float. Also, you should explictly convert yourInts intoFloatwhen applying(/)– Luis Morillo
Nov 26 '18 at 10:49
Possible duplicate of Haskell types frustrating a simple 'average' function
– Redu
Nov 26 '18 at 14:31