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When does Haskell complain about incorrect typing in functions?

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8

I am a Haskell newbie trying to wrap my head around type binding in functions and how Haskell enforces it. For example, even though the type for the fst function is fst :: (a, b) -> a, the compiler does not complain for the function fst'. But the compiler complains about type bindings for the function elem'.

fst' :: (a,a) -> a
fst' s = fst s

elem' :: (Eq a, Eq b) => a -> [b] -> Bool
elem' x xs = elem x xs

haskell types typeclass type-mismatch typechecking

share|improve this question

edited 8 hours ago

Will Ness

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• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  When you use a function with incorrect types. The reason why it does not complain about the first one is because you here define a "subset" type signature, that is fine. Whereas in the latter, that is not the case.
  
  – Willem Van Onsem
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The nicest way I've found to think about this is that you can always make a type more specific than it needs to be, but you can't always make it more general. A spoon can be used for many things, but my spoon I choose to use only on ice-cream ;)
  
  – AJFarmar
  6 hours ago
  

  
  
  
  

add a comment | 

8

I am a Haskell newbie trying to wrap my head around type binding in functions and how Haskell enforces it. For example, even though the type for the fst function is fst :: (a, b) -> a, the compiler does not complain for the function fst'. But the compiler complains about type bindings for the function elem'.

fst' :: (a,a) -> a
fst' s = fst s

elem' :: (Eq a, Eq b) => a -> [b] -> Bool
elem' x xs = elem x xs

haskell types typeclass type-mismatch typechecking

share|improve this question

edited 8 hours ago

Will Ness

49.1k44 gold badges7272 silver badges134134 bronze badges

asked 9 hours ago

user4132user4132

4322 bronze badges

New contributor

user4132 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  When you use a function with incorrect types. The reason why it does not complain about the first one is because you here define a "subset" type signature, that is fine. Whereas in the latter, that is not the case.
  
  – Willem Van Onsem
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The nicest way I've found to think about this is that you can always make a type more specific than it needs to be, but you can't always make it more general. A spoon can be used for many things, but my spoon I choose to use only on ice-cream ;)
  
  – AJFarmar
  6 hours ago
  

  
  
  
  

add a comment | 

I am a Haskell newbie trying to wrap my head around type binding in functions and how Haskell enforces it. For example, even though the type for the fst function is fst :: (a, b) -> a, the compiler does not complain for the function fst'. But the compiler complains about type bindings for the function elem'.

fst' :: (a,a) -> a
fst' s = fst s

elem' :: (Eq a, Eq b) => a -> [b] -> Bool
elem' x xs = elem x xs

haskell types typeclass type-mismatch typechecking

share|improve this question

edited 8 hours ago

Will Ness

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asked 9 hours ago

user4132user4132

4322 bronze badges

New contributor

user4132 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

fst' :: (a,a) -> a
fst' s = fst s

elem' :: (Eq a, Eq b) => a -> [b] -> Bool
elem' x xs = elem x xs

share|improve this question

edited 8 hours ago

Will Ness

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asked 9 hours ago

user4132user4132

4322 bronze badges

New contributor

user4132 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 8 hours ago

Will Ness

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asked 9 hours ago

user4132user4132

4322 bronze badges

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user4132 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 8 hours ago

Will Ness

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edited 8 hours ago

Will Ness

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asked 9 hours ago

user4132user4132

4322 bronze badges

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asked 9 hours ago

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12

fst has as type fst :: (a, b) -> a so that means it is fine to define a function:

fst' :: (a, a) -> a
fst' = fst

Your fst' function is more restrictive than the fst function. Regardless with what you replace a in your fst' function that is fine for fst. If for example a ~ Bool holds, then you call fst with signature fst :: (Bool, Bool) -> Bool. But since fst can deal with all as and b, it is fine that both elements of the tuple are Bool, so given fst can handle tuples for all possible types for both the first and the second item of the 2-tuple, it is defintely ok if the two items have the same type.

The latter is not ok, here you define:

elem' :: (Eq a, Eq b) => a -> [b] -> Bool
elem' = elem

but elem has type elem :: Eq a => a -> [a] -> Bool. The signatures you can make with the elem' function are not a subset of the ones of the elem function, since you can set a ~ Int and b ~ Bool. In that case you expect elem :: Int -> [Bool] -> Bool, but evidently that does not hold, since the Int and Bool type are two different types, and in the elem signature, these are both as.

share|improve this answer

answered 8 hours ago

Willem Van OnsemWillem Van Onsem

174k1818 gold badges175175 silver badges263263 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks for the explanation.
  
  – user4132
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Covers everything, nothing to add here, I wanted to answer adding something. But usually your answers are very good and complete.
  
  – Damian Lattenero
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @DamianLattenero: thank you :)
  
  – Willem Van Onsem
  4 hours ago
  

  
  
  
  

add a comment | 

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12

fst has as type fst :: (a, b) -> a so that means it is fine to define a function:

fst' :: (a, a) -> a
fst' = fst

Your fst' function is more restrictive than the fst function. Regardless with what you replace a in your fst' function that is fine for fst. If for example a ~ Bool holds, then you call fst with signature fst :: (Bool, Bool) -> Bool. But since fst can deal with all as and b, it is fine that both elements of the tuple are Bool, so given fst can handle tuples for all possible types for both the first and the second item of the 2-tuple, it is defintely ok if the two items have the same type.

The latter is not ok, here you define:

elem' :: (Eq a, Eq b) => a -> [b] -> Bool
elem' = elem

but elem has type elem :: Eq a => a -> [a] -> Bool. The signatures you can make with the elem' function are not a subset of the ones of the elem function, since you can set a ~ Int and b ~ Bool. In that case you expect elem :: Int -> [Bool] -> Bool, but evidently that does not hold, since the Int and Bool type are two different types, and in the elem signature, these are both as.

share|improve this answer

answered 8 hours ago

Willem Van OnsemWillem Van Onsem

174k1818 gold badges175175 silver badges263263 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks for the explanation.
  
  – user4132
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Covers everything, nothing to add here, I wanted to answer adding something. But usually your answers are very good and complete.
  
  – Damian Lattenero
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @DamianLattenero: thank you :)
  
  – Willem Van Onsem
  4 hours ago
  

  
  
  
  

add a comment | 

12

fst has as type fst :: (a, b) -> a so that means it is fine to define a function:

fst' :: (a, a) -> a
fst' = fst

Your fst' function is more restrictive than the fst function. Regardless with what you replace a in your fst' function that is fine for fst. If for example a ~ Bool holds, then you call fst with signature fst :: (Bool, Bool) -> Bool. But since fst can deal with all as and b, it is fine that both elements of the tuple are Bool, so given fst can handle tuples for all possible types for both the first and the second item of the 2-tuple, it is defintely ok if the two items have the same type.

The latter is not ok, here you define:

elem' :: (Eq a, Eq b) => a -> [b] -> Bool
elem' = elem

but elem has type elem :: Eq a => a -> [a] -> Bool. The signatures you can make with the elem' function are not a subset of the ones of the elem function, since you can set a ~ Int and b ~ Bool. In that case you expect elem :: Int -> [Bool] -> Bool, but evidently that does not hold, since the Int and Bool type are two different types, and in the elem signature, these are both as.

share|improve this answer

answered 8 hours ago

Willem Van OnsemWillem Van Onsem

174k1818 gold badges175175 silver badges263263 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks for the explanation.
  
  – user4132
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Covers everything, nothing to add here, I wanted to answer adding something. But usually your answers are very good and complete.
  
  – Damian Lattenero
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @DamianLattenero: thank you :)
  
  – Willem Van Onsem
  4 hours ago
  

  
  
  
  

add a comment | 

fst has as type fst :: (a, b) -> a so that means it is fine to define a function:

fst' :: (a, a) -> a
fst' = fst

Your fst' function is more restrictive than the fst function. Regardless with what you replace a in your fst' function that is fine for fst. If for example a ~ Bool holds, then you call fst with signature fst :: (Bool, Bool) -> Bool. But since fst can deal with all as and b, it is fine that both elements of the tuple are Bool, so given fst can handle tuples for all possible types for both the first and the second item of the 2-tuple, it is defintely ok if the two items have the same type.

The latter is not ok, here you define:

elem' :: (Eq a, Eq b) => a -> [b] -> Bool
elem' = elem

but elem has type elem :: Eq a => a -> [a] -> Bool. The signatures you can make with the elem' function are not a subset of the ones of the elem function, since you can set a ~ Int and b ~ Bool. In that case you expect elem :: Int -> [Bool] -> Bool, but evidently that does not hold, since the Int and Bool type are two different types, and in the elem signature, these are both as.

share|improve this answer

answered 8 hours ago

Willem Van OnsemWillem Van Onsem

174k1818 gold badges175175 silver badges263263 bronze badges

fst' :: (a, a) -> a
fst' = fst

elem' :: (Eq a, Eq b) => a -> [b] -> Bool
elem' = elem

share|improve this answer

answered 8 hours ago

Willem Van OnsemWillem Van Onsem

174k1818 gold badges175175 silver badges263263 bronze badges

answered 8 hours ago

Willem Van OnsemWillem Van Onsem

174k1818 gold badges175175 silver badges263263 bronze badges

answered 8 hours ago

Willem Van OnsemWillem Van Onsem

174k1818 gold badges175175 silver badges263263 bronze badges