I'm puzzled by how the Haskell compiler sometimes infers types that are less
polymorphic than what I'd expect, for example when using point-free definitions.
It seems like the issue is the "monomorphism restriction", which is on by default on
older versions of the compiler.
Consider the following Haskell program:
{-# LANGUAGE MonomorphismRestriction #-}
import Data.List(sortBy)
plus = (+)
plus' x = (+ x)
sort = sortBy compare
main = do
print $ plus' 1.0 2.0
print $ plus 1.0 2.0
print $ sort [3, 1, 2]
If I compile this with ghc I obtain no errors and the output of the executable is:
3.0
3.0
[1,2,3]
If I change the main body to:
main = do
print $ plus' 1.0 2.0
print $ plus (1 :: Int) 2
print $ sort [3, 1, 2]
I get no compile time errors and the output becomes:
3.0
3
[1,2,3]
as expected. However if I try to change it to:
main = do
print $ plus' 1.0 2.0
print $ plus (1 :: Int) 2
print $ plus 1.0 2.0
print $ sort [3, 1, 2]
I get a type error:
test.hs:13:16:
No instance for (Fractional Int) arising from the literal ‘1.0’
In the first argument of ‘plus’, namely ‘1.0’
In the second argument of ‘($)’, namely ‘plus 1.0 2.0’
In a stmt of a 'do' block: print $ plus 1.0 2.0
The same happens when trying to call sort twice with different types:
main = do
print $ plus' 1.0 2.0
print $ plus 1.0 2.0
print $ sort [3, 1, 2]
print $ sort "cba"
produces the following error:
test.hs:14:17:
No instance for (Num Char) arising from the literal ‘3’
In the expression: 3
In the first argument of ‘sort’, namely ‘[3, 1, 2]’
In the second argument of ‘($)’, namely ‘sort [3, 1, 2]’
Why does ghc suddenly think that plus isn't polymorphic and requires an Int argument?
The only reference to Int is in an application of plus, how can that matter
when the definition is clearly polymorphic?
Why does ghc suddenly think that sort requires a Num Char instance?
Moreover if I try to place the function definitions into their own module, as in:
{-# LANGUAGE MonomorphismRestriction #-}
module TestMono where
import Data.List(sortBy)
plus = (+)
plus' x = (+ x)
sort = sortBy compare
I get the following error when compiling:
TestMono.hs:10:15:
No instance for (Ord a0) arising from a use of ‘compare’
The type variable ‘a0’ is ambiguous
Relevant bindings include
sort :: [a0] -> [a0] (bound at TestMono.hs:10:1)
Note: there are several potential instances:
instance Integral a => Ord (GHC.Real.Ratio a)
-- Defined in ‘GHC.Real’
instance Ord () -- Defined in ‘GHC.Classes’
instance (Ord a, Ord b) => Ord (a, b) -- Defined in ‘GHC.Classes’
...plus 23 others
In the first argument of ‘sortBy’, namely ‘compare’
In the expression: sortBy compare
In an equation for ‘sort’: sort = sortBy compare
Why isn't ghc able to use the polymorphic type Ord a => [a] -> [a] for sort?
And why does ghc treat plus and plus' differently? plus should have the
polymorphic type Num a => a -> a -> a and I don't really see how this is different
from the type of sort and yet only sort raises an error.
Last thing: if I comment the definition of sort the file compiles. However
if I try to load it into ghci and check the types I get:
*TestMono> :t plus
plus :: Integer -> Integer -> Integer
*TestMono> :t plus'
plus' :: Num a => a -> a -> a
Why isn't the type for plus polymorphic?
This is the canonical question about monomorphism restriction in Haskell
as discussed in [the meta question](https://meta.stackoverflow.com/questions/294053/can-we-provide-a-canonical-questionanswer-for-haskells-monomorphism-restrictio).
What is the monomorphism restriction?
The monomorphism restriction as stated by the Haskell wiki is:
a counter-intuitive rule in Haskell type inference.
If you forget to provide a type signature, sometimes this rule will fill
the free type variables with specific types using "type defaulting" rules.
What this means is that, in some circumstances, if your type is ambiguous (i.e. polymorphic)
the compiler will choose to instantiate that type to something not ambiguous.
How do I fix it?
First of all you can always explicitly provide a type signature and this will
avoid the triggering of the restriction:
plus :: Num a => a -> a -> a
plus = (+) -- Okay!
-- Runs as:
Prelude> plus 1.0 1
2.0
Note that only normal type signatures on variables count for this purpose, not expression type signatures. For example, writing this would still result in the restriction being triggered:
plus = (+) :: Num a => a -> a -> a
Alternatively, if you are defining a function, you can avoid point-free style,
and for example write:
plus x y = x + y
Turning it off
It is possible to simply turn off the restriction so that you don't have to do
anything to your code to fix it. The behaviour is controlled by two extensions:
MonomorphismRestriction will enable it (which is the default) while
NoMonomorphismRestriction will disable it.
You can put the following line at the very top of your file:
{-# LANGUAGE NoMonomorphismRestriction #-}
If you are using GHCi you can enable the extension using the :set command:
Prelude> :set -XNoMonomorphismRestriction
You can also tell ghc to enable the extension from the command line:
ghc ... -XNoMonomorphismRestriction
Note: You should really prefer the first option over choosing extension via command-line options.
Refer to GHC's page for an explanation of this and other extensions.
A complete explanation
I'll try to summarize below everything you need to know to understand what the
monomorphism restriction is, why it was introduced and how it behaves.
An example
Take the following trivial definition:
plus = (+)
you'd think to be able to replace every occurrence of + with plus. In particular since (+) :: Num a => a -> a -> a you'd expect to also have plus :: Num a => a -> a -> a.
Unfortunately this is not the case. For example if we try the following in GHCi:
Prelude> let plus = (+)
Prelude> plus 1.0 1
We get the following output:
<interactive>:4:6:
No instance for (Fractional Integer) arising from the literal ‘1.0’
In the first argument of ‘plus’, namely ‘1.0’
In the expression: plus 1.0 1
In an equation for ‘it’: it = plus 1.0 1
You may need to :set -XMonomorphismRestriction in newer GHCi versions.
And in fact we can see that the type of plus is not what we would expect:
Prelude> :t plus
plus :: Integer -> Integer -> Integer
What happened is that the compiler saw that plus had type Num a => a -> a -> a, a polymorphic type.
Moreover it happens that the above definition falls under the rules that I'll explain later and so
he decided to make the type monomorphic by defaulting the type variable a.
The default is Integer as we can see.
Note that if you try to compile the above code using ghc you won't get any errors.
This is due to how ghci handles (and must handle) the interactive definitions.
Basically every statement entered in ghci must be completely type checked before
the following is considered; in other words it's as if every statement was in a separate
module. Later I'll explain why this matter.
Some other example
Consider the following definitions:
f1 x = show x
f2 = \x -> show x
f3 :: (Show a) => a -> String
f3 = \x -> show x
f4 = show
f5 :: (Show a) => a -> String
f5 = show
We'd expect all these functions to behave in the same way and have the same type,
i.e. the type of show: Show a => a -> String.
Yet when compiling the above definitions we obtain the following errors:
test.hs:3:12:
No instance for (Show a1) arising from a use of ‘show’
The type variable ‘a1’ is ambiguous
Relevant bindings include
x :: a1 (bound at blah.hs:3:7)
f2 :: a1 -> String (bound at blah.hs:3:1)
Note: there are several potential instances:
instance Show Double -- Defined in ‘GHC.Float’
instance Show Float -- Defined in ‘GHC.Float’
instance (Integral a, Show a) => Show (GHC.Real.Ratio a)
-- Defined in ‘GHC.Real’
...plus 24 others
In the expression: show x
In the expression: \ x -> show x
In an equation for ‘f2’: f2 = \ x -> show x
test.hs:8:6:
No instance for (Show a0) arising from a use of ‘show’
The type variable ‘a0’ is ambiguous
Relevant bindings include f4 :: a0 -> String (bound at blah.hs:8:1)
Note: there are several potential instances:
instance Show Double -- Defined in ‘GHC.Float’
instance Show Float -- Defined in ‘GHC.Float’
instance (Integral a, Show a) => Show (GHC.Real.Ratio a)
-- Defined in ‘GHC.Real’
...plus 24 others
In the expression: show
In an equation for ‘f4’: f4 = show
So f2 and f4 don't compile. Moreover when trying to define these function
in GHCi we get no errors, but the type for f2 and f4 is () -> String!
Monomorphism restriction is what makes f2 and f4 require a monomorphic
type, and the different behaviour bewteen ghc and ghci is due to different
defaulting rules.
When does it happen?
In Haskell, as defined by the report, there are two distinct type of bindings.
Function bindings and pattern bindings. A function binding is nothing else than
a definition of a function:
f x = x + 1
Note that their syntax is:
<identifier> arg1 arg2 ... argn = expr
Modulo guards and where declarations. But they don't really matter.
where there must be at least one argument.
A pattern binding is a declaration of the form:
<pattern> = expr
Again, modulo guards.
Note that variables are patterns, so the binding:
plus = (+)
is a pattern binding. It's binding the pattern plus (a variable) to the expression (+).
When a pattern binding consists of only a variable name it's called a
simple pattern binding.
The monomorphism restriction applies to simple pattern bindings!
Well, formally we should say that:
A declaration group is a minimal set of mutually dependent bindings.
Section 4.5.1 of the report.
And then (Section 4.5.5 of the report):
a given declaration group is unrestricted if and only if:
every variable in the group is bound by a function binding (e.g. f x = x)
or a simple pattern binding (e.g. plus = (+); Section 4.4.3.2 ), and
an explicit type signature is given for every variable in the group that
is bound by simple pattern binding. (e.g. plus :: Num a => a -> a -> a; plus = (+)).
Examples added by me.
So a restricted declaration group is a group where, either there are
non-simple pattern bindings (e.g. (x:xs) = f something or (f, g) = ((+), (-))) or
there is some simple pattern binding without a type signature (as in plus = (+)).
The monomorphism restriction affects restricted declaration groups.
Most of the time you don't define mutual recursive functions and hence a declaration
group becomes just a binding.
What does it do?
The monomorphism restriction is described by two rules in Section
4.5.5 of the report.
First rule
The usual Hindley-Milner restriction on polymorphism is that only type
variables that do not occur free in the environment may be generalized.
In addition, the constrained type variables of a restricted declaration
group may not be generalized in the generalization step for that group.
(Recall that a type variable is constrained if it must belong to some
type class; see Section 4.5.2 .)
The highlighted part is what the monomorphism restriction introduces.
It says that if the type is polymorphic (i.e. it contain some type variable)
and that type variable is constrained (i.e. it has a class constraint on it:
e.g. the type Num a => a -> a -> a is polymorphic because it contains a and
also contrained because the a has the constraint Num over it.)
then it cannot be generalized.
In simple words not generalizing means that the uses of the function plus may change its type.
If you had the definitions:
plus = (+)
x :: Integer
x = plus 1 2
y :: Double
y = plus 1.0 2
then you'd get a type error. Because when the compiler sees that plus is
called over an Integer in the declaration of x it will unify the type
variable a with Integer and hence the type of plus becomes:
Integer -> Integer -> Integer
but then, when it will type check the definition of y, it will see that plus
is applied to a Double argument, and the types don't match.
Note that you can still use plus without getting an error:
plus = (+)
x = plus 1.0 2
In this case the type of plus is first inferred to be Num a => a -> a -> a
but then its use in the definition of x, where 1.0 requires a Fractional
constraint, will change it to Fractional a => a -> a -> a.
Rationale
The report says:
Rule 1 is required for two reasons, both of which are fairly subtle.
Rule 1 prevents computations from being unexpectedly repeated.
For example, genericLength is a standard function (in library Data.List)
whose type is given by
genericLength :: Num a => [b] -> a
Now consider the following expression:
let len = genericLength xs
in (len, len)
It looks as if len should be computed only once, but without Rule 1 it
might be computed twice, once at each of two different overloadings.
If the programmer does actually wish the computation to be repeated,
an explicit type signature may be added:
let len :: Num a => a
len = genericLength xs
in (len, len)
For this point the example from the wiki is, I believe, clearer.
Consider the function:
f xs = (len, len)
where
len = genericLength xs
If len was polymorphic the type of f would be:
f :: Num a, Num b => [c] -> (a, b)
So the two elements of the tuple (len, len) could actually be
different values! But this means that the computation done by genericLength
must be repeated to obtain the two different values.
The rationale here is: the code contains one function call, but not introducing
this rule could produce two hidden function calls, which is counter intuitive.
With the monomorphism restriction the type of f becomes:
f :: Num a => [b] -> (a, a)
In this way there is no need to perform the computation multiple times.
Rule 1 prevents ambiguity. For example, consider the declaration group
[(n,s)] = reads t
Recall that reads is a standard function whose type is given by the signature
reads :: (Read a) => String -> [(a,String)]
Without Rule 1, n would be assigned the type ∀ a. Read a ⇒ a and s
the type ∀ a. Read a ⇒ String.
The latter is an invalid type, because it is inherently ambiguous.
It is not possible to determine at what overloading to use s,
nor can this be solved by adding a type signature for s.
Hence, when non-simple pattern bindings are used (Section 4.4.3.2 ),
the types inferred are always monomorphic in their constrained type variables,
irrespective of whether a type signature is provided.
In this case, both n and s are monomorphic in a.
Well, I believe this example is self-explanatory. There are situations when not
applying the rule results in type ambiguity.
If you disable the extension as suggest above you will get a type error when
trying to compile the above declaration. However this isn't really a problem:
you already know that when using read you have to somehow tell the compiler
which type it should try to parse...
Second rule
Any monomorphic type variables that remain when type inference for an
entire module is complete, are considered ambiguous, and are resolved
to particular types using the defaulting rules (Section 4.3.4 ).
This means that. If you have your usual definition:
plus = (+)
This will have a type Num a => a -> a -> a where a is a
monomorphic type variable due to rule 1 described above. Once the whole module
is inferred the compiler will simply choose a type that will replace that a
according to the defaulting rules.
The final result is: plus :: Integer -> Integer -> Integer.
Note that this is done after the whole module is inferred.
This means that if you have the following declarations:
plus = (+)
x = plus 1.0 2.0
inside a module, before type defaulting the type of plus will be:
Fractional a => a -> a -> a (see rule 1 for why this happens).
At this point, following the defaulting rules, a will be replaced by Double
and so we will have plus :: Double -> Double -> Double and x :: Double.
Defaulting
As stated before there exist some defaulting rules, described in Section 4.3.4 of the Report,
that the inferencer can adopt and that will replace a polymorphic type with a monomorphic one.
This happens whenever a type is ambiguous.
For example in the expression:
let x = read "<something>" in show x
here the expression is ambiguous because the types for show and read are:
show :: Show a => a -> String
read :: Read a => String -> a
So the x has type Read a => a. But this constraint is satisfied by a lot of types:
Int, Double or () for example. Which one to choose? There's nothing that can tell us.
In this case we can resolve the ambiguity by telling the compiler which type we want,
adding a type signature:
let x = read "<something>" :: Int in show x
Now the problem is: since Haskell uses the Num type class to handle numbers,
there are a lot of cases where numerical expressions contain ambiguities.
Consider:
show 1
What should the result be?
As before 1 has type Num a => a and there are many type of numbers that could be used.
Which one to choose?
Having a compiler error almost every time we use a number isn't a good thing,
and hence the defaulting rules were introduced. The rules can be controlled
using a default declaration. By specifying default (T1, T2, T3) we can change
how the inferencer defaults the different types.
An ambiguous type variable v is defaultable if:
v appears only in contraints of the kind C v were C is a class
(i.e. if it appears as in: Monad (m v) then it is not defaultable).
at least one of these classes is Num or a subclass of Num.
all of these classes are defined in the Prelude or a standard library.
A defaultable type variable is replaced by the first type in the default list
that is an instance of all the ambiguous variable’s classes.
The default default declaration is default (Integer, Double).
For example:
plus = (+)
minus = (-)
x = plus 1.0 1
y = minus 2 1
The types inferred would be:
plus :: Fractional a => a -> a -> a
minus :: Num a => a -> a -> a
which, by defaulting rules, become:
plus :: Double -> Double -> Double
minus :: Integer -> Integer -> Integer
Note that this explains why in the example in the question only the sort
definition raises an error. The type Ord a => [a] -> [a] cannot be defaulted
because Ord isn't a numeric class.
Extended defaulting
Note that GHCi comes with extended defaulting rules (or here for GHC8),
which can be enabled in files as well using the ExtendedDefaultRules extensions.
The defaultable type variables need not only appear in contraints where all
the classes are standard and there must be at least one class that is among
Eq, Ord, Show or Num and its subclasses.
Moreover the default default declaration is default ((), Integer, Double).
This may produce odd results. Taking the example from the question:
Prelude> :set -XMonomorphismRestriction
Prelude> import Data.List(sortBy)
Prelude Data.List> let sort = sortBy compare
Prelude Data.List> :t sort
sort :: [()] -> [()]
in ghci we don't get a type error but the Ord a constraints results in
a default of () which is pretty much useless.
Useful links
There are a lot of resources and discussions about the monomorphism restriction.
Here are some links that I find useful and that may help you understand or deep further into the topic:
Haskell's wiki page: Monomorphism Restriction
The report
An accessible and nice blog post
Sections 6.2 and 6.3 of A History Of Haskell: Being Lazy With Class deals with the monomorphism restriction and type defaulting
I have a composition of functions, many of them operators, to raise a constant to the power of a constant minus the size of the ith set in a list. Here's my code:
(k^).(n-).size.(phi !!) i
When I tried it on a test case I got
<interactive>:64:16-25:
Couldn't match expected type ‘a -> Set a0’
with actual type ‘Set a1’
Relevant bindings include
it :: a -> c (bound at <interactive>:64:1)
Possible cause: ‘phi !!’ is applied to too many arguments
In the second argument of ‘(.)’, namely ‘(phi !!) 3’
In the second argument of ‘(.)’, namely ‘size . (phi !!) 3’
However,
(k^)$(n-)$size$(phi !!)$i
works. What's wrong? Why does composition not work but application work? Also, is composing the operators in parentheses the most idiomatic way to do it? It feels weird.
Rather than f . g . h x, you want f . g . h $ x: the former calls h x immediately and then composes that, as a function, with f and g The latter composes together f, g, and h into a new function and then calls that on i.
Note that you can use parenthesis to group function compositions,
((k^).(n-).size.(phi !!)) i should work.
print $ concat ["abc", "bde"]
prints
abcbde
whereas,
print . concat ["abc", "bde"]
The error thrown in second case is,
Couldn't match expected type ‘a -> b0’ with actual type ‘[Char]’
Relevant bindings include
it :: a -> IO () (bound at <interactive>:3:1)
Possible cause: ‘concat’ is applied to too many arguments
. (function-composition operator) is used because i thought it will take the output of concat function and pass it to the preceding function print? What is wrong in the code?
Just precedence, really; this will work fine:
(print . concat) ["abc", "bde"]
. composes two functions to create a new function, whereas $ is just a way of avoiding parentheses in this context.
I have defined the following function
let repl x n = [x | _ <- [1..n]]
which imitates the built-in replicate function.
While experimenting with it, I noticed a strange thing: repl 10 0 evaluates to [], while repl 10 -1 produces an error:
No instance for (Show (t10 -> [t0])) arising from a use of ‘print’
In a stmt of an interactive GHCi command: print it
On the other hand, both [1 .. 0] and [1 .. -1] evaluate to [] without producing any errors.
Moreover, both [42 | _ <- [1 .. 0]] and [42 | _ <- [1 .. -1]] evaluate to [] without errors.
So why does my function call result in an error where the explicit substitution doesn't? And more importantly, where does the apparent difference between [1 .. 0] and [1 .. -1] stem from?
And a final question: when I write:
repl 42 -1
the error is exactly the same as with repl 10 -1, i.e. it still has the (Show (t10 -> [t0])) bit in it. I was expecting it to have something like ((Show (t42 -> [t0]))). What's this 10?
Other answers have pointed out that you need to wrap -1 in parentheses. This is an odd corner of the Haskell 98 spec that jumps out to bite unexpectedly. It's not the case that you can never write a negative number without parentheses: -1 * 5 is fine. It's just that the unary prefix operator doesn't have higher precedence than the binary infix operator, so a - is frequently parsed as the latter. Whitespace around operators is not significant in Haskell.
And the incomprehensible typeclass error doesn't help. Incidentally, t10 and t0 are just placeholder type variables made up by the compiler; I don't think it has anything to do with the actual numeric literals you use. And informally, errors like Could not deduce (Num (a0 -> t)) usually indicate to me that a function is applied to too few arguments.
Alternatively, the (undocumented?) NegativeLiterals language extension in GHC 7.8 changes the meaning of -1 to address this problem.
> :set -XNegativeLiterals
> :t repl 10 -1
repl 10 -1 :: Num t => [t]
You have not included the full error message in your question, and if you had, you'd see that repl 10 -1 is parsed as (repl 10) - (1) which is not what you intended.
You'd get the same error with repl 10 +1.
You can often find clues as to how a program code is parsed, by looking closely into the error message. There's no harm in overusing parentheses, while you learn, either.
The program:
repl x n = [x | _ <- [1..n]] -- line 1
main = print (repl 10 -1) -- line 3
The message:
prog.hs:3:8:
No instance for (Show (t1 -> [t0])) arising from a use of `print'
Possible fix: add an instance declaration for (Show (t1 -> [t0]))
In the expression: print (repl 10 - 1)
In an equation for `main': main = print (repl 10 - 1)
prog.hs:3:15:
No instance for (Num t1) arising from a use of `repl'
The type variable `t1' is ambiguous
Possible fix: add a type signature that fixes these type variable(s)
Note: there are several potential instances:
instance Num Double -- Defined in `GHC.Float'
instance Num Float -- Defined in `GHC.Float'
instance Integral a => Num (GHC.Real.Ratio a)
-- Defined in `GHC.Real'
...plus three others
In the first argument of `(-)', namely `repl 10' --------- NB!
In the first argument of `print', namely `(repl 10 - 1)'
In the expression: print (repl 10 - 1)
prog.hs:3:20:
No instance for (Num t0) arising from the literal `10'
The type variable `t0' is ambiguous
Possible fix: add a type signature that fixes these type variable(s)
Note: there are several potential instances:
instance Num Double -- Defined in `GHC.Float'
instance Num Float -- Defined in `GHC.Float'
instance Integral a => Num (GHC.Real.Ratio a)
-- Defined in `GHC.Real'
...plus three others
In the first argument of `repl', namely `10'
In the first argument of `(-)', namely `repl 10' --------- NB!
In the first argument of `print', namely `(repl 10 - 1)'
prog.hs:3:23:
No instance for (Num (t1 -> [t0])) arising from a use of `-'
Possible fix: add an instance declaration for (Num (t1 -> [t0]))
In the first argument of `print', namely `(repl 10 - 1)'
In the expression: print (repl 10 - 1)
In an equation for `main': main = print (repl 10 - 1)
Did you try [1..(-1)]? In Haskell, you cannot write negative numbers like -1 directly. You need to put them in parentheses. The reason is that Haskell doesn't have prefix unary operators because operators in Haskell are always infix. Hence -1 is parsed as [operator (-)] [numeric 1] and not [numeric -1].
This is what causes the problem. To avoid this problem, negative numbers must always be put in parentheses. This ensures that (-1) is parsed as [numeric -1]. It's one of the few corner cases which gives migraines to newcomers in Haskell.
I've got this simple function:
bombplaces::Int->[(Int,Int)]->[(Int,Int)]
bombplaces bombCount listOfPossiblePoints = nub (map (take bombCount) (perms listOfPossiblePoints))
bombs are (x,y) (carthesian points)
i need to get an all permutations and take only first few (bombCount) points.
I'm getting following error:
Couldn't match expected type `(Int,Int)' with actual type `[a0]'
Expected type: [a0] -> (Int,Int)
Actual type: [a0] -> [a0]
In the return type of a call of `take'
In the first argument of `map', namely `(take liczbaBomb)'
If you remove the type signature and ask GHCi for the type, your problem will be obvious:
> :t bombplaces
bombplaces :: Eq a => Int -> [a] -> [[a]]
That is, bombplaces wants to return a list of lists whereas you want it to return a plain list. You need to either change the type signature, or change the definition of the function, depending on what you want the behaviour to be.
N.B. You didn't tell us what definition of perms you are using, so I assumed the obvious one.