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I need to verify the type for the lambda expression:
My method gives me:
Im trying to define it in Haskell (on Hugs) like this:
h= \f x -> f (f x)
When i call the :type comamnd it gives me:
(a -> a) -> a -> a
Is mi function correctly defined in Haskell?, or my method gives me a wrong result?
820
Note that lambda expressions can only be used to implement functional interfaces. In the above example also, the lambda expression implements Consumer Functional Interface. A Java program to demonstrate working of lambda expression with two arguments. // data type for x and y is optional.
The lambda expressions have a very simple, precise syntax and provide flexibility to specify the datatypes for the function parameters. Its return type is a parameter -> expression body to understand the syntax, we can divide it into three parts.
A lambda expression is a function or subroutine without a name that can be used wherever a delegate is valid. Lambda expressions can be functions or subroutines and can be single-line or multi-line.
Note that f gets called with both x and f x as its argument—this immediately means that the type of x and the type of f x must be the same[1]. Carrying this argument onward, we see that since x is the input to f and f x is the output of f, the input and output of f must be the same[2].
Finally, we examine the lambda term
\f x -> f (f x)
it has two inputs, f (a function) and x, and it returns whatever the return type of f is[3]. Putting all of this information together we have
(a -> b) -> c -> d
where:
b ~ c by [1]
a ~ b by [2]
d ~ b by [3]
thus the type which Haskell inferred is correct
h :: (a -> a) -> a -> a
h f x = f (f x)
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