I'm sorry to bother you so often.
I have a question about Haskell.
& expressionTrue&&_=_
False &_= False
The problem is to express with a conditional expression.
(bb)::Bool->a->b
x bby = if x == True then else x
When I compiled the above program, I got the following error
enzan.hs:1:1:error:
Invalid type signature: (bb)::...
Should be of form <variable>::<type>
|
1 | (bb)::Bool->a->b||Bool
| ^^^^
I'm not sure what's wrong with the program, but please let me know.
haskell
bb is used like an intermediate operator, but the intermediate operator must be a symbol.
For example, modifying to define .&. eliminates errors.
(.&.)::Bool->Bool->Bool
x.&&.y = if x then else x
This is an answer to the question of the commenter.
The error message is
I expected type b, but I couldn't match it because it was actually Bool.
b is a type variable bound by type signature (.&.)::forall a b.Bool->a->b.
I mean, actually
x.&.y = if x then else x
By definition, (.&.) has no other type than Bool->Bool->Bool.
Conversely, consider whether you can define an operator with the first type signature.
Type Signature (.&.): According to Bool->a->b, in the formula x.&y, the value of x is of any type, the type of y is of any type, and the type of whole expression (x.&y) is of any type (even if it is different from x).
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