无法将预期类型'Integer - > t'与实际类型'Bool'匹配

Question

In Haskell,

This works perfectly fine: (mod 9) 7 . It gives the expected result: remainder when 9 is divided by 7 ( 2 ).

Similarly, this works too: (mod 9) 9 . It returns 0 .

This led me to think that (mod 9 == 0) 9 should return True . However, that hasn't been the case: it threw up an error instead.

THE ERROR:

<interactive>:62:1: error:
    • Couldn't match expected type ‘Integer -> t’
                  with actual type ‘Bool’
    • The function ‘mod 9 == 0’ is applied to one argument,
      but its type ‘Bool’ has none
      In the expression: (mod 9 == 0) 9
      In an equation for ‘it’: it = (mod 9 == 0) 9
    • Relevant bindings include it :: t (bound at <interactive>:62:1)

Please help me understand why (mod 9 == 0) 9 wouldn't return True .

PS: I'm convinced that my usage of "return" in Haskell's context is flawed. However, I am just starting out, so please excuse me. (Would be nice if you could correct me if I am, indeed, wrong.)

Answer 1

As I mentioned in a comment, it appears that you expect mod 9 == 0 to be a function that takes an argument, passes it to mod 9 , then returns the result of the comparison. You can write such an expression, but it's a little more complicated.

>>> ((== 0) . (mod 9)) 9
True

Here, (== 0) . (mod 9) (== 0) . (mod 9) is the composition of two functions, (== 0) and mod 9 . The composed function takes its argument, applies mod 9 to it, then applies (== 0) to the result. (Where (== 0) is a short form for \\x -> x == 0 .)