functional programming - Haskell, Don't know why this has a *parse error on input ‘if’* -

functional programming - Haskell, Don't know why this has a *parse error on input ‘if’* -

this take number, factorial , double it, because of base of operations case if input 0 gives 2 reply in order bypass used if statement, error parse error on input ‘if’. appreciate if guys help :)

fact :: int -> int fact 0 = 1 fact n = n * fact(n-1) doub :: int -> int doub r = 2 * r factorialdouble :: io() factorialdouble = putstr "enter value: " x <- getline allow num = (read x) :: int if (num == 0) error "factorial of 0 0" else allow y = doub (fact num) putstrln ("the double of factorial of " ++ x ++ " " ++ (show y))

i've spotted 2 issues should addressed

you have let has no continuation: (else allow y = doub (fact num) ...). because you're not within do, want alter let ... in statement. your if indented far in. should under let.

i've corrected mentioned , code works me...

fact :: int -> int fact 0 = 1 fact n = n * fact(n-1) doub :: int -> int doub r = 2 * r factorialdouble :: io () factorialdouble = putstr "enter value: " x <- getline allow num = (read x) :: int if num == 0 (error "factorial of 0 0") else allow y = doub (fact num) in putstrln ("the double of factorial of " ++ x ++ " " ++ (show y))

haskell functional-programming