Haskell: parse error on input ‘=’ in where
I am new in Haskell and practicing Algorithm in Haskell follow the book "Pearls of Functional Algorithm Design"
This is the algo to find the smallest natural number not in a given finite set X of natural numbers
import Data.List
import Data.Array
minfree xs = if null ([0..b-1] \ us)
then head ([b..] \ vs)
else head ([0..b-1] \ us)
where (us, vs) = (partition (<b) xs)
b = div (length xs) 2
Pff the compiler error like this
Prelude Data.Array Data.List> :load 01_the_smallest_free_number.hs
[1 of 1] Compiling Main ( 01_the_smallest_free_number.hs, interpreted )
01_the_smallest_free_number.hs:11:29: error:
parse error on input ‘=’
Perhaps you need a 'let' in a 'do' block?
e.g. 'let x = 5' instead of 'x = 5'
|
11 | b = div (length xs) 2
| ^
Failed, no modules loaded.
Well obveriously add let
before b
is not the right answer I've tried.
Then I replaced all b
to div (length xs) 2
that works, so seems here is the problem to be, but I don't get it
haskell
add a comment |
I am new in Haskell and practicing Algorithm in Haskell follow the book "Pearls of Functional Algorithm Design"
This is the algo to find the smallest natural number not in a given finite set X of natural numbers
import Data.List
import Data.Array
minfree xs = if null ([0..b-1] \ us)
then head ([b..] \ vs)
else head ([0..b-1] \ us)
where (us, vs) = (partition (<b) xs)
b = div (length xs) 2
Pff the compiler error like this
Prelude Data.Array Data.List> :load 01_the_smallest_free_number.hs
[1 of 1] Compiling Main ( 01_the_smallest_free_number.hs, interpreted )
01_the_smallest_free_number.hs:11:29: error:
parse error on input ‘=’
Perhaps you need a 'let' in a 'do' block?
e.g. 'let x = 5' instead of 'x = 5'
|
11 | b = div (length xs) 2
| ^
Failed, no modules loaded.
Well obveriously add let
before b
is not the right answer I've tried.
Then I replaced all b
to div (length xs) 2
that works, so seems here is the problem to be, but I don't get it
haskell
2
Bad identation causes the problem I guess.
– Sereja Bogolubov
Nov 11 at 8:35
add a comment |
I am new in Haskell and practicing Algorithm in Haskell follow the book "Pearls of Functional Algorithm Design"
This is the algo to find the smallest natural number not in a given finite set X of natural numbers
import Data.List
import Data.Array
minfree xs = if null ([0..b-1] \ us)
then head ([b..] \ vs)
else head ([0..b-1] \ us)
where (us, vs) = (partition (<b) xs)
b = div (length xs) 2
Pff the compiler error like this
Prelude Data.Array Data.List> :load 01_the_smallest_free_number.hs
[1 of 1] Compiling Main ( 01_the_smallest_free_number.hs, interpreted )
01_the_smallest_free_number.hs:11:29: error:
parse error on input ‘=’
Perhaps you need a 'let' in a 'do' block?
e.g. 'let x = 5' instead of 'x = 5'
|
11 | b = div (length xs) 2
| ^
Failed, no modules loaded.
Well obveriously add let
before b
is not the right answer I've tried.
Then I replaced all b
to div (length xs) 2
that works, so seems here is the problem to be, but I don't get it
haskell
I am new in Haskell and practicing Algorithm in Haskell follow the book "Pearls of Functional Algorithm Design"
This is the algo to find the smallest natural number not in a given finite set X of natural numbers
import Data.List
import Data.Array
minfree xs = if null ([0..b-1] \ us)
then head ([b..] \ vs)
else head ([0..b-1] \ us)
where (us, vs) = (partition (<b) xs)
b = div (length xs) 2
Pff the compiler error like this
Prelude Data.Array Data.List> :load 01_the_smallest_free_number.hs
[1 of 1] Compiling Main ( 01_the_smallest_free_number.hs, interpreted )
01_the_smallest_free_number.hs:11:29: error:
parse error on input ‘=’
Perhaps you need a 'let' in a 'do' block?
e.g. 'let x = 5' instead of 'x = 5'
|
11 | b = div (length xs) 2
| ^
Failed, no modules loaded.
Well obveriously add let
before b
is not the right answer I've tried.
Then I replaced all b
to div (length xs) 2
that works, so seems here is the problem to be, but I don't get it
haskell
haskell
asked Nov 11 at 7:47
Aries_is_there
1518
1518
2
Bad identation causes the problem I guess.
– Sereja Bogolubov
Nov 11 at 8:35
add a comment |
2
Bad identation causes the problem I guess.
– Sereja Bogolubov
Nov 11 at 8:35
2
2
Bad identation causes the problem I guess.
– Sereja Bogolubov
Nov 11 at 8:35
Bad identation causes the problem I guess.
– Sereja Bogolubov
Nov 11 at 8:35
add a comment |
1 Answer
1
active
oldest
votes
This is bad indentation:
where (us, vs) = (partition (<b) xs)
b = div (length xs) 2
Since the second equality is indented more than the first, it continues the first one, as if we wrote
where (us, vs) = (partition (<b) xs) b = div (length xs) 2
triggering the error.
You instead want:
where (us, vs) = (partition (<b) xs)
b = div (length xs) 2
so that both equations are indented in the same way.
add a comment |
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1 Answer
1
active
oldest
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oldest
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active
oldest
votes
This is bad indentation:
where (us, vs) = (partition (<b) xs)
b = div (length xs) 2
Since the second equality is indented more than the first, it continues the first one, as if we wrote
where (us, vs) = (partition (<b) xs) b = div (length xs) 2
triggering the error.
You instead want:
where (us, vs) = (partition (<b) xs)
b = div (length xs) 2
so that both equations are indented in the same way.
add a comment |
This is bad indentation:
where (us, vs) = (partition (<b) xs)
b = div (length xs) 2
Since the second equality is indented more than the first, it continues the first one, as if we wrote
where (us, vs) = (partition (<b) xs) b = div (length xs) 2
triggering the error.
You instead want:
where (us, vs) = (partition (<b) xs)
b = div (length xs) 2
so that both equations are indented in the same way.
add a comment |
This is bad indentation:
where (us, vs) = (partition (<b) xs)
b = div (length xs) 2
Since the second equality is indented more than the first, it continues the first one, as if we wrote
where (us, vs) = (partition (<b) xs) b = div (length xs) 2
triggering the error.
You instead want:
where (us, vs) = (partition (<b) xs)
b = div (length xs) 2
so that both equations are indented in the same way.
This is bad indentation:
where (us, vs) = (partition (<b) xs)
b = div (length xs) 2
Since the second equality is indented more than the first, it continues the first one, as if we wrote
where (us, vs) = (partition (<b) xs) b = div (length xs) 2
triggering the error.
You instead want:
where (us, vs) = (partition (<b) xs)
b = div (length xs) 2
so that both equations are indented in the same way.
answered Nov 11 at 8:04
chi
72.8k280135
72.8k280135
add a comment |
add a comment |
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2
Bad identation causes the problem I guess.
– Sereja Bogolubov
Nov 11 at 8:35