Getting parse error on input ‘=’parser, even if my indentation seems correct-CodePudding

I am very beginner with Haskell, and I m trying to create some small algorithms to understand the syntax of Haskell, I m trying to write and compile a quicksort algorithm, but it gives me an error at the compilation.

quicksort [] = []
 quicksort (x:xs) = quicksort small    (x : quicksort large)
   where small = [y | y <- xs, y <= x]
         large = [y | y <- xs, y > x]

Can someone help me to understand why I get this error at the 2nd line of my code. parse error on input ‘=’parser.

Thanks,

I m also new at this platform so if I m not clear enough or I need to improve my question let me know at the comments.

CodePudding user response：

Intendation matters in haskell. I can reproduce you error with

quicksort [] = []
 quicksort (x:xs) = quicksort small    (x : quicksort large)
   where small = [y | y <- xs, y <= x]
         large = [y | y <- xs, y > x]

.code.tio.hs:2:19: error: parse error on input ‘=’ Perhaps you need a 'let' in a 'do' block?

But when you remove the space before the second quicksort, it compiles fine.

Technically, you can use any intendation when you use {, } and ;. This is not recommended, because you will have a difficult time finding the exact places where you need braces, and examples are rare. ("Rare" as in "endangered species" / "almost extinct")

{
quicksort [] = [];
 quicksort (x:xs) = quicksort small    (x : quicksort large) -- no `;` before `where`
   where {small = [y | y <- xs, y <= x];
         large = [y | y <- xs, y > x]};
main = print . quicksort $ "hello"
}