Haskell syntax for a case expression in a do block

Question

I can't quite figure out this syntax problem with a case expression in a do block.

What is the correct syntax?

If you could correct my example and explain it that would be the best.

module Main where   main = do         putStrLn "This is a test"      s <- foo      putStrLn s    foo = do     args <- getArgs      return case args of                 [] -> "No Args"                 [s]-> "Some Args" 

A little update. My source file was a mix of spaces and tabs and it was causing all kinds of problems. Just a tip for any one else starting in Haskell. If you are having problems check for tabs and spaces in your source code.

Answer 1

return is an (overloaded) function, and it's not expecting its first argument to be a keyword. You can either parenthesize:

module Main where  import System(getArgs)  main = do         putStrLn "This is a test"      s <- foo      putStrLn s    foo = do     args <- getArgs      return (case args of                 [] -> "No Args"                 [s]-> "Some Args") 

or use the handy application operator ($):

foo = do     args <- getArgs      return $ case args of                 [] -> "No Args"                 [s]-> "Some Args" 

Stylewise, I'd break it out into another function:

foo = do     args <- getArgs      return (has_args args)  has_args [] = "No Args" has_args _  = "Some Args" 

but you still need to parenthesize or use ($), because return takes one argument, and function application is the highest precedence.