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What is the difference between the dot (.) and the dollar sign ($)?
As I understand it, they are both syntactic sugar for not needing to use parentheses.
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The dollar sign, $ , is a controversial little Haskell operator. Semantically, it doesn't mean much, and its type signature doesn't give you a hint of why it should be used as often as it is. It is best understood not via its type but via its precedence.
The dot ( . ) is the Haskell operator for function composition. Function composition comes from mathematics but actually, it can be really useful to make music. Haskell was originally designed by mathematicians and computer magicians. Its syntax borrowed quite a lot from mathematical notation.
It's merely an infix synonym for fmap , so you can write e.g. Prelude> (*2) <$> [1.. 3] [2,4,6] Prelude> show <$> Just 11 Just "11" Like most infix functions, it is not built-in syntax, just a function definition. But functors are such a fundamental tool that <$> is found pretty much everywhere.
Haskell provides special syntax to support infix notation. An operator is a function that can be applied using infix syntax (Section 3.4), or partially applied using a section (Section 3.5).
The $ operator is for avoiding parentheses. Anything appearing after it will take precedence over anything that comes before.
For example, let's say you've got a line that reads:
putStrLn (show (1 + 1)) If you want to get rid of those parentheses, any of the following lines would also do the same thing:
putStrLn (show $ 1 + 1) putStrLn $ show (1 + 1) putStrLn $ show $ 1 + 1 The primary purpose of the . operator is not to avoid parentheses, but to chain functions. It lets you tie the output of whatever appears on the right to the input of whatever appears on the left. This usually also results in fewer parentheses, but works differently.
Going back to the same example:
putStrLn (show (1 + 1)) (1 + 1) doesn't have an input, and therefore cannot be used with the . operator.show can take an Int and return a String.putStrLn can take a String and return an IO ().You can chain show to putStrLn like this:
(putStrLn . show) (1 + 1) If that's too many parentheses for your liking, get rid of them with the $ operator:
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