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parsingoperatorsoperator-precedenceassociativity
Robert Bean
Robert Bean

Reputation: 859

Left-associative operators vs Right-associative operators

If we have an expression:

a $ b @ c

$ is a left-associative operator, @ is right-associative. They have the same precedence.

How is this expression parsed? As (a $ b) @ c or as a $ (b @ c)?

Upvotes: 12

Views: 5272

Answers (2)

Matt Fenwick
Matt Fenwick

Reputation: 49095

This is an excellent question. While Dipstick is correct that in many languages, operator precedences and associativities are defined to avoid such a problem, there are languages in which such a situation may arise.

Haskell is such a language. It allows you to define your own infix operators, and their precedences (integer from 0 to 9) and associativity (left, right, non). It's easy to create the preconditions for the scenario you described:

infixl 5 $$
($$) :: Int -> Int -> Int
a $$ b = a + b

infixr 5 @@
(@@) :: Int -> Int -> Int
a @@ b = a * b

And then the situation itself:

uhoh = 1 $$ 2 @@ 3

This results in this error message:

Precedence parsing error
    cannot mix `$$' [infixl 5] and `@@' [infixr 5] in the same infix expression

Of course, Haskell's solution -- aborting with a parse error -- is not the only way to deal with this problem, but it is certainly a reasonable one.

For more information about operator parsing in Haskell, please see section 4.4.2 of the Haskell report.

Upvotes: 12

Dipstick
Dipstick

Reputation: 10129

Operators at the same precedence are all either right associative or all left associative so the problem doesn't arise.

Upvotes: 6

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