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POHaskell operator vs function precedence
 

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    <p>I am trying to verify something for myself about operator and function precedence in Haskell. For instance, the following code </p> <pre><code>list = map foo $ xs </code></pre> <p>can be rewritten as </p> <pre><code>list = (map foo) $ (xs) </code></pre> <p>and will eventually be</p> <pre><code>list = map foo xs </code></pre> <p>My question used to be, why the first formulation would not be rewritten as </p> <pre><code>list = (map foo $) xs </code></pre> <p>since function precedence is always higher than operator precedence, but I think that I have found the answer: operators are simply not allowed to be arguments of functions (except of course, if you surround them with parentheses). Is this right? If so, I find it odd, that there is no mention of this mechanic/rule in RWH or Learn you a Haskell, or any of the other places that I have searched. So if you know a place, where the rule is stated, please link to it.</p> <p>-- edit: Thank you for your quick answers. I think my confusion came from thinking that an operator literal would somehow evaluate to something, that could get consumed by a function as an argument. It helped me to remember, that an infix operator can be mechanically translated to a prefix functions. Doing this to the first formulation yields </p> <pre><code>($) (map foo) (xs) </code></pre> <p>where there is no doubt that ($) is the consuming function, and since the two formulations are equivalent, then the $ literal in the first formulation cannot be consumed by map. </p>
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