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3322306
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2010-07-23T20:59:20.530
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This works because of operator precedence. The function application operator, juxtaposition or <code></code> (the space), has the highest precedence, so <code>take 34 fibseq |> filter even</code> parses as <code>((take 34) fibseq) |> (filter even)</code>, which is equivalent to <code>(filter even) ((take 34) fibseq)</code>; since function application is left-associative, this is then equivalent to <code>filter even (take 34 fibseq)</code>. In general, any binary operator can be given a precedence with a fixity declaration, such as <pre><code>infixl 0 |> infixr 9 . </code></pre> The <code>l</code> or <code>r</code> says whether the operation is left- or right-associative (that is, whether <code>a • b • c</code> groups as <code>(a • b) • c</code> or <code>a • (b • c)</code>); the number—an integer between 0 and 9—specifies the precedence level. Higher numbers mean higher precedence (with application having an effective precedence of ∞); for instance, <code>*</code> and <code>/</code> have precedence 7, and <code>+</code> and <code>-</code> have precedence 6. To check the precedence of an operator in ghci, just type <code>:info $</code> (or whichever operator) at the prompt. And just as a note: your code will work, but it's not how I would typically write it. If you're curious, in Haskell, I would write that code with the <code>$</code> operator, which just performs function application but is low precedence: <code>filter even $ take 34 fibseq</code>. If I had more functions to apply, I would use the composition operator: <code>fun1 arg1 . fun2 . fun3 arg2 arg3 . filter even $ take 34 fibseq</code>. It reads the other way, but it's what you typically find in Haskell.
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1. POHaskell function composition (forward pipe) - why does this work?
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1. USAntal Spector-Zabusky
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