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by tome
1866 days ago
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`($x)` could equivalently have been written `\f -> f x`, which would have been a bit clearer. In general in Haskell if you have an infix operator `OP` then `(OP x)` means the same thing as `\y -> y OP x`. (Additionally in this case you need to know that `$` is defined by `f $ x = f x`.) |
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And I just re-read my example in JS and I've made a TERRIBLE mistake ^^
`.every()` would be `foldl1 (&&)`, I would need `.some()` instead to have the correct behavior.