Re: [RFC] Function pipe operator

On Tue, Apr 21, 2020 at 4:44 PM Larry Garfield <larry@garfieldtech.com> wrote: > > On Mon, Apr 20, 2020, at 11:20 PM, Stanislav Malyshev wrote: > > Just a small pedantry note - in a comparison section, the RFC compares > > this syntax to function composition. But this is not function > > composition. This is a syntax sugar for calling two functions one after > > another, not operator that produces a function. It sounds pedantic but > > it's rather important distinction - if |> is composition, than $foo |> > > $bar is a new callable provided $foo and $bar are callable (but no > > function is actually being called here!). If |> is call syntax, it's > > actually the result of calling $bar($foo). > > > > So comparing it to function composition is a bit confusing. Otherwise it > > looks OK to me, except the syntax for calling functions and methods is a > > bit awkward, but it's not the problem of this RFC I imagine. > > I'm not sure I follow. The only place composition is mentioned is in the F# section, > where it calls out specifically that we're implementing "pipe forward" and *not* the > composition operators ( >> ). Is that unclear? Actually it's also mentioned in the Haskell section, but as "function concatenation" (which adds to the confusion I guess). Speaking of Haskell, that reminded me of http://learnyouahaskell.com/a-fistful-of-monads#walk-the-line where the author defines a custom -: operator such that x -: f desugars to f x (equivalent to how $x |> $f would desugar to $f($x) with the RFC), which allows to write e.g. xs -: sort -: reverse instead of reverse (sort xs) or (reverse . sort) xs. -- Guilliam Xavier