adrien changed the topic of #ocaml to: Discussions about the OCaml programming language | http://www.ocaml.org | OCaml 4.09 release notes: https://caml.inria.fr/pub/distrib/ocaml-4.09/notes/Changes | Try OCaml in your browser: http://try.ocamlpro.com | Public channel logs at http://irclog.whitequark.org/ocaml
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<laokz> how can understand this: let ($) x y = x * y <br> 2 $ 3 + 2 ==> 10 <br> ($) 2 3 + 2 ==> 8 ?
<d_bot> <Et7f3> `let ($) x y = x * y` we declare operator `$` to be the same has `*`
<d_bot> <Et7f3> so `2 $ 3 + 2` is like `2 * (3 + 2)` because `$` has lower precedence than `+`
<sonologico> but `$` has different precedence, so you get different results for an expression without parens if you just replace `*` with `$`
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<d_bot> <Et7f3> on the other side it is not a operator but a function call
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<laokz> no mater how ($) declared, its evaluation precedence is according $'s original precedence ?
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<sonologico> laokz: Yes. In OCaml, operator precedence is determined by the first character of the operator. (http://caml.inria.fr/pub/docs/manual-ocaml/expr.html#ss:precedence-and-associativity)
<laokz> even for eventually new operator? sonologico
<sonologico> yes, so all operators that you define starting with `$` will have the same precedence
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<laokz> sonologico: cleared. thanks.
<laokz> d_bot：thanks.
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<d_bot> <EduardoRFS> what if two operators starts with the same token?
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<octachron> Then they have the same precedence and associativity. Thus it is the associativity that is used to determine how they are grouped together
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<d_bot> <Aaron Christianson> > how can understand this: let ($) x y = x * y <br> 2 $ 3 + 2 ==> 10 <br> ($) 2 3 + 2 ==> 8 ?
<d_bot> <Aaron Christianson> @laokz#0000 Is this real code or just learning material?
<d_bot> <Aaron Christianson> because I would go on a murderous rampage if I found this in the wild.
<companion_cube> that's probably an example
<companion_cube> but you can find infix operators in the wild, like, in cmdliner
<d_bot> <Aaron Christianson> Well, I like my monad operators as much as anyone, but defining `$` as `*` is a bridge too far! (but you're probably right and it's just an example)
<companion_cube> ah, yeah
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