For example, if I have a function “f(x, y)” and want to abbreviate it with “x S y”, how can it be done?

You can create a string macro and then do anything at all with the contents of that string…but it’s really not worth the effort. I’d recommend just living within the set of syntax supported by the Julia parser.

[image] How to define a new infix operator? New to Julia see in base\operators.jl that the built-in operators are defined concisely like: <(x, y) = isless(x, y) however, seems like I can’t define a custom infix operator like: julia> (α)(x, y) = x + y α (generic function…

If you really want this enough that wrapping your code in a macro is worthwhile you can get it… @withSop ...mycode a S b... you’d need to look for the S symbol, and convert it to S(a,b) manually.

No, that won’t work. A macro can only be applied to syntactically correct code.

From reading this thread , I conclude that some symbols are parsed as infix, but some aren’t, for example julia> function ↑(x, y) return x - y + 3 end ↑ (generic function with 1 method) julia> 9 ↑ 2 10 julia> function f(x, y) return x - y + 3 end f (generic func…

[image] Lucas_Queiroz: I don’t know if it is possible to make Julia parse f as an infix. It’s not possible… but using the macro approach above, if you did :f it would be parsed as “the symbol f” and then you could write your macro to convert 9 :f 2 into f(9,2)

I would need an introduction to Julia macros, because I am not very familiar with them.

[image] dlakelan: and then you could write your macro to convert 9 :f 2 into f(9,2) That’s still not valid syntax.

hmm… my lisp background is showing through. (9 :f 2) is valid lisp, and results in a list of three items… I guess in Julia there’s no equivalent where you just have a list of values? The closes equivalent of that is a tuple (9,:f,2) and that’s no help. is there any hope for something equivalent to …

9 f 2 is three expressions, which if you have sufficient patience perhaps you could make a macro understand, in some contexts: julia> macro pr(exs...) @show exs; nothing end; julia> @pr x = 9 f 2 exs = (:(x = 9), :f, 2) julia> @pr x = (9 f 2) ERROR: syntax: missing comma or ) in ar…

How do I create syntactic sugar for Julia functions?

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nilshg June 30, 2020, 4:20pm 2

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How do I create syntactic sugar for Julia functions?

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