So, I would like use a macro to convert a struct into a tuple of its members let boo be the following struct boo a b c end and collect defined as follows macro collect(x) :( tuple($x.a,$x.b,$x.c) ) end @collect (macro with 1 method) things seem to work if invoked from the REPL P = bo…

You need to escape x inside the macro with esc(x), this is also explained in the manual on macro hygiene: https://docs.julialang.org/en/v1/manual/metaprogramming/#Hygiene Note that in this particular example, there is not really a reason to prefer a macro over a regular function. Functions OTOH, a…

many thanks! I will look that up! the reason for using macros here is on one hand to experiment, on the other hand I thought it might be faster since macros are substituted before compile, so there is really no overhead due to argument passing etc.

The macro is just generating the code tuple(y.a, y.b, y.c), which will then still have to be run at runtime, so it won’t be any faster than a function.

I got it to work in this fashion, macro collect(x) :($(esc(:($x.a, $x.b, $x.c)))) end @collect (macro with 1 method) P = boo(1,2,3) a,b,c = @collect P (1, 2, 3) a,b,c = let y = P @collect y end (1, 2, 3) but I definitely need to ponder it some more. you are probably right about speed, bu…

Note that you can do this with a generated function which will be just as fast as the macro version and usable on any type: julia> @generated fieldvalues(x) = Expr(:tuple, (:(x.$f) for f=fieldnames(x))...) julia> P = boo(1,2,3); julia> fieldvalues(P) (1,2,3) If you look at e.g. @code_llvm (P -> …

this is very nice, thanks! Did not know about generated functions.

thanks, I’ll look into that. I need to learn more about the metaprogramming side of Julia.

Using macros to explode a struct into a tuple of its members

General Usage

stevengj April 11, 2021, 12:25am 7

See also this discourse thread, which discussed conversion to a NamedTuple and also recommended a generated function.

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Using macros to explode a struct into a tuple of its members

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