How to use an expression in function from other function in julia

Question

When I try those code below:

function f(x)
    Meta.parse("x -> x " * x) |> eval
end

function g(x)
    findall(Base.invokelatest(f,x),[1,2,3]) |> println
end

g("<3")

Julia throws "The applicable method may be too new" error.

If I tried these code below:

function f(x)
    Meta.parse("x -> x " * x) |> eval
end

findall(f("<3"),[1,2,3]) |> println

Julia could give me corrected result: [1, 2]

How can I modify the first codes to use an String to generate function in other function, Thx!

Test in Julia 1.6.7

Instead evaling strings (which is frowned upon), you can pass anonymous functions instead: `findall(<(3), [1,2,3])`. This is a better approach in general. — DNF, Jan 09 '23 at 13:01
Fully agreed. However, for some reason recently many users on SO ask about parsing strings in Julia (as @DNF noted - this should be done normally). — Bogumił Kamiński, Jan 09 '23 at 13:41

score 3 · Accepted Answer · answered Jan 09 '23 at 08:18

3

Do

function g(x)
    h = f(x)
    findall(x -> Base.invokelatest(h, x) ,[1,2,3]) |> println
end

g("<3")

The difference in your code is that when you write:

Base.invokelatest(f, x)

you invoke f, but f is not redefined. What you want to do is invokelatest the function that is returned by f instead.

answered Jan 09 '23 at 08:18

Thanks very much for helping me understand the 'Base.invokelatest' – kaji331 Jan 11 '23 at 01:32
There's another point confused me. In my f(x), I parsed "x -> x". Why I have to add "x ->" befor "Base.invokelatest..." ? – kaji331 Jan 11 '23 at 02:07
Because `findall` takes a predicate as a first argument. In my solution I pass an anonymous function that can be compiled (i.e. is not dynamically defined) to `findall`. The dynamically defined function is only invoked in its body. This way compiler has a known function to dispatch to in `findall`. – Bogumił Kamiński Jan 11 '23 at 06:53

score 1 · Answer 2 · answered Jan 09 '23 at 15:34

1

Use a macro instead of function:

macro f(expr)
    Meta.parse("x -> x " * expr)
end

Now you can just do:

julia> filter(@f("<3"), [1,2,3])
2-element Vector{Int64}:
 1
 2

answered Jan 09 '23 at 15:34

Przemyslaw Szufel

2 Answers2