Stumped on eval of function with parameters

New to Julia
1

I am trying to provide a feature to allow a user to supply a function expression and have that expression evaluated at runtime.

The function needs two arguments, m and n. Both are assigned values before the evaluation.

function calculate(e::Expr)
   m = rand(1:4, 5, 5)
   n = rand(0:1, 5, 5)
   return eval(e)
end

Why does the following not work?

d = :(rank(m * n))
julia> calculate(d)
ERROR: UndefVarError: m not defined
Stacktrace:
 [1] top-level scope at none:0
 [2] eval at ./boot.jl:330 [inlined]
 [3] calculate(::Expr) at ./none:6
 [4] top-level scope at none:0

Thanks!

2

Why not use a function or function closure like that:

function calculate(f::Function)
    m = rand(1:4, 5, 5)
    n = rand(0:1, 5, 5)
    f(m, n) 
end

julia> using LinearAlgebra

julia> d(m,n) = rank(m*n)
d (generic function with 1 method)

julia> calculate(d)
4

julia> calculate((m,n)->rank(m*n))
4

wouldn’t that be faster, simpler and more idiomatic?

4 Likes
3

The technical reason is that eval works in global scope. But as already been said, passing higher order functions is almost always the better API in cases like this.

4

Yes! Thank you. Using Function directly as the type did the trick.

calculate(reducef::Function = (m,n,f)->rank(m * n), ...) # this is the default function
…then…
reducef(m, n, f)

@kristoffer.carlsson - I’m still not clear why eval() fails to work, even if I use eval(mymodule, expr). Are there built-in functions of Julia that can clarify the context in which code is evaluated at runtime? Very powerful stuff, this metaprogramming, but powerfully complicated as well. :wink: