CODEWITHSUNDEEP
julia
dj_a
dj_a

Reputation: 141

Conditionally define a function inside another function in Julia

Slow to learning newer julia syntax and scoping.

In Julia v1.1.1

what is the explanation for why the MWE below throws error "ff not defined" ?

N = 5;
typp = "b";
mu = 2;

function bigfun()

  function f(u,mu)
    ee = mu*u;
    return ee
  end

  function g(uv,mu)
    ee = (mu^2)*uv
    return ee;
  end

  while 1 == 1

    u = ones(N);
    if typp == "a"
      ff(u) = f(u,mu);
    elseif typp == "b"
      ff(u) = g(u,mu);
    end
    fu = ff(u);
    break;

  end

end

bigfun();

Upvotes: 8

Views: 1144

Answers (2)

giordano
giordano

Reputation: 8344

This is a known bug in Julia: https://github.com/JuliaLang/julia/issues/15602. You can't define an inner function conditionally. There a few ways around this issue:

  1. Define ff as an anonymous function: 
        if typp == "a"
      ff = u -> f(u,mu)
    elseif typp == "b"
      ff = u -> g(u,mu)
    end
    fu = ff(u)
  2. Define ff once, and add the conditional inside of it: 
        function ff(u, typp)
      if typp == "a"
        f(u,mu)
      elseif typp == "b"
        g(u,mu)
      end
    end
    fu = ff(u, typp)
  3. Don't define the function ff at all. You don't need to, in the example you provided, just assign fu conditionally 
        if typp == "a"
      fu = f(u,mu)
    elseif typp == "b"
      fu = g(u,mu)
    end

Upvotes: 14

logankilpatrick
logankilpatrick

Reputation: 14521

As mention in the Julia Docs,

Julia uses lexical scoping, meaning that a function's scope does not inherit from its caller's scope, but from the scope in which the function was defined.

It is worth reading some examples in the above link but the gist is that the function is defined in the local scope/context of the if/else statement and does not exist outside of that scope (at the end of that function).

Upvotes: 0

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