Javascript: can I return a reference to a function defined outside the function I'm calling?

Question

Let's say I have this code:

function QueCalculamos(valor) {

    function CalculaADeB() {
      console.log( "Calculando A de B");
      }

    function CalculaBDeA() {
      console.log( "Calculando B de A");
      }


   if (valor =="A") { return CalculaADeB; }
    else if (valor == "B") { return CalculaBDeA; }
}

let bvla="B";

var exe = QueCalculamos(bvla);
exe();

Depending on the value I pass to "QueCalculamos", it will execute one function or the other.

But let's suppose that I don't have to choose between two functions, but among many, and that they are a bit large. For code readability purposes, I would like to define "CalculaADeB" and "CalculaBDeA" separately, outside "QueCalculamos".

How would I return a reference to those functions in that case? And if it's not possible, how would you organize the code to make it more readable then? (Maybe I don't have enough experience with modern Javascript, but this whole "define a function inside a function" thing always looks messy to me).

Define them wherever. Put them into an array. Pass the array as a variable. — John
– John, Commented Jul 14, 2020 at 10:07

Mitya · Accepted Answer · 2020-07-14 10:11:37Z

3

You can define your functions as methods of an object.

let funcs = {
    func1() { /* ... */ },
    func2() { /* ... */ },
    /* ... */
}

Then reference these from your router function via something like a switch statement.

function QueCalculamos(valor) {
    switch (valor) {
        case 'A': return funcs.func1;
        case 'B': return funcs.func2;
        /* ... */
    }
}

answered Jul 14, 2020 at 10:11

Mitya

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Comments

Bergi · Accepted Answer · 2020-07-14 10:13:21Z

You can define them wherever you want, as long as they are in scope in QueCalculamos so that you can access (get a reference to) them. The only difference is that the functions will no longer be closures that could access the valor - but they don't do that anyway.

It will just work if you move them outside:

function CalculaADeB() {
  console.log("Calculando A de B");
}
function CalculaBDeA() {
  console.log("Calculando B de A");
}

function QueCalculamos(valor) {
  if (valor =="A") { return CalculaADeB; }
  else if (valor == "B") { return CalculaBDeA; }
}

let bvla="B";

var exe = QueCalculamos(bvla);
exe();

You might be able to further simplify the code by putting them on an object with property names to select them:

const calcs = {
  A() {
    console.log("Calculando A de B");
  },
  B() {
    console.log("Calculando B de A");
  },
};
function QueCalculamos(valor) {
  return calcs[valor];
}

var exe = QueCalculamos("B");
exe();

ibrahim mahrir · Accepted Answer · 2020-07-14 10:20:47Z

Group them into an object instead of making them separate functions, and access them using a key:

let functions = {                                          // keys are the possible values of valor
  "A": function() { /* ... */ },                           // you can either define the functions here
  "B": CalculaBDeA,                                        // ... or assign a reference to an already existing function
  "C": function() { /* ... */ },
  "D": function() { /* ... */ },
  /* ... */
}

function QueCalculamos(valor) {
  if(functions.hasOwnProperty(valor)) {                     // if the functions object contains a function for valor
    return functions[valor];                                // return it
  }

  throw "there is no function for '" + valor + "'";        // otherwise throw an error or return something else to signal failure
}

Demo:

let functions = {
  "A": function() {
    console.log("Function A");
  },
  "B": function() {
    console.log("Function B");
  },
  "C": function() {
    console.log("Function C");
  },
  "D": function() {
    console.log("Function D");
  },
  /* ... */
}

function QueCalculamos(valor) {
  if (functions.hasOwnProperty(valor)) {
    return functions[valor];
  }

  throw "there is no function for '" + valor + "'";
}


let exe = QueCalculamos("C");
exe();

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