Javascript calling function reference

Question

function (type, a, b) {
  var types = {
    'add' : this.add(a,b),
    'subtract' : this.subtract(a,b),
    'multiply' : this.multiply(a,b)
  };

  if(types[type])
    return types[type];
}

This might be simple, but I can't seem to figure it out. I want a map of strings to function references (with parameters pre populated), and if the type is valid, I want to call the specified function in the map.

this setup works, but it's actually calling the methods when it's initializing the array which isn't what i want. i only want to call the function if the type is valid. I just want to keep a reference to the function call I want to make when I set up the function. Is this possible? Thanks.

Where is this.add defined? What is this supposed to be here? — Felix Kling
– Felix Kling, Commented Feb 2, 2014 at 0:21
The expression this.add(a,b) is evaluated during the object construction. You probably want to use a closure (if using a/b), for whatever operation contract you have in mind. — user2864740
– user2864740, Commented Feb 2, 2014 at 0:21
If you want only the function reference, then remove the brackets and parameters. — C3roe
– C3roe, Commented Feb 2, 2014 at 0:21
make no sense to re-initialize types hash each time when code enters function — Eugene P.
– Eugene P., Commented Feb 2, 2014 at 0:22

Felix Kling · Accepted Answer · 2014-02-02 00:22:37Z

4

It looks like you want:

function (type, a, b) {
  var types = {
    'add' : this.add, // assign reference
    'subtract' : this.subtract,
    'multiply' : this.multiply
  };

  if(types[type])
    return types[type](a, b); // call function
}

But you could also avoid creating the object entirely:

function (type, a, b) {
  if(['add', 'subtract', 'multiply'].indexOf(type) > -1)
    return this[type](a, b);
}

answered Feb 2, 2014 at 0:22

Felix Kling

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