call or access a local function by name

Question

function f1 () {
    console.log('f1')
}

var s = 'f1'

runLocalFunctionByName(s)

Is this possible at all to write runLocalFunctionByName() or just call f1 without typing f1 in the source code, but using a variable holding its name? I mean without modifying f1 into a method, that answer is obvious: just make myobj.f1 = function or declare it globally like f1= function(). I am talking about normal local functions declared with function keyword only, not as vars, global vars or some other object property.

Why do you need to do this? I suspect this is an XY problem, and there's a better way to solve it. — Barmar, Dec 24 '15 at 11:43
Somewhat related: http://stackoverflow.com/questions/2146173/how-to-get-local-variable-by-its-name-in-js, http://stackoverflow.com/questions/598878/how-can-i-access-local-scope-dynamically-in-javascript. — , Dec 24 '15 at 13:24
`f1` is not a local function here--by "local" do you mean that it is defined inside some other function? — , Dec 24 '15 at 13:32

score 4 · Answer 1 · answered Dec 24 '15 at 11:42

Not without the use of eval, which is evil (when used for this purpose for sure!).

Global functions could be called as attributes of the window object, but if they are in a local/closure scope that's not possible.

If you need to call functions by name, the only proper solution is storing them as attributes on an object, and then using obj[s]() for calling them.

Soubhik Mondal · Answer 2 · 2015-12-24T12:06:08.337

0

One way of doing this is by using the instantiating a Function object.

var runLocalFunctionByName = function(fName) { 
    return (new Function('','return '+fName+'();'))(); 
};

Calling runLocalFunctionByName with a name of a function will now return the output of the named function.

EDIT

In case of local functions, the scope has to be mentioned, so we can modify the code, such that:

var runLocalFunctionByName = function(fName, scope) { 
    return (new Function('','return '+(scope?scope:'this')+'.'+fName+'();'))(); 
};

edited Dec 24 '15 at 12:06

answered Dec 24 '15 at 11:48

Soubhik Mondal

2,666
1
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That will only work for global functions, not local ones. – Bergi Dec 24 '15 at 11:52
Yes. In that case, we can modify this code to include the scope for the local functions. – Soubhik Mondal Dec 24 '15 at 11:55
How? Show that please, it is what was asked for. – Bergi Dec 24 '15 at 11:56
1

And what would you pass as `scope` in OPs example? – Bergi Dec 24 '15 at 12:06
1

It's a poor idea to propose this semi-eval-ish solution without understanding what the OP is trying to accomplish. – Dec 24 '15 at 13:09

score 0 · Answer 3 · answered Dec 24 '15 at 12:27

0

You can't access a local function outside its scope until unless you assign it to a variable having an accessible scope. MDN: A function defined by a function expression inherits the current scope. That is, the function forms a closure. On the other hand, a function defined by a Function constructor does not inherit any scope other than the global scope (which all functions inherit)

answered Dec 24 '15 at 12:27

mak

71
3

1

He's not worried about scoping. He's trying to call a function when what he has is its name in string form. – Dec 24 '15 at 13:10

call or access a local function by name

3 Answers3