Javascript: cached function property vs new value

Question

A basic function:

function myStuff(a,b,c){
    var _c = _c || c || {};
    console.log(_c);
}

The idea is to use a cached value if used before, or a new value if new/different.

The question is: if the function invoke does not include the third property why the console.log(_c) shows undefined or how can I write this line var _c = _c || c || {}; better?

Thanks

score 2 · Answer 1 · edited May 23 '17 at 12:06

2

the scope of the variable you defined is the function itself, therefore, it is redefined every time you call the function.

the correct way to cache a variable, is to define it in the parent scope of the function:

var _c;
function myStuff(a,b,c){
    _c = c || _c || {};
    console.log(_c);
}

read more about js scopes here: What is the scope of variables in JavaScript?

edited May 23 '17 at 12:06

Community

1
1

answered Aug 20 '15 at 15:03

MoLow

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only in the first call. after that it gets cached – MoLow Aug 20 '15 at 15:05
but the part of ` || {}` is ignored when function is used without third property. – thednp Aug 20 '15 at 15:13
mabe in the console you see undefined, becase calling the function return undefined – MoLow Aug 20 '15 at 15:16

score 1 · Answer 2 · answered Aug 20 '15 at 15:07

1

You can store the cache inside the function itself

function myStuff(a,b,c){
    if (!myStuff.cache) {
        myStuff.cache = {};
    }

    var cache = myStuff.cache;
    c = cache.c = cache.c || c || {};

    console.log(c);
}

answered Aug 20 '15 at 15:07

axelduch

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score 1 · Answer 3 · answered Aug 20 '15 at 15:08

1

If yout want to access _c external value you have to use the function like this:

function myStuff(a,b,c){
    var _c = window._c || c || {};
    console.log(_c);
}

answered Aug 20 '15 at 15:08

deFreitas

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thednp · Accepted Answer · 2015-08-20T15:55:29.110

1

I figured. Having this code if ( c === undefined ) { c = {}; } solves the issue with undefined c:

function myStuff(a,b,c){
  if ( c === undefined ) { c = {}; }
  var _c = _c || c || {};
  console.log(_c);
}

because

_c = c || _c || {};

does not count the last case for {}.

edited Aug 20 '15 at 15:55

answered Aug 20 '15 at 15:30

thednp

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Javascript: cached function property vs new value

4 Answers4