Parsing a variable datatype to a function which contains a conditional check on the datatype in JavaScript

Question

To pass a datatype that varies (from an array to an integer) to the same function, and then check against the datatype before changing its value, looking at the below method using instanceof Array is there a better/more efficient way?

function foo(x) { 
  if (x instanceof Array) {
    for(i=0;i<x.length;i++){    
      x[i].bar = '1';
      x[i].baz = '2';       
    }   
  }
  else{
    x.bar = '1';
    x.baz = '2';    
  }
}

Thanks :)

Object.prototype.toString.call(x) === "[object Array]", this is an alternative method, but even yours is good — Niccolò Campolungo
– Niccolò Campolungo, Commented Jun 16, 2013 at 17:26

Nagarjun · Accepted Answer · 2013-06-16 18:07:28Z

2

An alternative (using ECMAScript standard)

if( Object.prototype.toString.call( x ) === '[object Array]' ) {

    for(i=0;i<x.length;i++) {  

      x[i].bar = '1';
      x[i].baz = '2';       
    }  
}

See ECMAScript

Or if you always want it as an array, but this is not recommended

x = [].concat( x );

edited Jun 16, 2013 at 18:07

answered Jun 16, 2013 at 17:30

Nagarjun

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1 Comment

Nagarjun Over a year ago

The first method is what is given in ECMAScript to find the class of an object. the second method, as I said is not recommended because of performance reasons.

basilikum · Accepted Answer · 2013-06-16 18:07:52Z

1

A more efficient way could also be to split up your function (if that is possible for you):

function fooArray(x) { 
    for(i = 0; i < x.length; i++){    
        foo(x[i]);   
    }   
}

function foo(x) {
    x.bar = '1';
    x.baz = '2'; 
}

This would also apply the DRY principle ("don't repeat yourself") because you don't need to code the same logic (i.e. x.bar = '1';) twice.

edited Jun 16, 2013 at 18:07

answered Jun 16, 2013 at 17:59

basilikum

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