Array of digit as string to simple int conversion problem

Question

Console debug shows me that array is ex. ["2"], but I need [2].

Why casting doesnt'work?

function filterElements(deals) {

    var result = deals,
        categories= $('#deals-categories').data('current_category');
        if (categories != undefined && categories.length > 0) {
            for (var i; i < categories.length; i++) {
                categories[i] = parseInt(categories[i]);
            }
            console.log(categories, 'cats');
                result = $.grep(result, function(e) {
                    return $.inArray(e.category_id, categories) != -1;
     });                
        }
    return result;
}

score 4 · Accepted Answer · answered Aug 29 '11 at 13:10

You need to initialize var i = 0 in the loop declaration.

Full code cleanup:

function filterElements(deals) {

    var result = deals,
        categories = $('#deals-categories').data('current_category');

        if (categories && categories.length) {
            for (var i=0; i<categories.length; i++) {
                categories[i] = parseInt(categories[i], 10);
            }
            console.log(categories, 'cats');
            result = $.grep(result, function(e) {
                return $.inArray(e.category_id, categories) !== -1;
            });                
        }

    return result;
}

The only weirdness I see now is using `$.inArray()` inside of `$.grep()`. — Matt Ball, Aug 29 '11 at 13:16

Michael Sagalovich · Answer 2 · 2011-08-29T13:16:47.743

0

use categories[i] * 1 to cast

parseInt works in a bit unexpected way sometimes :)

parseInt("010") will return 8 in some browsers, and 10 in others: http://www.w3schools.com/jsref/jsref_parseInt.asp

edited Aug 29 '11 at 13:16

answered Aug 29 '11 at 13:09

Michael Sagalovich

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I think Matt Ball explains the issue, while parseInt is still not recommended when dealing with strings which are for sure ints. – Michael Sagalovich Aug 29 '11 at 13:15
1

@Michael why not? Ohhh, w3schools... `/facepalm`. This is why JSLint recommends that you **always specify the radix in `parseInt` calls** (until ECMAScript 5). A **much** better resource than w3schools: https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/parseInt – Matt Ball Aug 29 '11 at 13:17
yeah, `radix` solves the issue, but requires too much letters :) – Michael Sagalovich Aug 29 '11 at 13:23
The better alternatives to `parseInt()` are `Number` or unary `+`, not `*1`. – Matt Ball Aug 29 '11 at 13:33
See http://stackoverflow.com/questions/4090518/string-to-int-use-parseint-or-number and http://stackoverflow.com/questions/850341/workarounds-for-javascript-parseint-octal-bug and http://stackoverflow.com/questions/594625/parseint-alternative – Matt Ball Aug 29 '11 at 13:40

score 0 · Answer 3 · answered Aug 29 '11 at 13:39

Are you sure? This is an example similar to yours:

var strings = ["1", "2", "3"];
var valueAsInt = 0;

for(var i = 0; i < strings.length; i++){
   valueAsInt = parseInt(strings[i]);
   if(typeof(valueAsInt) == 'number'){
      alert('Is an integer');
   }
}

The message 'Is an integer' is shown three times. I think in your code the parser works, but maybe later, the value is converted to string by comparing with other string or, maybe some concatenation.

Array of digit as string to simple int conversion problem

3 Answers3