JavaScript beginner: ParseInt() misunderstanding

Question

I'm self teaching myself JavaScript from videos I've dug up from around the Internet. Most tend to gloss over things quickly and leave tons of questions. So I test things within Firefox 21 using Firebug 2.0.2 to see how they behave. That's the only purpose of the following code.

Console.log output

>>> var n=parseInt("34","22","18",10);console.log(n);
70

My initial thoughts was that it parsed each of the supplied strings and then returned the sum. Then I did the actual math and realized the sum should have came out to be 74 if that were the case.

Even more interesting, and what actually prompted me to do the math on my own instead of just trusting that to be the sum without thinking. When I reduced the value of one of the numeric strings (22 to 21 in my experience anyways) returned 67 instead of the expected 69.

Console.log output

in:10>>> var n=parseInt("34","21","18",10);console.log(n);
67

What is actually happening here? Is parseInt just not designed to accommodate multiple numeric strings?

OJay · Accepted Answer · 2014-08-01 04:10:27Z

4

No parseInt does not take multiple strings, it take one string and a optional radix (i.e. what base the number is in)

Please refer to the MDN reference here

Whats happening in your last example

parseInt("34","21","18",10)

is that you are telling javascript to parse 34 in base 21 (the other parameters 18 and 10 are ignored)

34 in base 21 is 67 (i.e. 3 x 21 + 4)

EDIT: and 34 in base 22 is 70 (3 * 22 + 4). See this answer for an informative approach to convert bases: https://answers.yahoo.com/question/index?qid=20090508092544AAxAokR

If you want to add up numbers, parse them individually first, then add them

i.e.

var n1 = parseInt("34",10);
var n2 = parseInt("22",10);
var n3 = parseInt("18",10);
var result = n1 + n2 + n3;

Also if you are positive that the strings only ever contain numbers (i.e. "22", not something like "22 widgets", because parseInt will parse everything up to the first non numeric character so parseInt("22 widgets") will return 22), then you can cheat a bit, and just put a + at the start

i.e.

var n1 = +"34";
var n2 = +"22";
var n3 = +"18";
var result = n1 + n2 + n3;

same outcome

edited Aug 1, 2014 at 4:10

answered Aug 1, 2014 at 3:54

OJay

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1 Comment

SkItZ Over a year ago

Awesome thanks. Having never messed with radixes(radi?) I was was unable to recognize what it was doing. Without any error I never considered that it would interpret the second string any way other than another string to parse. Instead of 'parsing' then using it in the parseInt function lol.

user513951 · Accepted Answer · 2014-08-01 03:59:54Z

1

parseInt takes one required argument and one optional argument, and ignores everything else. The optional (second) argument is the radix, which tells the function which numerical base to use when converting the string.

From the MDN docs on parseInt:

The following examples all return 15:

parseInt(" 0xF", 16)
parseInt(" F", 16)
parseInt("17", 8)
parseInt(021, 8)
parseInt("015", 10)
parseInt(15.99, 10)
parseInt("1111", 2)
parseInt("12", 13)

So, what's happening in your example is that your first function call is interpreting the characters "34" in base 22 and returning 70, and the second function call is interpreting the characters "34" in base 21 and returning 67.

edited Aug 1, 2014 at 3:59

answered Aug 1, 2014 at 3:54

user513951

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Comments

Manjesh V · Accepted Answer · 2014-08-01 04:03:28Z

0

parseInt takes 2 arguements: 1. Numuric String 2. Radix - Number System (2 to 36)

parseInt("50",10);// Decimal equivalant 50
parseInt("10",2); // Binary equivalant 2 
parseInt("50",8); // Octal equivalant 40
parseInt("50",16); // Hexadecimal Equivalant 80

See in the above example parseInt("10",2);, (10)Binary === (2)Decimal

Refer: MDN DOCS

answered Aug 1, 2014 at 4:03

Manjesh V

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