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javascripthexoctal
theking963
theking963

Reputation: 2216

Converting octal and hexadecimal numbers to base 10

I am trying to understand javascript octal and hexadecimal computations. I know I can use parseInt(string, radix) to get the Integer value.

For example, when I try this why are the values different? 

var octal = parseInt('026', 8); 
var octal1 = parseInt('030', 8); 
alert(octal); //22
alert(octal1); //24    

var hex = parseInt('0xf5', 16); 
var hex1 = parseInt('0xf8', 16); 
alert(hex); //245
alert(hex1); //248

But if I try to save it in an array why are the answers different and incorrect?

var x = new Array(026, 5, 0xf5, "abc");
var y = new Array(030, 3, 0xf8, "def");

alert('026 is ' + parseInt(x[0],8)); //18
alert('0xf5 is ' + parseInt(x[2],16)); //581
alert('030 is ' + parseInt(y[0],8)); //20
alert('0xf8 is ' + parseInt(y[2],16)); //584

Upvotes: 9

Views: 10598

Answers (3)

Anderson Green
Anderson Green

Reputation: 31800

I've written a function that can convert a number from one base to another, where the starting base and the new base are specified as parameters.

function convertFromBaseToBase(str, fromBase, toBase){
    var num = parseInt(str, fromBase);
    return num.toString(toBase);
}

alert(convertFromBaseToBase(10, 2, 10));

http://jsfiddle.net/jarble/p9ZLa/

Upvotes: 3

Phrogz
Phrogz

Reputation: 303146

0xf5 seen outside a string is the integer 245; it is not a string that needs to be parsed:

console.log( "0xf5" );              //-> 0xf5
console.log( parseInt("0xf5",16) ); //-> 245
console.log( 0xf5 );                //-> 245
console.log( parseInt(0xf5,16)  );  //-> 581
console.log( parseInt("245",16) );  //-> 581
console.log( parseInt(245,16)   );  //-> 581

Similarly, 026 outside a string is the integer 22:

console.log( 026 );                 //-> 22

What you're doing is passing already-converted-to-base10 integers to parseInt(), which then converts them to strings and interprets the base10 as base16, performing the base conversion a second time.

Upvotes: 0

Esailija
Esailija

Reputation: 140210

parseInt converts the argument to string before parsing it:

  1. Let inputString be ToString(string)

So:

0x35.toString() // "53"
parseInt( "53", 16 ); //83
parseInt( 0x35, 16 ); //83

What you have in your arrays are mostly already numbers so there is no need to parse them. You will get expected results if you change them to strings:

var x = new Array("026", "5", "0xf5", "abc");
var y = new Array("030", "3", "0xf8", "def");

Upvotes: 7

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