# converting binary to decimal

Page 1 of 1## 10 Replies - 6639 Views - Last Post: 19 August 2008 - 02:18 AM

### #1

# converting binary to decimal

Posted 06 August 2008 - 11:01 PM

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**Replies To:** converting binary to decimal

### #2

## Re: converting binary to decimal

Posted 06 August 2008 - 11:29 PM

### #3

## Re: converting binary to decimal

Posted 07 August 2008 - 07:45 AM

String binaryString = "1010"; System.out.println(Integer.parseInt(binaryString, 2));

Post "convert decimal to binary"

int i = 10; String binaryString = Integer.toBinaryString(i); System.out.println(binaryString);

This code is not what you want but I guess but I guess it would help

This post has been edited by **lordms12**: 07 August 2008 - 07:47 AM

### #4

## Re: converting binary to decimal

Posted 07 August 2008 - 01:13 PM

Anyway:

Each bit in a binary number is a power of 2, starting with 2^0. So the nTH bit represents the value of 2^n, where n>=0 and we count n from right to left.

Therefore:

0101 = 0*2^3 + 1*2^2 + 0*2^1 + 1*2^0 = 0 + 4 + 0 + 1 = 5

Now surely you can turn this into an algorithm?

### #5

## Re: converting binary to decimal

Posted 13 August 2008 - 07:57 AM

This post has been edited by **eradcoil**: 13 August 2008 - 08:02 AM

### #6

## Re: converting binary to decimal

Posted 13 August 2008 - 11:15 AM

If so, then I hope this will prove helpful...its cool either way:

in order to convert a decimal number into a binary one, simply divide by two repeatedly, take the remainders, and write them in reverse order.

10 / 2 = 5 remainder =

**0**

5 / 2 = 2 remainder =

**1**

2/2 = 1 remainder =

**0**

1/2 = 0 remainder =

**1**

when you reach zero, you are done, and the answer is the remainders from bottom up :

**1010**

### #7

## Re: converting binary to decimal

Posted 18 August 2008 - 05:45 AM

dobbersp, on 13 Aug, 2008 - 11:15 AM, said:

If so, then I hope this will prove helpful...its cool either way:

in order to convert a decimal number into a binary one, simply divide by two repeatedly, take the remainders, and write them in reverse order.

10 / 2 = 5 remainder =

**0**

5 / 2 = 2 remainder =

**1**

2/2 = 1 remainder =

**0**

1/2 = 0 remainder =

**1**

when you reach zero, you are done, and the answer is the remainders from bottom up :

**1010**

42 / 2 = 21, remainder =

**0**

21 / 2 = 10, remainder =

**1**

10 / 2 = 5, remainder =

**0**

5 / 2 = 2, remainder =

**1**

2/2 = 1, remainder =

**0**

1/2 = 0, remainder =

**1**

42 = 101010

Well what do you know, it works :-)

Let's see if I can do the same for octal:

42 / 8 = 5, remainder =

**2**

5 / 8 = 0, remainder =

**5**

42 = 52

And for hexadecimal:

42 / 16 = 2, remainder =

**10**=

**A**

2 / 16 = 0, remainder =

**2**

42 = 2A

---

However, the original post was about converting binary to decimal, not the other way around (still good to know though).

I also wrote a tutorial about it:

http://www.dreaminco...wtopic59949.htm

This post has been edited by **JeroenFM**: 18 August 2008 - 05:48 AM

### #8

## Re: converting binary to decimal

Posted 18 August 2008 - 07:56 AM

patjoy_24, on 6 Aug, 2008 - 11:01 PM, said:

Do you mean mathematically as in the manual process of pen and paper?

or for java iprogramming itself

sites

www.java-tutorials.com

www. java-coffebreak.com something like that just search thru good sites that give programing tips

### #9

## Re: converting binary to decimal

Posted 18 August 2008 - 08:03 AM

patjoy_24, on 6 Aug, 2008 - 11:01 PM, said:

Oh not bad

manually guys here have given you a great way of solving it

just divide by base 2 for Dec-Bin

for Bin-Dec ^2 i.e 0011 ----------> 1*2^0 + 1*2^1 + 0*2^2 etc

or the alternative is use the conversion table which is the best short cut once mastered

### #10

## Re: converting binary to decimal

Posted 18 August 2008 - 08:08 AM

dobbersp, on 13 Aug, 2008 - 11:15 AM, said:

If so, then I hope this will prove helpful...its cool either way:

in order to convert a decimal number into a binary one, simply divide by two repeatedly, take the remainders, and write them in reverse order.

10 / 2 = 5 remainder =

**0**

5 / 2 = 2 remainder =

**1**

2/2 = 1 remainder =

**0**

1/2 = 0 remainder =

**1**

when you reach zero, you are done, and the answer is the remainders from bottom up :

**1010**

just to simplify you forgot to add remainders >=5 = 1

<=5 = 0

good stuff

### #11

## Re: converting binary to decimal

Posted 19 August 2008 - 02:18 AM

Ak-Emm, on 18 Aug, 2008 - 08:08 AM, said:

<=5 = 0

good stuff

Clearly you didn't understand his method, the remainder of

**any positive number**divided by 2 is always either 0 or 1, so it's always less than 5.

Try this:

public int remainderOfDiv2(int number) { return number % 2; }

Pass any positive number, if you find one that yields something other than 1 or 0 let me know.

This post has been edited by **JeroenFM**: 19 August 2008 - 02:22 AM