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# how can I change 10 bit data into 2 bit data

it means change a dec number into a bin number
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OdinXp2000
1 Solution

Commented:
We're not allowed to answer homework questions, unfortunately, so I doubt you're going to get a reply to this one.
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Commented:
What do you mean by change 10 bit data into 2 bit data?
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Commented:
throw 8 bits away. But why?
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Commented:
keep the most significant 2 bits. This can be done for images, for example. Convert 256 to 16 color depth, just keep the top bits and its a rough way to do that...
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Commented:
Hello OdinXP2000

(There's some spacing in this posting, I hope it'll survive. Otherwise copy/paste and read it in notepad or something)

Homework or no homework assignment, it's a question of understanding different radix-systems. (i.e. decimal, hex, binary etc.) and that's pretty important to learn, no matter where. So please excuse this rather long post...

The way numbers usually work when we deal with the decimal system is the following:

********************
** Decimal system **
********************

We have ten symbols, digits, called 0-9 and any given number is made up as a sequence of these digits.

A (positive, whole) number, can be considered as a sequence of digits

[ a_n, a_n-1, ... , a4,a3,a2,a1,a0 ]

Now, every digit in itself only ranges from 0 to 9 but what's important is -where- the digit is placed in the sequence. The -decimal- system assigns each digit a power of 10, where the actual value of the number can be computed as the sum of

digits[i] * 10^i
^
|
base of number system

where i runs from n downto 0.

example:

the number 14332 (d(ecimal))

has the digit sequence, n=4..0

[1,4,3,3,2]

We calculate the value by performing

1 * 10^4 + 4 * 10^3 + 3 * 10^2 + 3 * 10^1 + 2 * 10^0
^      ^
digit  exponent

= 14332 d

**************************
** What's binary then? ***
**************************

The binary system is absolutely the same, except for the usage of powers of 2 instead of 10, and instead of digits 0-9  we only use 0-1, now simply called "bits" instead of digits.

So we just sum

bits[i] * 2^i

from i = n downto 0

Example:

1011001 b(inary)

1 * 2^6 + 0*2^5 + 1*2^4 + 1*2^3 + 0*2^2 + 0*2^1 + 1*2^0

= 89 d

-------------

So, that'll give you the possibility to make your computer program read and understand any given number in ANY radix, be it binary, hex or decimal.

**********************************************************
* I know that! I need to convert a number the OTHER way **
**********************************************************

The way you need to go about is using division and
modulus with the given power of the desired number-system.

Example: You need to write the number

123 d

in binary.

First, you'd would retrieve the lowest power by dividing the number by 2, i.e. the base of the binary number system.

123 div 2 = 61

and the modulus (the remainder of the division) would be

123 mod 2 = 1

There. That's your (least significant) digit. We'll put it in the end of the sequence... 1---+
|

.......1        there :)

And now we proceed to handle the next bit by performing division again (this time with the quotient left by the last division)

61 div 2 = 30, 61 mod 2 = 1
|
we get another digit       +-----+
|
......11

go again...

30 div 2 = 15, 30 mod 2 = 0
|
+----+
|
.....011

and so forth, until you have zero as quotient, where the algorithm stops to get the number

1111011 b

--algorithm-- DEC->BIN ----

n is given number

digitlist = [0,0,0,...,0] (just some integer list)
0->index

as long as n > 0
[
divide n by 2 and store in "q", the remainder in "r"
place "r" as a digit in the list index 0
increase index by 1
set n = q
] repeat loop

-------------------------

**************
** Sidenode **
**************

(Mathematically, it is not relevant how you perform divisions and such, but for a computer efficient routines can be written. Especially when dealing with binary numbers where division (by 2, 4, 8, etc.) can be replaced by BIT-SHIFTS. Don't bother for now.)

I hope this was of some help to you,
Anders Rasmussen.
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Commented:
example:

#include <stdio.h>

void main () {
char InputString[] = "1047";
char OutputString[5];
int Value;
int Ret;

/* Convert decimal string to internal format*/
/* internaly its binary*/
Ret = sscanf ( InputString, "%d", &Value);
if ( Ret != 1 ) { exit -1; }

/* Convery internal representation binary to binary string */

}

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Author Commented:
Thank all of you very much!
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