librrd4.symbols: Updated for 1.4~rc2+200909281.

[pkg-rrdtool.git] / doc / bin_dec_hex.1

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.\"
.IX Title "BIN_DEC_HEX 1"
.TH BIN_DEC_HEX 1 "2009-07-12" "1.3.999" "rrdtool"
.\" For nroff, turn off justification.  Always turn off hyphenation; it makes
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.nh
.SH "NAME"
bin_dec_hex \- How to use binary, decimal, and hexadecimal notation.
.SH "DESCRIPTION"
.IX Header "DESCRIPTION"
Most people use the decimal numbering system. This system uses ten
symbols to represent numbers. When those ten symbols are used up, they
start all over again and increment the position to the left. The
digit 0 is only shown if it is the only symbol in the sequence, or if
it is not the first one.
.PP
If this sounds cryptic to you, this is what I've just said in numbers:
.PP
.Vb 10
\&     0
\&     1
\&     2
\&     3
\&     4
\&     5
\&     6
\&     7
\&     8
\&     9
\&    10
\&    11
\&    12
\&    13
.Ve
.PP
and so on.
.PP
Each time the digit nine is incremented, it is reset to 0 and the
position before (to the left) is incremented (from 0 to 1). Then
number 9 can be seen as \*(L"00009\*(R" and when we should increment 9, we
reset it to zero and increment the digit just before the 9 so the
number becomes \*(L"00010\*(R". Leading zeros we don't write except if it is
the only digit (number 0). And of course, we write zeros if they occur
anywhere inside or at the end of a number:
.PP
.Vb 1
\& "00010" \-> " 0010" \-> " 010" \-> "  10", but not "  1 ".
.Ve
.PP
This was pretty basic, you already knew this. Why did I tell it?
Well, computers usually do not represent numbers with 10 different
digits. They only use two different symbols, namely \*(L"0\*(R" and \*(L"1\*(R". Apply
the same rules to this set of digits and you get the binary numbering
system:
.PP
.Vb 10
\&     0
\&     1
\&    10
\&    11
\&   100
\&   101
\&   110
\&   111
\&  1000
\&  1001
\&  1010
\&  1011
\&  1100
\&  1101
.Ve
.PP
and so on.
.PP
If you count the number of rows, you'll see that these are again 14
different numbers. The numbers are the same and mean the same as in
the first list, we just used a different representation. This means
that you have to know the representation used, or as it is called the
numbering system or base.  Normally, if we do not explicitly specify
the numbering system used, we implicitly use the decimal system. If we
want to use any other numbering system, we'll have to make that
clear. There are a few widely adopted methods to do so. One common
form is to write 1010(2) which means that you wrote down a number in
its binary representation. It is the number ten. If you would write
1010 without specifying the base, the number is interpreted as one
thousand and ten using base 10.
.PP
In books, another form is common. It uses subscripts (little
characters, more or less in between two rows). You can leave out the
parentheses in that case and write down the number in normal
characters followed by a little two just behind it.
.PP
As the numbering system used is also called the base, we talk of the
number 1100 base 2, the number 12 base 10.
.PP
Within the binary system, it is common to write leading zeros. The
numbers are written down in series of four, eight or sixteen depending
on the context.
.PP
We can use the binary form when talking to computers
(...programming...), but the numbers will have large
representations. The number 65'535 (often in the decimal system a ' is
used to separate blocks of three digits for readability) would be
written down as 1111111111111111(2) which is 16 times the digit 1.
This is difficult and prone to errors. Therefore, we usually would use
another base, called hexadecimal. It uses 16 different symbols. First
the symbols from the decimal system are used, thereafter we continue
with alphabetic characters. We get 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,
A, B, C, D, E and F. This system is chosen because the hexadecimal
form can be converted into the binary system very easily (and back).
.PP
There is yet another system in use, called the octal system. This was
more common in the old days, but is not used very often anymore. As
you might find it in use sometimes, you should get used to it and
we'll show it below. It's the same story as with the other
representations, but with eight different symbols.
.PP
.Vb 4
\& Binary      (2)
\& Octal       (8)
\& Decimal     (10)
\& Hexadecimal (16)
\&
\& (2)    (8) (10) (16)
\& 00000   0    0    0
\& 00001   1    1    1
\& 00010   2    2    2
\& 00011   3    3    3
\& 00100   4    4    4
\& 00101   5    5    5
\& 00110   6    6    6
\& 00111   7    7    7
\& 01000  10    8    8
\& 01001  11    9    9
\& 01010  12   10    A
\& 01011  13   11    B
\& 01100  14   12    C
\& 01101  15   13    D
\& 01110  16   14    E
\& 01111  17   15    F
\& 10000  20   16   10
\& 10001  21   17   11
\& 10010  22   18   12
\& 10011  23   19   13
\& 10100  24   20   14
\& 10101  25   21   15
.Ve
.PP
Most computers used nowadays are using bytes of eight bits. This means
that they store eight bits at a time. You can see why the octal system
is not the most practical for that: You'd need three digits to represent
the eight bits and this means that you'd have to use one complete digit
to represent only two bits (2+3+3=8). This is a waste. For hexadecimal
digits, you need only two digits which are used completely:
.PP
.Vb 2
\& (2)      (8)  (10) (16)
\& 11111111 377  255   FF
.Ve
.PP
You can see why binary and hexadecimal can be converted quickly: For
each hexadecimal digit there are exactly four binary digits.  Take a
binary number: take four digits from the right and make a hexadecimal
digit from it (see the table above). Repeat this until there are no
more digits. And the other way around: Take a hexadecimal number. For
each digit, write down its binary equivalent.
.PP
Computers (or rather the parsers running on them) would have a hard
time converting a number like 1234(16). Therefore hexadecimal numbers
are specified with a prefix. This prefix depends on the language
you're writing in. Some of the prefixes are \*(L"0x\*(R" for C, \*(L"$\*(R" for
Pascal, \*(L"#\*(R" for \s-1HTML\s0.  It is common to assume that if a number starts
with a zero, it is octal. It does not matter what is used as long as
you know what it is. I will use \*(L"0x\*(R" for hexadecimal, \*(L"%\*(R" for binary
and \*(L"0\*(R" for octal.  The following numbers are all the same, just their
representation (base) is different: 021 0x11 17 \f(CW%00010001\fR
.PP
To do arithmetics and conversions you need to understand one more thing.
It is something you already know but perhaps you do not \*(L"see\*(R" it yet:
.PP
If you write down 1234, (no prefix, so it is decimal) you are talking
about the number one thousand, two hundred and thirty four. In sort of
a formula:
.PP
.Vb 4
\& 1 * 1000 = 1000
\& 2 *  100 =  200
\& 3 *   10 =   30
\& 4 *    1 =    4
.Ve
.PP
This can also be written as:
.PP
.Vb 4
\& 1 * 10^3
\& 2 * 10^2
\& 3 * 10^1
\& 4 * 10^0
.Ve
.PP
where ^ means \*(L"to the power of\*(R".
.PP
We are using the base 10, and the positions 0,1,2 and 3.
The right-most position should \s-1NOT\s0 be multiplied with 10. The second
from the right should be multiplied one time with 10. The third from
the right is multiplied with 10 two times. This continues for whatever
positions are used.
.PP
It is the same in all other representations:
.PP
0x1234 will be
.PP
.Vb 4
\& 1 * 16^3
\& 2 * 16^2
\& 3 * 16^1
\& 4 * 16^0
.Ve
.PP
01234 would be
.PP
.Vb 4
\& 1 * 8^3
\& 2 * 8^2
\& 3 * 8^1
\& 4 * 8^0
.Ve
.PP
This example can not be done for binary as that system only uses two
symbols. Another example:
.PP
\&\f(CW%1010\fR would be
.PP
.Vb 4
\& 1 * 2^3
\& 0 * 2^2
\& 1 * 2^1
\& 0 * 2^0
.Ve
.PP
It would have been easier to convert it to its hexadecimal form and
just translate \f(CW%1010\fR into 0xA. After a while you get used to it. You will
not need to do any calculations anymore, but just know that 0xA means 10.
.PP
To convert a decimal number into a hexadecimal you could use the next
method. It will take some time to be able to do the estimates, but it
will be easier when you use the system more frequently. We'll look at
yet another way afterwards.
.PP
First you need to know how many positions will be used in the other
system. To do so, you need to know the maximum numbers you'll be
using. Well, that's not as hard as it looks. In decimal, the maximum
number that you can form with two digits is \*(L"99\*(R". The maximum for
three: \*(L"999\*(R". The next number would need an extra position. Reverse
this idea and you will see that the number can be found by taking 10^3
(10*10*10 is 1000) minus 1 or 10^2 minus one.
.PP
This can be done for hexadecimal as well:
.PP
.Vb 4
\& 16^4 = 0x10000 = 65536
\& 16^3 =  0x1000 =  4096
\& 16^2 =   0x100 =   256
\& 16^1 =    0x10 =    16
.Ve
.PP
If a number is smaller than 65'536 it will fit in four positions.
If the number is bigger than 4'095, you must use position 4.
How many times you can subtract 4'096 from the number without going below
zero is the first digit you write down. This will always be a number
from 1 to 15 (0x1 to 0xF). Do the same for the other positions.
.PP
Let's try with 41'029. It is smaller than 16^4 but bigger than 16^3\-1. This
means that we have to use four positions.
We can subtract 16^3 from 41'029 ten times without going below zero.
The left-most digit will therefore be \*(L"A\*(R", so we have 0xA????.
The number is reduced to 41'029 \- 10*4'096 = 41'029\-40'960 = 69.
69 is smaller than 16^3 but not bigger than 16^2\-1. The second digit
is therefore \*(L"0\*(R" and we now have 0xA0??.
69 is smaller than 16^2 and bigger than 16^1\-1. We can subtract 16^1
(which is just plain 16) four times and write down \*(L"4\*(R" to get 0xA04?.
Subtract 64 from 69 (69 \- 4*16) and the last digit is 5 \-\-> 0xA045.
.PP
The other method builds up the number from the right. Let's try 41'029
again.  Divide by 16 and do not use fractions (only whole numbers).
.PP
.Vb 4
\& 41\*(Aq029 / 16 is 2\*(Aq564 with a remainder of 5. Write down 5.
\& 2\*(Aq564 / 16 is 160 with a remainder of 4. Write the 4 before the 5.
\& 160 / 16 is 10 with no remainder. Prepend 45 with 0.
\& 10 / 16 is below one. End here and prepend 0xA. End up with 0xA045.
.Ve
.PP
Which method to use is up to you. Use whatever works for you.  I use
them both without being able to tell what method I use in each case,
it just depends on the number, I think. Fact is, some numbers will
occur frequently while programming. If the number is close to one I am
familiar with, then I will use the first method (like 32'770 which is
into 32'768 + 2 and I just know that it is 0x8000 + 0x2 = 0x8002).
.PP
For binary the same approach can be used. The base is 2 and not 16,
and the number of positions will grow rapidly. Using the second method
has the advantage that you can see very easily if you should write down
a zero or a one: if you divide by two the remainder will be zero if it
is an even number and one if it is an odd number:
.PP
.Vb 10
\& 41029 / 2 = 20514 remainder 1
\& 20514 / 2 = 10257 remainder 0
\& 10257 / 2 =  5128 remainder 1
\&  5128 / 2 =  2564 remainder 0
\&  2564 / 2 =  1282 remainder 0
\&  1282 / 2 =   641 remainder 0
\&   641 / 2 =   320 remainder 1
\&   320 / 2 =   160 remainder 0
\&   160 / 2 =    80 remainder 0
\&    80 / 2 =    40 remainder 0
\&    40 / 2 =    20 remainder 0
\&    20 / 2 =    10 remainder 0
\&    10 / 2 =     5 remainder 0
\&     5 / 2 =     2 remainder 1
\&     2 / 2 =     1 remainder 0
\&     1 / 2 below 0 remainder 1
.Ve
.PP
Write down the results from right to left: \f(CW%1010000001000101\fR
.PP
Group by four:
.PP
.Vb 4
\& %1010000001000101
\& %101000000100 0101
\& %10100000 0100 0101
\& %1010 0000 0100 0101
.Ve
.PP
Convert into hexadecimal: 0xA045
.PP
Group \f(CW%1010000001000101\fR by three and convert into octal:
.PP
.Vb 8
\& %1010000001000101
\& %1010000001000 101
\& %1010000001 000 101
\& %1010000 001 000 101
\& %1010 000 001 000 101
\& %1 010 000 001 000 101
\& %001 010 000 001 000 101
\&    1   2   0   1   0   5 \-\-> 0120105
\&
\& So: %1010000001000101 = 0120105 = 0xA045 = 41029
\& Or: 1010000001000101(2) = 120105(8) = A045(16) = 41029(10)
\& Or: 1010000001000101(2) = 120105(8) = A045(16) = 41029
.Ve
.PP
At first while adding numbers, you'll convert them to their decimal
form and then back into their original form after doing the addition.
If you use the other numbering system often, you will see that you'll
be able to do arithmetics directly in the base that is used.
In any representation it is the same, add the numbers on the right,
write down the right-most digit from the result, remember the other
digits and use them in the next round. Continue with the second digit
from the right and so on:
.PP
.Vb 1
\&    %1010 + %0111 \-\-> 10 + 7 \-\-> 17 \-\-> %00010001
.Ve
.PP
will become
.PP
.Vb 10
\&    %1010
\&    %0111 +
\&     ||||
\&     |||+\-\- add 0 + 1, result is 1, nothing to remember
\&     ||+\-\-\- add 1 + 1, result is %10, write down 0 and remember 1
\&     |+\-\-\-\- add 0 + 1 + 1(remembered), result = 0, remember 1
\&     +\-\-\-\-\- add 1 + 0 + 1(remembered), result = 0, remember 1
\&            nothing to add, 1 remembered, result = 1
\& \-\-\-\-\-\-\-\-
\&   %10001 is the result, I like to write it as %00010001
.Ve
.PP
For low values, try to do the calculations yourself, then check them with
a calculator. The more you do the calculations yourself, the more you'll
find that you didn't make mistakes. In the end, you'll do calculi in
other bases as easily as you do them in decimal.
.PP
When the numbers get bigger, you'll have to realize that a computer is
not called a computer just to have a nice name. There are many
different calculators available, use them. For Unix you could use \*(L"bc\*(R"
which is short for Binary Calculator. It calculates not only in
decimal, but in all bases you'll ever want to use (among them Binary).
.PP
For people on Windows:
Start the calculator (start\->programs\->accessories\->calculator)
and if necessary click view\->scientific. You now have a scientific
calculator and can compute in binary or hexadecimal.
.SH "AUTHOR"
.IX Header "AUTHOR"
I hope you enjoyed the examples and their descriptions. If you do, help
other people by pointing them to this document when they are asking
basic questions. They will not only get their answer, but at the same
time learn a whole lot more.
.PP
Alex van den Bogaerdt  <alex@vandenbogaerdt.nl>