Espresso: A Concentrated Introduction to Java

**Summary:** We consider the underlying binary representation
of numbers.

**Prerequisites:**
Numeric Values in Java.

**Important Classes:**

**Contents:**

- Background: Representing Values
- Basic Concepts Behind the Binary System
- Binary Addition
- Binary Multiplication
- Binary Division
- Decimal to Binary
- Negation in the Binary System
- Representing Integers in Java
- Revealing Representations
- Bitwise Operations
- Answers

As you may have noted from the lab on numeric values in Java, you get some fairly odd results when you add to the largest integer (or long) or subtract from the smallest integer (or long). Why? Because Java has chosen an interesting way to represent these values.

Behind the scenes on almost every computer, every value is represented
as a series of bits (0's and 1's). Computers use such a two-value system
because it is easy to represent physically or electronically. However,
the two-value system means that designers of languages and computers must
figure out how to convert from and to other represenations (such as the
traditional

representation of integers). In the next
few sections, we will consider some basic issues of binary numbers and
the representation of integers in binary. We will then return to the
details of these representations in Java.

To understand binary numbers, begin by recalling elementary school math. When we first learned about numbers, we were taught that, in the decimal system, things are organized into columns:

H | T | O 1 | 9 | 3

such that "H" is the hundreds column, "T" is the tens column, and "O"
is the ones column. So the number "193" is 1-hundreds plus 9-tens
plus 3-ones. Years later, we learned that the ones column meant 10^{0},
the tens column meant 10^{1}, the hundreds column 10^{2} and so on, such that

10^{2}|10^{1}|10^{0}1^{ }| 9^{ }| 3^{ }

the number 193 is really [(1*10^{2})+(9*10^{1})+(3*10^{0})].

As you know, the decimal system uses the digits 0-9 to represent numbers.
If we wanted to put a larger number in column 10^{n} (e.g.,
10), we would have to multiply 10*10^{n}, which would give
10^{(n+1)}, and be carried a column to the left. For example,
putting ten in the 10^{0} column is impossible, so we put
a 1 in the 10^{1} column, and a 0 in the 10^{0}
column, thus using two columns. Twelve would be 12*10^{0},
or 10^{0}(10+2), or 10^{1}+2*10^{0}, which also
uses an additional column to the left (12).

The binary system works under the exact same principles as the decimal system, only it operates in base 2 rather than base 10. In other words, instead of columns being

10^{2}|10^{1}|10^{0}

they are

2^{2}|2^{1}|2^{0}

Instead of using the digits 0-9, we only use 0-1 (again, if we used anything larger it would be like multiplying 2*2^{n} and getting 2^{(n+1)}, which would not fit in the 2^{n} column. Therefore, it would shift you one column to the left. For example, "3" in binary cannot be put into one column. The first column we fill is the right-most column, which is 2^{0}, or 1. Since 3>1, we need to use an extra column to the left, and indicate it as "11" in binary (1*2^{1}) + (1*2^{0}).

**What would the binary number 1011 be in decimal notation?**

**Try converting these numbers from binary to decimal**

- 10
- 111
- 10101
- 11110

Remember:

2^{4}| 2^{3}| 2^{2}| 2^{1}| 2^{0}^{ }|^{ }|^{ }| 1^{ }| 0^{ }|^{ }| 1^{ }| 1^{ }| 1 1^{ }| 0^{ }| 1^{ }| 0^{ }| 1 1^{ }| 1^{ }| 1^{ }| 1^{ }| 0

Consider the addition of decimal numbers:

23 +48 ___

We begin by adding 3+8=11. Since 11 is greater than 10, a one is put into the 10's column (carried), and a 1 is recorded in the one's column of the sum. Next, add [(2+4) +1] (the one is from the carry)=7, which is put in the 10's column of the sum. Thus, the answer is 71.

Binary addition works on the same principle, but the numerals are different. Begin with one-bit binary addition:

0 0 1 +0 +1 +0 ___ ___ ___ 0 1 1

1+1 carries us into the next column. In decimal form, 1+1=2. In binary,
any digit higher than 1 puts us a column to the left (as would 10 in
decimal notation). The decimal number "2" is written in binary notation
as "10" (1*2^{1})+(0*2^{0}). Record the 0 in the ones column, and carry the
1 to the twos column to get an answer of "10." In our vertical notation,

1 +1 ___ 10

The process is the same for multiple-bit binary numbers:

1010 +1111 ______

- Step one:

Column 2^{0}: 0+1=1.

Record the 1.

Temporary Result: 1; Carry: 0 - Step two:

Column 2^{1}: 1+1=10.

Record the 0, carry the 1.

Temporary Result: 01; Carry: 1 - Step three:

Column 2^{2}: 1+0=1 Add 1 from carry: 1+1=10.

Record the 0, carry the 1.

Temporary Result: 001; Carry: 1 - Step four:

Column 2^{3}: 1+1=10. Add 1 from carry: 10+1=11.

Record the 11.

Final result: 11001

Alternately:

11 (carry) 1010 +1111 ______ 11001

**Always remember**

- 0+0=0
- 1+0=1
- 1+1=10

Try a few examples of binary addition:

111 101 111 +110 +111 +111 ______ _____ _____

Click here to see the answer

Multiplication in the binary system works the same way as in the decimal system

For one-digit values:

- 1*1=1
- 1*0=0
- 0*1=0

- For multiple digit values, we use our standard technique
101 * 11 ____ 101 1010 _____ 1111

Note that multiplying by two is extremely easy. To multiply by two, just add a 0 on the end.

Follow the same rules as in decimal division. For the sake of simplicity, throw away the remainder.

For Example: 111011/11

10011 r 10 _______ 11)111011 -11 ______ 101 -11 ______ 101 11 ______ 10

Converting from decimal to binary notation is slightly more difficult
conceptually, but can easily be done once you know how through the use
of algorithms. Begin by thinking of a few examples. We can easily see
that the number 3= 2+1. and that this is equivalent to (1*2^{1})+(1*2^{0}).
This translates into putting a "1" in the 2^{1} column and a "1" in the
2^{0} column, to get "11". Almost as intuitive is the number 5: it is
obviously 4+1, which is the same as saying [(2*2) +1], or 2^{2}+1. This can
also be written as [(1*2^{2})+(1*2^{0})]. Looking at this in columns,

2^{2}| 2^{1}| 2^{0}1^{ }| 0^{ }| 1

or 101.

What we're doing here is finding the largest power of two within the number (2^{2}=4 is the largest power of 2 in 5), subtracting that from the number (5-4=1), and finding the largest power of 2 in the remainder (2^{0}=1 is the largest power of 2 in 1). Then we just put this into columns. This process continues until we have a remainder of 0. Let's take a look at how it works. We know that:

2^{0}=1 2^{1}=2 2^{2}=4 2^{3}=8 2^{4}=16 2^{5}=32 2^{6}=64 2^{7}=128

and so on. To convert the decimal number 75 to binary, we would find the largest power of 2 less than 75, which is 64. Thus, we would put a 1 in the 2^{6} column, and subtract 64 from 75, giving us 11. The largest power of 2 in 11 is 8, or 2^{3}. Put 1 in the 2^{3} column, and 0 in 2^{4} and 2^{5}. Subtract 8 from 11 to get 3. Put 1 in the 2^{1} column, 0 in 2^{2}, and subtract 2 from 3. We're left with 1, which goes in 2^{0}, and we subtract one to get zero. Thus, our number is 1001011.

Making this algorithm a bit more formal gives us:

Let D=number we wish to convert from decimal to binary Repeat until D=0 a. Find the largest power of two in D. Let this equal P. b. Put a 1 in binary column P. c. Subtract P from D. Put zeros in all columns which don't have ones.

This algorithm is a bit awkward. Particularly step 3, "filling in the zeros." Therefore, we should rewrite it such that we ascertain the value of each column individually, putting in 0's and 1's as we go:

1. Let D= the number we wish to convert from decimal to binary 2. Find P, such that 2^{P}is the largest power of two smaller than D. 3. Repeat until P<0 a. If 2^{P}<=D then i. Put 1 into column P ii. Subtract 2^{P}from D b. Else i. Put 0 into column P c. Subtract 1 from P

Now that we have an algorithm, we can use it to convert numbers from decimal to binary relatively painlessly. Let's try the number D=55.

- Our first step is to find P. We know that 2
^{4}=16, 2^{5}=32, and 2^{6}=64. Therefore, P=5. - 2
^{5}<=55, so we put a 1 in the 2^{5}column:`1-----`

. - Subtracting 55-32 leaves us with 23. Subtracting 1 from P gives us 4.
- Following step 3 again, 2
^{4}<=23, so we put a 1 in the 2^{4}column:`11----`

. - Next, subtract 16 from 23, to get 7. Subtract 1 from P gives us 3.
- 2
^{3}>7, so we put a 0 in the 2^{3}column:`110---`

- Next, subtract 1 from P, which gives us 2.
- 2
^{2}<=7, so we put a 1 in the 2^{2}column:`1101--`

- Subtract 4 from 7 to get 3. Subtract 1 from P to get 1.
- 2
^{1}<=3, so we put a 1 in the 2^{1}column:`11011-`

- Subtract 2 from 3 to get 1. Subtract 1 from P to get 0.
- 2
^{0}<=1, so we put a 1 in the 2^{0}column:`110111`

- Subtract 1 from 1 to get 0. Subtract 1 from P to get -1.
- P is now less than zero, so we stop.

However, this is not the only approach possible. We can start at the right, rather than the left.

All binary numbers are in the form

a[n]*2^{n}+ a[n-1]*2^{(n-1)}+...+a[1]*2^{1}+ a[0]*2^{0}

where each a[i] is either a 1 or a 0 (the only possible digits for the binary system). The only way a number can be odd is if it has a 1 in the 2^{0} column, because all powers of two greater than 0 are even numbers (2, 4, 8, 16...). This gives us the rightmost digit as a starting point.

Now we need to do the remaining digits. One idea is to "shift" them. It is also easy to see that multiplying and dividing by 2 shifts everything by one column: two in binary is 10, or (1*2^{1}). Dividing (1*2^{1}) by 2 gives us (1*2^{0}), or just a 1 in binary. Similarly, multiplying by 2 shifts in the other direction: (1*2^{1})*2=(1*2^{2}) or 10 in binary. Therefore

{a[n]*2^{n}+ a[n-1]*2^{(n-1)}+ ... + a[1]*2^{1}+ a[0]*2^{0}}/2

is equal to

a[n]*2^{(n-1)}+ a[n-1]*2^{(n-2)}+ ... + a[1]2^{0}

Let's look at how this can help us convert from decimal to binary. Take the number 163. We know that since it is odd, there must be a 1 in the 2^{0} column (a[0]=1). We also know that it equals 162+1. If we put the 1 in the 2^{0} column, we have 162 left, and have to decide how to translate the remaining digits.

Two's column: Dividing 162 by 2 gives 81. The number 81 in binary would also have a 1 in the 2^{0} column. Since we divided the number by two, we "took out" one power of two. Similarly, the statement a[n-1]*2^{(n-1)} + a[n-2]*2^{(n-2)} + ... + a[1]*2^{0} has a power of two removed. Our "new" 2^{0} column now contains a1. We learned earlier that there is a 1 in the 2^{0} column if the number is odd. Since 81 is odd, a[1]=1. Practically, we can simply keep a "running total", which now stands at 11 (a[1]=1 and a[0]=1). Also note that a1 is essentially "remultiplied" by two just by putting it in front of a[0], so it is automatically fit into the correct column.

Four's column: Now we can subtract 1 from 81 to see what remainder we still must place (80). Dividing 80 by 2 gives 40. Therefore, there must be a 0 in the 4's column, (because what we are actually placing is a 2^{0} column, and the number is not odd).

Eight's column: We can divide by two again to get 20. This is even, so we put a 0 in the 8's column. Our running total now stands at a[3]=0, a[2]=0, a[1]=1, and a[0]=1.

We can continue in this manner until there is no remainder to place.

Let's formalize this algorithm:

1. Let D= the number we wish to convert from decimal to binary. 2. Repeat until D=0 a. If D is even, put "0" in the leftmost open column. a. Else if D is odd, put "1" in the leftmost open column, and subtract 1 from D. c. Divide D by 2.

For the number 163, this works as follows:

1. Let D=163 2. b. D is odd Put a 1 in the 2^{0}column. Subtract 1 from D to get 162. c. Divide D=162 by 2. Temporary Result: 1 New D=81 D does not equal 0, so we repeat step 2. 2. b. D is odd Put a 1 in the 2^{1}column. Subtract 1 from D to get 80. c. Divide D=80 by 2. Temporary Result: 11 New D=40 D does not equal 0, so we repeat step 2. 2. a. D is even Put a 0 in the 2^{2}column. c. Divide D by 2. Temporary Result:011 New D=20 D does not equal 0, so we repeat step 2. 2. a. D is even Put a 0 in the 2^{3}column. c. Divide D by 2. Temporary Result: 0011 New D=10 D does not equal 0, so we repeat step 2. 2. a. D is even Put a 0 in the 2^{4}column. c. Divide D by 2. Temporary Result: 00011 New D=5 D does not equal 0, so we repeat step 2. 2. b. D is odd Put a 1 in the 2^{5}column. Subtract 1 from D to get 4. c. Divide D by 2. Temporary Result: 100011 New D=2 D does not equal 0, so we repeat step 2. 2. a. D is even Put a 0 in the 2^{6}column. c. Divide D by 2. Temporary Result: 0100011 New D=1 D does not equal 0, so we repeat step 2. 2. b. D is odd Put a 1 in the 2^{7}column. Subtract 1 from D to get D=0. c. Divide D by 2. Temporary Result: 10100011 New D=0 D=0, so we are done Conclusion: the decimal number 163 is equivalent to the binary number 10100011.

Since we already knew how to convert from binary to decimal, we can easily verify our result.

- 10100011
- = (1*2
^{0})+(1*2^{1})+(1*2^{5})+(1*2^{7}) - = 1+2+32+128
- = 163.

The techniques discussed above. work well for non-negative integers, but how do we indicate negative numbers in the binary system? Before we investigate negative numbers, we note that the computer uses a fixed number of "bits" or binary digits. An 8-bit number is 8 digits long. For this section, we will work with 8 bits.

The simplest way to indicate negation is signed magnitude. In signed
magnitude, the left-most bit is not actually part of the number, but
is just the equivalent of a +/- sign. "0" indicates that the number
is positive, "1" indicates negative. In 8 bits, 00001100 would be 12
(break this down into (1*2^{3}) + (1*2^{2}) ). To indicate -12, we would
simply put a "1" rather than a "0" as the first bit: 10001100.

In one's complement, positive numbers are represented as usual in regular binary. However, negative numbers are represented differently. To negate a number, replace all zeros with ones, and ones with zeros - flip the bits. Thus, 12 would be 00001100, and -12 would be 11110011. As in signed magnitude, the leftmost bit indicates the sign (1 is negative, 0 is positive). To compute the value of a negative number, flip the bits and translate as before.

Two's complement is an interesting variant of one's complement that is more procedural than anything. You can negate a number by flipping all the bits and then adding 1 with the techniques of binary addition. (A little math will tell you that this technique works correctly when you doubly negate a number.) Begin with the number in one's complement. Add 1 if the number is negative.

In this notation, twelve would be represented as 00001100, and -12 as 11110100. To verify this, let's subtract 1 from 11110100, to get 11110011. If we flip the bits, we get 00001100, or 12 in decimal.

In excess 2^{(m-1)}, "m" indicates the total number of bits. For us (working with 8 bits), it would be excess 2^{7}. To represent a number (positive or negative) in excess 2^{7}, begin by taking the number in unsigned binary representation. Then add 2^{7} (=128) to that number. For example, 7 would be 128 + 7=135, or 2^{7}+2^{2}+2^{1}+2^{0}, and, in binary,10000111. We would represent -7 as 128-7=121, and, in binary, 01111001.

- Unless you know which representation has been used, you cannot figure out the value of a sequence of bits.
- A number in excess 2
^{(m-1)}is the same as that number in two's complement with the leftmost bit flipped.

To see the advantages and disadvantages of each method, let's try working with them.

Click here to see the answers.

Java uses two's complement to represent the various forms of integers,
with different numbers of bits for the different forms. A `byte`

has eight bits, a `short`

sixteen, an `int`

thirty-two,
and a `long`

sixty-four. Why does Java use two's complement?
Because it provides the advantage that addition is simple (you can use
the standard algorithm for positive and negative numbers), subtraction
is simple (subtraction can be implemented as negate and add

), and
because you can easily tell the sign of a number (if the leftmost bit
is 0, the number is non-negative; if the leftmost bit is 1, the number
is negative).

Now, why do we get a negative number when we add to the largest value
in each? Let's consider the largest integer,
`01111111111111111111111111111111`

. All addition is done
using the simple additional algorithm. Let's consider what happens
when we add 2.

111111111111111111111111111111(carry) 01111111111111111111111111111111 + 10 --------------------------------- 10000000000000000000000000000001

What number is that? Well, we know it's negative because it starts with a 1. Hence, we flip all the bits and then add 1 to find the corresponding positive number.

01111111111111111111111111111110 + 1 --------------------------------- 01111111111111111111111111111111

So, in this notation, 2^{31} + 2 = -2^{31}

Since the underlying representation can be important or useful, Java
provides methods in the `Integer`

class to convert

values to and from strings of 0's and 1's.
__int__

You can get the sequence of bits (without leading 0's) from
an

using
__int__`Integer.toBinaryString`

. For example,

inti = Integer.MIN_VALUE; pen.print("The representation of"); pen.print(i) pen.print("is"); pen.println(Integer.toBinaryString(i));

In the lab, we'll explore how to add the leading bits.

You can convert a string representing a sequence of bits to an
`int`

with the two-parameter `Integer.parseInt`

,
using a string of 0's and 1's as the first parameter and the integer
2 as the second parameter. For example,

String code = "00011101"; pen.print(code); pen.print("represents"); pen.println(Integer.parseInt(code, 2));

Since some programmers use the underlying bits in different ways, Java provides a wide variety of infix binary operations that manipulate those bits, including

- shift the bits left
*n*places:*i*<<*n* - shift the bits right
*n*places:*i*>>*n* -
*and*the individual bits in*x*and*y*:

. The and of 1 and 1 is 1, and the and of any other combination is 0.*x*&*y* -
*inclusive or*the individual bits in*x*and*y*:

. The inclusive or of 0 and 0 is 0, and the inclusive or of any other combination is 1.*x*|*y* -
*exclusive or*the individual bits in*x*and*y*:

. The exclusive or of two different bits is 1 and the exclusive or of two identical bits is 0. is 0, and the inclusive or of any other combination is 1.*x*|*y*

**What would the binary number 1011 be in decimal notation?**

1011=(1*2^{3})+(0*2^{2})+(1*2^{1})+(1*2^{0}) = (1*8) + (0*4) + (1*2) + (1*1) = 11 (in decimal notation)

Try converting these numbers from binary to decimal.

10=(1*2^{1}) + (0*2^{0}) = 2+0 = 2 111 = (1*2^{2}) + (1*2^{1}) + (1*2^{0}) = 4+2+1=7 10101= (1*2^{4}) + (0*2^{3}) + (1*2^{2}) + (0*2^{1}) + (1*2^{0})=16+0+4+0+1=21 11110= (1*2^{4}) + (1*2^{3}) + (1*2^{2}) + (1*2^{1}) + (0*2^{0})=16+8+4+2+0=30

**Try a few examples of binary addition:**

1 1 111 111 111 +110 +110 +110 ______ ______ _____ 1 01 1101 1 11 1 101 101 101 +111 +111 +111 _____ _____ ______ 0 00 1100 1 1 1 111 111 111 +111 +111 +111 _____ _____ _____ 0 10 1110

Click here to return to the question

Signed Magnitude: 5+12 -5+12 -12+-5 12+-12 00000101 10000101 10001100 00001100 00001100 00001100 10000101 10001100 __________ ________ ________ _________ 00010001 10010001 00010000 10011000 17 -17 16 -24 One' Complement: 00000101 11111010 11110011 00001100 00001100 00001100 11111010 11110011 _________ ________ ________ ________ 00010001 00000110 11101101 11111111 17 6 -18 0 Two's Complement: 00000101 11111011 11110100 00001100 00001100 00001100 11111011 11110100 ________ ________ ________ ________ 00010001 00000111 11101111 00000000 17 7 -17 0 Signed Magnitude: 10000101 01111011 01110100 00001100 10001100 10001100 01111011 01110100 ________ ________ ________ ________ 00010001 00000111 11101111 01111100 109 119 111 124

Click here to return to the question

Circa 1995 [Christine R. Wright]

- Original version, entitled
The Binary System: A pretty damn clear guide to a quite confusing concept by Christine R. Wright with some help from Samuel A. Rebelsky.

Monday, 6 February 2006 [Samuel A. Rebelsky]

- Reformatted.
- Added sections on Java.
- Minor editing changes.

Sunday, 5 September 2008 [Samuel A. Rebelsky]

- Corrected small typo in two's complement section.

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