1. Module 10 Operations on Bits

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What’s a bit/byte?
Computer memory is made up of hardware that can represent a 0 or a 1.
Each individual 0 or 1 is a bit (Binary digIT)
C provides operators for manipulating these individual bits
Bits are commonly group in sets of 8 known as a byte
Bits in a byte are numbered from right-to-left starting with bit 0
Bit 0 is known as the “least significant bit” (lsb)
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Bits and Integers
Each bit in a byte represents a power of 2
Bit 0 represents 20 = 1
Bit 1 represents 21 = 2
etc
Example: What’s the value of
= = = 3810 (in binary!!) 27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 1
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Binary Exercises
ex. 1 = 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
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5. Binary Exercises – Round 2=
117 1 86 2
158 3
201 4
136 5
250 6
179 7 23 8
100 9
200 10
231
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6. Another way of converting a Decimal to Binary
Another way of Solving
Routinely done in programming
Another way of converting a Decimal to Binary
68 ÷ 2 = 34 R 0
34 ÷ 2 = 17 R 0
17 ÷ 2 = 8 R 1
8 ÷ 2 = 4 R 0
4 ÷ 2 = 2 R 0
2 ÷ 2 = 1 R 0
1 ÷ 2 = 0 R 1
= 68 (base 10) …
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Octal
Octal = base 8
A group of 3 bits is one octal digit
Octal constants start with 0
Binary
Octal
Decimal
000
001 1
010 2
011 3
100 4
101 5
110 6
111 7
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Octal to Decimal
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Another way of Solving
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Octal in Applications
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Exercises
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Hexadecimal
Hexadecimal = base 16
A group of 4 bits is one hex digit
Hex constants start with 0x
Binary
Hex
Decimal
0000
0001 1
0010 2
0011 3
0100 4
0101 5
0110 6
0111 7
Binary
Hex
Decimal
1000 8
1001 9
1010 A 10
1011 B 11
1100 C 12
1101 D 13
1110 E 14
1111 F 15
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13. Hexadecimal to Decimal
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14. Hexadecimal in an Application
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Another way of Solving
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Exercises
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17. Why Binary , Octal and Hex?
Translate the following 32-bit integer Octal Hexadecimal A D A 5
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Two’s Complement
To store both positive and negative values, the left-most bit is the designated sign bit
Sign bit = 0 indicates a positive value
Sign bit = 1 indicates a negative value
To store a negative number, store its absolute value, add 1, then complement the result.
To store –5, first store =
Then complement =
Then add 1, =
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19. Two’s Compliment Visually
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20. Solving for Two’s Compliment
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More two’s complement
Largest value in a byte: = 127
Smallest value in a byte: = -128
-1 in a byte =
Smallest value in a 4-byte (32-bit) integer
-232 – 1 = -2,147,483,648
Largest value in 4-byte (32 bit) integer
231 – 1 = 2, 147, 483, 647
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Exercises
Integer Representation Exercises
Submit it to your “submission” folder
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Bitwise Operators
Apply to any integer type or to characters
Result is 0 or 1
Apply to an integer bit-by-bit
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Bitwise AND
& (AND)
short w1 = 25, w2 = 77;
short w3 = w1 & w2;
w1 =
w2 =
w3 = = 9 A B
A & B 1
Notes:
a & 1 = a ANDing a bit with 1 retains the value of the bit
a & 0 = 0 ANDing a bit with 0 sets the bit to 0
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Masking
Using the AND operator to select bits.
char c = 45;
c = c & 0x7; //0x means Octal
c =
0x7 =
c & 0x7 =
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Let’s Take a Look
C::B
Program 12-1
GUESS the output FIRST
Then go ahead and run to confirm your answer
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Bitwise OR
| (OR)
short w1 = 0431, w2 = 0152;
short w3 = w1 | w2;
w1 =
w2 =
w3 = = 0573 A B
A | B 1
Notes:
a | 1 = 1 ORing a bit with 1 sets the bit to 1
a | 0 = a ORing a bit with 0 retains the value of the bit
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Turning Bits “On”
Bits may be turned “on” (set to 1) by ORing
short w1 = 0431;
w1 = w1 | 07;
w1 =
07 =
w1 | 7 =
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29. Bitwise XOR
^ (XOR)
short w1 = 0536, w2 = 0266;
short w3 = w1 ^ w2;
w1 =
w2 =
w3 = = 0750 A B
A ^ B 1
Notes:
1 ^ 1 = 0 and 0 ^ 1 = 1 XORing a bit with 1 changes the bit
0 ^ 0 = 0 and 1 ^ 0 = 1 XORing a bit with 0 retains the value of the bit
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Bitwise Complement
~ (1’s complement)
short w1 = ;
short w2 = ~w1;
w1 = =
w2 = = A ~A 1
Notes:
~0 = -1 regardless of the number of bits in an int
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Let’s Take a Look
C::B- Program12-2
Predict output first, run it to confirm
Note: warnings from line 15 – suggest parens
Change to use parens around 1st two operands – different results!
Note precedence of bitwise operators
Pg 440 -> Kochan book
Redo parens around last 2 operand to get original results3
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Exercise
PP2-14.docx
Complete
Submit in your “submit” folder
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Left-Shift Operator
The << (left shift) operator literally moves bits in a value to the left. The left-most (high-order) bits are lost and 0’s replace the right most (low-order) bits which are shifted.
char c = 0x43;
char d = c << 2;
c =
d =
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Notes on Shifting
Left-shift is used to multiply by 2N
Shifting with a count that is more than 32 bits (number of bits in an int) is undefined
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Right-shift Operator
The >> (right shift) operator literally moves bits in a value to the right. The right-most (low-order) bits are lost.
If the value is “signed” (the default) then copies of the sign-bit fill in the high-order bits.
If the value is “unsigned”, then 0s fill in the high-order bits.
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Signed Right-Shift
signed short w1 = 0x9013, w2 = 0x0322; signed short w3 = w1 >> 3; signed short w4 = w2 >> 2; w1 = w3 = w2 = w4 =
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Notes on Shifting
Right-shift is used to divide by 2N
Shifting with a negative count is undefined
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Unsigned Right-Shift
unsigned short w1 = 0x9013, w2 = 0x0322;
unsigned short w3 = w1 >> 3;
unsigned short w4 = w2 >> 2;
w1 =
w3 =
w2 =
w4 =
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Let’s take a look
C::B
Program 12-3
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Rotating Bits
No “rotate” operator
Need to write a function
Similar to shifting
Left-Rotate
High-order bits that were lost when shifting become the low-order bits
Right-Rotate
Low-order bits that were lost when shifting become the high-order bits
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Rotate Examples
unsigned short w1 = 0x4356; unsigned short w2 = rightRotate(w1, 3); Unsigned short w3 = 0x9234; unsigned short w4 = leftRotate(w1, 2); w1 = w2 = w3 = w4 =
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Let’s Take a Look
C::B – Program 12-4
Notice that shift operators have higher precedence than bit-wise ops
Would prefer separate functions
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Bit Fields
Multiple data values can be stored in an int, short
Values are set and extracted via mask & shift 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
day
month
1 - 12
year 0 -99
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March 11, 2033 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
day
month
1 - 12
year 0 -99 1 15 14 13 12 11 10 9 8 7 6 5 4 3 2
day = 11
month = 3
year = 33
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45. Extracting Day from the Byte
int ymd = 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 1 1 1 1 1 1 1
year = 33
month = 3
day = 11
0x1F = 1 1 1 1 1
ymd & 0x1F = 1 1 1
day = ymd & 0x1F;
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Extracting Month 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
int ymd = 1 1 1 1 1 1 1
year = 33
month = 3
day = 11
0x1E0 = 1 1 1 1
ymd & 0x1E0 = 1 1
ymd & 0x1E0 >> 5 = 1 1
month = ymd & 0x1E0 >> 5;
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Extracting Year 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
ymd = 1 1 1 1 1 1 1
year = 33
month = 3
day = 11
0x7E00 = 1 1 1 1 1 1 1 1
ymd & 0x7E00 =
ymd & 0x7E00 >> 9 = 1 1
year = ymd & 0x7E00 >> 9;
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48. Packing Bits short ymd = (year << 9) | (month << 5) | day;15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
// March 11, 2033 short day = 11; short month = 3; short year = 33; 1 1 1 1 1 1
day = 11 1 1
month = 3 1 1
year = 33 1 1 1 1 1 1 1 1 1
year = 33
month=3
day = 11
short ymd = (year << 9) | (month << 5) | day;
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Struct Bit Fields
struct ymd {
unsigned int year:6;
unsigned int month:4;
unsigned int day: 5; };
struct ymd birthday; // in main()
Access fields using dot notation
Fields automatically converted to integers
birthday.year = 44;
printf(“ %d\n”, birthday.year);
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Exercises
Text (page 297)
#1 (12.1, 12.2, 12.3 in HEX, 12.4 in octal)
# Text #5 + 7 – Ex12-5_7.docx
bitTest and bitSet – 2 lines of code
bigGet - extract bits from an unsigned int. e.g. extractBits( x, 3, 5) extracts bits 3-7
bit – is lsb
Ex12-6.docx – bitPatternMatch( x, pattern, n ) – are the rightmost N bits of pattern in x
Change bit orientation so that bit 0 is lsb
Assume 32-bit integers
Ex1-Bytes.docx -Extract byte N
Challenge - replace value in byte N
Ex2-Fields.docx – breaking apart packed data
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