CS 325: CS Hardware and Software Organization and Architecture
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Transcript of CS 325: CS Hardware and Software Organization and Architecture
+ CS 325: CS Hardware and SoftwareOrganization and Architecture
Integers and Arithmetic
Part 4
+Outline
Binary Multiplication Booth’s Algorithm
Number Representations
+2’s Complement Binary Multiplication – Booth’s Algorithm Multiplication by bit shifting and addition.
Removes the need for multiply circuit Requires:
A way to compute 2’s Complement Available as fast hardware instructions
X86 assembly instruction: NEG A way to compare two values for equality
How to do this quickly?Exclusive Not OR (NXOR) Gate
Compare all sequential bits of bit string A and bit string B. Values are equal if the comparison process produces all 1s.
A way to shift bit strings. Arithmetic bit shift, which preserves the sign bit when
shifting to the right.10110110 arithmetic shift right 11011011
x86 assembly instruction: SAR
A B A NXOR B
0 0 1
0 1 0
1 0 0
1 1 1
+2’s Complement Binary Multiplication – Booth’s Algorithm Example: 5 x -3
First, convert to 2s comp bin: 5 = 0101 -3 = 1101
If we add 0 to the right of both values, there are 4 0-1 or 1-0 switches in 0101, and 3 in 1101. Pick 1101 as X value, and 0101 as Y value
Next, 2s Comp of Y: 1011 for bin subtraction.
Next, set 2 registers, U and V, to 0. Make a table using U, V, and 2 additional registers X, and X-1.
+2’s Complement Binary Multiplication – Booth’s Algorithm Register X is set to the predetermined value of x, and X-1 is
set to 0
Rules: Look at the LSB of X and the number in the X-1 register.
If the LSB of X is 1, and X-1 is 0, we subtract Y from U. If LSB of X is 0, and X-1 is 1, then we add Y to U. If both LSB of X and X-1 are equal, do nothing and skip to shifting
stage.
U V X X-1
0000 0000 1101 0
+2’s Complement Binary Multiplication – Booth’s Algorithm In our case, the LSB of X is one, and X-1 is zero, so we subtract Y from
U.
Next, we do an arithmetic right shift on U and V 1011 1101, 0000 1000
Copy the LSB of X into X-1
And then perform a circular right shift on X 1101 1110
Repeat the process three more times.
U V X X-1
0000 0000 1101 0
+1011
1011
1101 1000 1110 1
+2’s Complement Binary Multiplication – Booth’s Algorithm The LSB of X is zero, and X-1 is one, so we add Y to U.
Next, we do an arithmetic right shift on U and V 0010 0001, 1000 0100
Copy the LSB of X into X-1
And then perform a circular right shift on X 1110 0111
Repeat the process two more times.
U V X X-1
1101 1000 1110 1
+0101
0010
0001 0100 0111 0
+2’s Complement Binary Multiplication – Booth’s Algorithm The LSB of X is one, and X-1 is zero, so we subtract Y from U.
Next, we do an arithmetic right shift on U and V 1100 1110, 0100 0010
Copy the LSB of X into X-1
And then perform a circular right shift on X 0111 1011
Repeat the process one more time.
U V X X-1
0001 0100 0111 0
+1011
1100
1110 0010 1011 1
+2’s Complement Binary Multiplication – Booth’s Algorithm The LSB of X is one, and X-1 is one, begin shifts.
Next, we do an arithmetic right shift on U and V 1110 1111, 0010 0001
Copy the LSB of X into X-1
And then perform a circular right shift on X 1011 1101
U V X X-1
1110 0010 1011 1
1111 0001 1101 1
+2’s Complement Binary Multiplication – Booth’s Algorithm The result is stored in U followed by V.
This result is stored in 2’s complement notation. Convert to decimal:
11110001 00001111 -1510
This gives the correct result of 3 x -5
U V X X-1
1111 0001 1101 1
11110001
+2’s Complement Binary Multiplication – Booth’s Algorithm Another Example: 7 x -4
First, convert to 2s comp bin: 7 0111, add zero to right gives 01110, 2 switches -4 1100, add zero to right gives 11000, 1 switch
X = 1100 Y = 0111 -Y = 1001, for easy bin subtract
+2’s Complement Binary Multiplication – Booth’s Algorithm U V X X-1
0: 0000 0000 1100 0
1: 0000 0000 0110 0
2: 0000 0000 0011 0
+1001
1001
3: 1100 1000 1001 1
4: 1110 0100 1100 1
Result of 7 x -4: UV11100100 00011100 -2810
+2’s Complement Binary Multiplication – Booth’s Algorithm
Try: -9 x 7
+Numbers are stored at addressesMemory is a place to store bits
A word is a fixed number of bitsEx: 32 bits, or 4 bytes
An address is also a fixed number of bits
Represented as unsigned numbers
+Numbering Bits and Bytes
Need to choose order for:Storage in physical memory systemTransmission over serial/parallel medium
(data network)
Bit orderHandled by hardwareUsually hidden from programmer
Byte orderAffects multi-byte data items such as
integersVisible and important to programmers
+Possible Byte Orders
Least significant byte of integer in lowest memory location Little endian
Most Significant byte of integer in lowest memory location. Big endian
+Byte Order Illustration
Note: Difference is especially important when transferring data between computers for which the byte ordering differs.
+Sign Extension
Convert 2’s comp number using N bits to more than N bits (int to long int): Replicate the MSB (sign bit) of the smaller
number to fill new bits. 2’s comp positive number has infinite 0s 2’s comp negative number has infinite 1s
Ex: 16bit -410 to 32-bit:
1111 1111 1111 11001111 1111 1111 1111 1111 1111 1111 1100
+Conclusion
We represent “things” in computers as particular bit patterns: N bits 2N
Decimal for human calculations, binary for computers, hex for convenient way to write binary
2’s comp universal in computing: so make sure to learn!
Number are infinite, computers are not, so errors can occur (overflow, underflow)
Know the powers of 2.