CS 325: CS Hardware and Software Organization and Architecture

1. CS 325: CS Hardware and SoftwareOrganization and Architecture Integers and Arithmetic Part 4

2. Outline • Binary Multiplication • Booth’s Algorithm • Number Representations

3. 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

4. 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.

5. 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.

6. 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.

7. 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.

8. 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.

9. 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

10. 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

11. 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

12. 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: UV 11100100  00011100  -2810

13. 2’s Complement Binary Multiplication – Booth’s Algorithm • Try: -9 x 7

14. Numbers are stored at addresses • Memory is a place to store bits • A word is a fixed number of bits • Ex: 32 bits, or 4 bytes • An address is also a fixed number of bits • Represented as unsigned numbers

15. Numbering Bits and Bytes • Need to choose order for: • Storage in physical memory system • Transmission over serial/parallel medium (data network) • Bit order • Handled by hardware • Usually hidden from programmer • Byte order • Affects multi-byte data items such as integers • Visible and important to programmers

16. 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

17. Byte Order Illustration • Note: Difference is especially important when transferring data between computers for which the byte ordering differs.

18. 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 1100 1111 1111 1111 1111 1111 1111 1111 1100

19. 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.