chapter 3 – digital logic and binary numbers

Richard P. Paul, SPARC Architecture, Assembly Language Programming, and C

Chapter 3 – Digital Logic and Binary Numbers

These are lecture notes to accompany the book SPARC Architecture, Assembly Language

Programming, and C,

by Richard P. Paul, 2nd edition, 2000.

By Michael Weeks


Binary

• A computer is a “bistable” device

• A bistable device:– Easy to design and build– Has 2 states: 0 and 1

• One Binary digit (bit) represents 2 possible states (0, 1)


• With 2 bits, 4 states are possible (22 = 4)

• With n bits, 2n states are possible

• With 3 bits, 8 states are possible (23 = 8)

Bit1 Bit0 State

0 0 1

0 1 2

1 0 3

1 1 4

Bit2 Bit1 Bit0 State

0 0 01

0 0 12

0 1 03

0 1 14

1 0 05

1 0 16

1 1 07

1 1 18


Binary Coded Decimal (BCD)

• Why not use 4 bits to represent decimal?• Let 0000 represent 0

• Let 0001 represent 1

• Let 0010 represent 2

• Let 0011 represent 3, etc.

– This is called BCD– Only uses 10 of the 16 possibilities


Binary Number System

• From left to right, the position of the digit indicates its magnitude (in decreasing order)– E.g. in decimal, 123 is less than 321– In binary, 011 is less than 100

• A subscript indicates the number’s base– E.g. is 100 decimal or binary? We don’t know!

– But 1410 = 11102 is clear


Bytes

• A group of 8 bits is a byte

• A byte can represent 28 = 256 possible states

• Registers are usually a multiple of bytes

• SPARC registers have 32 bits (4 bytes)

• 232 = 4,294,967,296


Memory Addresses

• Memory addresses are in binary – often 32 bits, these days– if each memory address maps to 1 byte:

• 232 bytes = 4 GB

• K = kilo = thousand, • but 1KB actually means 1024 bytes• 1MB = 1024 x 1024 bytes• 1GB = 1024 x 1024 x 1024 bytes


Octal and Hexadecimal

• It is difficult for a human to work with long strings of 0’s and 1’s

• Octal and Hexadecimal are ways to group bits together

• Octal: base 8

• Hexadecimal: base 16


Hexadecimal

• With 4 bits, there are 16 possibilities

• Use 0, 1, 2, 3, …9 for the first 10 symbols

• Use a, b, c, d, e, and f for the last 6

Bit3 Bit2 Bit1 Bit0 Symbol0 0 0 0

00 0 0 1

10 0 1 0

20 0 1 1

30 1 0 0

40 1 0 1

50 1 1 0

60 1 1 1

71 0 0 0

81 0 0 1

91 0 1 0

a1 0 1 1

b1 1 0 0

c1 1 0 1

d1 1 1 0

e1 1 1 1

f


Binary to Hexadecimal

• 01010110101100112 = ? in hex• Group into 4 bits, from the right:• 0101, 0110, 1011, 00112

• Now translate each (see previous table):01012 => 5, 01102 => 6, 10112 => b, 00112 => 3

So this is 56b316

• What if there are not enough bits? – Pad with 0’s on the left


Hexadecimal to Binary

• f0e516 = ? in binary

• Translate each into a group of 4 bits:

f16 => 11112, 016 => 00002, e16 => 11102, 516 => 01012

So this is 11110000111001012


Decimal to Any Number Base

• Take the decimal number, and divide by the new number base

• Keep track of the quotient and remainder

• Repeat until quotient = 0

• Read number from the bottom to the top


Decimal to Binary

• Binary is base 2• Example: convert 35 (decimal) to binary

Quotient Remainder

35 / 2 = 17 117 / 2 = 8 18 / 2 = 4 04 / 2 = 2 02 / 2 = 1 01 / 2 = 0 1

• So 3510 = 1000112


Any Number Base to Decimal

• From right to left, multiply the digit of the number-to-convert by its baseposition

• Sum all results


Binary to Decimal

• Binary is base 2

• Example: convert 10110 (binary) to decimal

101102 = 1x24 + 0x23 + 1x22 + 1x21 + 0x20

= 1x16 + 0x8 + 1x4 + 1x2 + 0x1

= 16 + 0 + 4 + 2 + 0

= 22

• So 101102 = 2210


Hexadecimal to Decimal

• Hexadecimal is base 16• Example: convert 16 (hex) to decimal

1616 = 1x161 + 6x160

= 1x16 + 6x1= 16 + 6= 22

• So 1616 = 2210

• Not surprising, since 1616 = 0001, 01102– If one of the hex digits had been > 9, say c, then we would have

used 12 in its place.


ASCII

• American Standard Code for Information Interchange

• Use byte values to represent characters • The assembler allows double-quotes

mov 0x4d, %r3 ! Moves capital M to register 3

mov “M”, %r3 ! This command does the same


ASCII chart


Bitwise Logical Operations

• There are several binary operations:– NOT– AND– OR– XOR– NAND– NOR


NOT

• The NOT operation simply complements a binary value– not (a) – a’

a not(a)

0 1

1 0


AND

• The AND operation uses 2 binary values– a and b

a b a and b

0 0 0

0 1 0

1 0 0

1 1 1


OR

• The OR operation uses 2 binary values– a or b

a b a or b

0 0 0

0 1 1

1 0 1

1 1 1


XOR• The XOR (exclusive-or) operation uses 2

binary values

• True when only one input is true.– a xor b a b a xor b

0 0 0

0 1 1

1 0 1

1 1 0


NAND• The NAND (Not-AND) operation uses 2

binary values

• Take the AND function, and complement it.– a nand b a b a nand b

0 0 1

0 1 1

1 0 1

1 1 0


NOR• The NOR (Not-OR) operation uses 2 binary values

• Take the OR function, and complement it.

• NAND and NOR are easy to make on a chip. Why? Take CSc 4250 and find out!

a b a nor b

0 0 1

0 1 0

1 0 0

1 1 0


Possible Logic Functions

• Suppose you have 2 binary digits: a, b• Imagine that some function operates on them to

create c.• What could this function be?

– There are only 16 possibilities

– And some of these are not useful!

ab

csome

function


Logic OperationsA 0011 Logical SparcB 01010 0000 false1 0001 a and b and2 0010 a and (not b) andn3 0011 a4 0100 b and (not a)5 0101 b6 0110 a xor b xor7 0111 a or b or8 1000 a nor b9 1001 a xor (not b) xnorA 1010 not bB 1011 a or (not b) ornC 1100 not aD 1101 b or (not a)E 1110 a nand bF 1111 true


Bitwise

• Each of these logic functions is a bitwise operation, meaning that the result is independent of the bits to the left or right

e.g. 1 0 1

or 0 1 1

1 1 1

• compare this with addition


Logic Instruction Examples

mov 0x21, %l0

and %l0, 0x3c, %l1

mov 0x21, %l0 or %l0, 0x3c, %l1

mov 0x55, %l0 xnor %l0, 0x3c, %l1

mov 0x55, %l0 xor %l0, 0x3c, %l1

mov 0x47, %l0 and %l0, 0xca, %l1

mov 0x47, %l0 andn %l0, 0xca, %l1

mov 0x47, %l0 or %l0, 0xca, %l1

mov 0x47, %l0 orn %l0, 0xca, %l1

mov 0x55, %l0 not %l0


SPARC Instruction Format• These commands are in the form:

command source register 1, source register 2, destination register

command source register 1, immediate value, destination register

• command can be any of the following:

and, andn, xor, or, xnor, orn

andcc, andncc, xorcc, orcc, xnorcc, orncc

• the cc means “set condition codes”

• andn means a and (not b)

• orn means a or (not b)


SPARC Logical Instruction Example

cmp %a_r, 0ble nextnopadd %b_r, 1, %b_r

next:

This is equivalent to:

if (a > 0)

b++;

%a_r and %b_r will be replaced by the actual registers,

such as %r2 and %r3


Synthetic Instructions• The cmp command is a synthetic one. It is a macro

that uses %g0. The above cmp command will be expanded to:

subcc %a_r, %g0, %g0

• Also, the tst command compares a register to 0:tst %a_r

which the assembler turns into:orcc %a_r, %g0, %g0

• Since %g0 ignores any updates, only the condition codes are affected.


Flags

• Since individual bits are used to represent boolean flags, a word may contain 32 flags.

• Common flag operations and mnemonics– set: bset ( done with or )– clear: bclr ( done with andn )– toggle: btog ( done with xor )


Testing FlagsThis command will see if one or more flags is set

btst reg_or_imm, regrs1

it expands to:

andcc regrs1, reg_or_imm, %g0(notice how the operands are switched)example: test if flag 0x02 is set btst 0x02, %a_r be clear nop

set:clear:

chapter 3 – digital logic and binary numbers

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