Lecture Objectives:
1)
2)
3)
4)
5)
6)
7)
8)
9)
Define the terms least significant bit and most significant bit.
Explain how unsigned integer numbers are represented in memory
Understand the limitations of using sign and magnitude to represent signed
integer numbers
Explain how signed integer numbers are represented in memory using twos
complement notation.
Convert twos complement numbers into decimal.
Convert decimal numbers into twos complement format.
Explain the concept of sign extension
Convert a binary number into hexadecimal
Explain how instructions are represented in memory as machine code
Numeric values can be represented in any
base. People commonly use base 10
(decimal).
Given a decimal (base 10) number, convert it
to binary:
• 51ten= ?two
• -7ten= ?two
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Given a binary number, convert it to
decimal format
• 1 1011two= ?ten
Processors use each bit of a word to
represent a binary digit of a numeric value
• Consider the integer value 305,419,896
represented with a MIPS 32-bit word:
00010010001101000101011001111000
In this representation, each bit
represents the value xi*2i, where i
is the bit number, and x is either
0 or 1.
Bit 0
(LSB)
Bit 31
(MSB)
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Signed numbers: How do we represent
negative values (+/-)?
• First approach
– Add a sign bit to the number
– 0 indicates positive
– 1 indicates negative
• Example:
-11ten= ?two
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Dual zeros: the problem with sign and
magnitude
• 1 00000000 == 0 00000000
Sign bit
Sign bit
Sign bits were used in early computers, but were soon abandoned.
No modern computers use sign bits!
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Solution: 2’s complement numbers


Leading 0’s mean positive, leading 1’s mean
negative
Example for 32 bits:
1111 1111 1111 1111 1111 1111 1111 11002
= –1×231 + 1×230 + … + 1×22 +0×21 +0×20
= –2,147,483,648 + 2,147,483,644
= –410

32 bits can represent the values
–2,147,483,648 to +2,147,483,647
Range: –2n – 1 to +2n – 1 – 1
x  xn12n1  xn2 2n2   x121  x0 20
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Shortcut: Converting a number into 2’s
complement
1. Convert the absolute value of the number
into a binary number
2. Complement the bits (1-> 0, 0->1)
3. Add 1 to the value
Example: Convert -51 (decimal) to binary
using 2’s complement.
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Sign extension
• Representing a number using more bits
– Preserve the numeric value
• Replicate the sign bit to the left
– c.f. unsigned values: extend with 0s
• Examples: 8-bit to 16-bit
– +2: 0000 0010 => 0000 0000 0000 0010
– –2: 1111 1110 => 1111 1111 1111 1110
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Example
• Convert -51 to a 16 bit signed binary (base 2)
representation based on the 2’s complement
value from before
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Converting binary to hex (CE1900 review)
• Hexadecimal – Base 16
– Group binary digits into sets of 4
– Convert each group into a hex digit
0000
0
1000
8
0001
1
1001
9
0010
2
1010
A
0011
3
1011
B
0100
4
1100
C
0101
5
1101
D
0110
6
1110
E
0111
7
1111
F
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Example
• Convert the number 10100101 to hex
• Convert the number 1100100 to hex
– What decimal number is this?
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Representing Instructions
• Instructions are represented in binary
– Called machine code
• MIPS instructions (e.g. add $t0, $s1, $s2)
– Encoded as 32-bit words
– Small number of formats
– Regularity in the pattern
• Register numbers
$t0 – $t7 are registers 8 – 15
$t8 – $t9 are registers 24 – 25
$s0 – $s7 are registers 16 – 23
§2.5 Representing Instructions in the Computer
MIPS R[egister]-format Instructions
op
rs
rt
rd
shamt
funct
6 bits
5 bits
5 bits
5 bits
5 bits
6 bits
• Instruction fields
–
–
–
–
–
–
op: operation code (opcode)
rs: first source register number
rt: second source register number
rd: destination register number
shamt: shift amount (00000 for now)
funct: function code (extends opcode)
• For add, $t0, $s1, $s2 (see the green card in the textbook)
–
–
–
–
–
–
op = 0
rs = $s1, register 17
rt = $s2, register 18
rd = $t0, register 8
shamt = 0
funct = 32
MIPS R-format for the add instruction
op
rs
rt
rd
shamt
funct
6 bits
5 bits
5 bits
5 bits
5 bits
6 bits
• For add, $t0, $s1, $s2
–
–
–
–
–
–
op = 0 (for add)
rs = $s1, register 17
rt = $s2, register 18
rd = $t0, register 8
shamt = 0 (for add)
funct = 32 (for add)
add instruction syntax:
add rd, rs, rt
(See the green card in the textbook)
Q1: why 5 bits for register values???
0
17
18
8
0
32
6 bits
5 bits
5 bits
5 bits
5 bits
6 bits
000000
10001
10010
01000
00000
100000
6 bits
5 bits
5 bits
5 bits
5 bits
6 bits
Q2: What is the hex equivalent?
MIPS I[mmediate]-format Instructions
op
rs
rt
Constant or address
6 bits
5 bits
5 bits
16 bits
• Instruction fields
–
–
–
–
op: operation code (opcode)
rs: first source register number
rt: second source register number
Last field: constant value or 16-bit base address offset
• For addi, $t0, $s1, 100
–
–
–
–
op = 8
rt = $t0, register 8
rs = $s1, register 17
Constant =100
0010 00
6 bits
# t0 = s1 + 100
addi instruction syntax:
addi rt, rs, value (sign-extended)
(See the green card in the textbook)
10 001 0 1000
5 bits
5 bits
0000 0000 0110 0100
16 bits
MIPS I[mmediate]-format Instructions
op
rs
rt
Constant or address
6 bits
5 bits
5 bits
16 bits
• Instruction fields
–
–
–
–
op: operation code (opcode)
rs: first source register number
rt: second source register number
Last field: constant value or 16-bit base address offset
• For lw $t0, 1200($t1) #load value from memory ref’d by t2 into t1
–
–
–
–
op = 35
rt = $t0, register 8
rs = $t1, register 9
Address offset=1200 (bytes)
1000 11
6 bits
lw instruction syntax:
lw rt, (offset)rs,
(See the green card in the textbook)
01 001 0 1000
5 bits
5 bits
0000 0100 1011 0000
16 bits