Download Lecture 4 - Representing Negative Numbers

Lecture 4: Representing
Negative Numbers
CS 2011
Spring 2016, Dr. Rozier
Recall from Last Time
• Unsigned integer representation and
mathematics.
Unsigned Addition
Operands: w bits
True Sum: w+1 bits
Discard Carry: w bits
u
•••
+ v
•••
u+v
•••
UAddw(u , v)
•••
• Standard Addition Function
– Ignores carry output
• Implements Modular Arithmetic
s = UAddw(u , v) = u + v mod 2w
 u  v
UAdd w (u,v)  
w
u  v  2
u  v  2w
u  v  2w
Visualizing (Mathematical) Integer
Addition
• Integer Addition
–4-bit integers u,
v
–Compute true
sum Add4(u , v)
–Values increase
linearly with u
and v
–Forms planar
surface
Add4(u , v)
Integer Addition
32
28
24
20
16
14
12
12
10
8
8
4
6
0
4
0
2
u
4
2
6
8
10
12
0
14
v
Visualizing Unsigned Addition
• Wraps Around
Overflow
– If true sum ≥ 2w
– At most once
True Sum
2w+1
UAdd4(u , v)
16
14
Overflow
12
10
8
2w
14
6
12
10
4
8
2
0
6
0
Modular Sum
4
0
u
2
4
2
6
8
10
12
0
14
v
Unsigned Binary Integers
• Given an n-bit number


Range: 0 to +2n – 1
Example


0000 0000 0000 0000 0000 0000 0000 10112
= 0 + … + 1×23 + 0×22 +1×21 +1×20
= 0 + … + 8 + 0 + 2 + 1 = 1110
Using 32 bits

0 to +4,294,967,295
How can we change our
representation to include signed
numbers?
How can we represent a negative
sign in a computer?
• Key point – we need to represent it in a way
which makes arithmetic easy!
Naïve Solution: Signed Integers
• Given n bits, we have 2n values we can
represent.
– Assign half to positive integers.
– Assign other half to negative integers.
• Use most significant bit as negative sign.
Number is negative is the most significant bit
is a 1, positive otherwise.
Naïve Solution: Signed Integers
• Advantage: Signed integers don’t change.
– 0001 is still 1.
• Disadvantage
– 0000 is 0
– 1000 is also 0
Binary
Unsigned
Signed Magnitude
0000
0
0
0001
1
1
0010
2
2
0011
3
3
0100
4
4
0101
5
5
0110
6
6
0111
7
7
1000
-0
-0
1001
-1
-1
1010
-2
-2
1011
-3
-3
1100
-4
-4
1101
-5
-5
1110
-6
-6
1111
-7
-7
Problem: How do we do
arithmetic?
+
0
0
1
0
1
1
0
1
1
1
1
1
2 + -3 = -7
OOPS!
Want a method which makes
arithmetic simple!
• Challenge: For any positive number X, find a
representation for –X such that:
X + (-X) = 0
+
0
1
0
1
?
?
?
?
0
0
0
0
Ignoring Carry Out
Want a method which makes
arithmetic simple!
• Challenge: For any positive number X, find a
representation for –X such that:
X + (-X) = 0
+
0
1
0
1
1
0
1
1
0
0
0
0
Ignoring Carry Out
2’s-Complement
• This representation, the number we can add
to X to get 0, is known as the two’s
complement.
• Short-cut to find it:
– Flip the bits, 0 -> 1, 1 -> 0.
– Add one to the result
Binary Pattern
Unsigned
Integer
Flipped Bits
Flipped Bits + 1 2’scomplement
Integer
0000
0
1111
0000
0
0001
1
1110
1111
-1
0010
2
1101
1110
-2
0011
3
1100
1101
-3
0100
4
1011
1100
-4
0101
5
1010
1011
-5
0110
6
1001
1010
-6
0111
7
1000
1001
-7
1000
8
0111
1000
8
Advantages of 2’s complement
• Only one number for 0
• Addition is still addition.
• Subtraction is just addition.
• No new hardware needed.
2s-Complement Signed Integers
• Given an n-bit number


Range: –2n – 1 to +2n – 1 – 1
Example


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
Using 32 bits

–2,147,483,648 to +2,147,483,647
2s-Complement Signed Integers
• Bit 31 is sign bit
– 1 for negative numbers
– 0 for non-negative numbers
• –(–2n – 1) can’t be represented
• Non-negative numbers have the same unsigned and
2s-complement representation
• Some specific numbers
–
–
–
–
0: 0000 0000 … 0000
–1: 1111 1111 … 1111
Most-negative: 1000 0000 … 0000
Most-positive: 0111 1111 … 1111
Signed Negation
• Complement and add 1
– Complement means 1 → 0, 0 → 1
x  x  1111...1112  1
x  1  x

Example: negate +2


+2 = 0000 0000 … 00102
–2 = 1111 1111 … 11012 + 1
= 1111 1111 … 11102
Mapping Signed  Unsigned
Bits
Signed
Unsigned
0000
0
0
0001
1
1
0010
2
2
0011
3
3
0100
4
4
0101
5
5
0110
6
6
0111
7
7
1000
-8
8
1001
-7
9
1010
-6
10
1011
-5
11
1100
-4
12
1101
-3
13
1110
-2
14
1111
-1
15
Unsigned & Signed Numeric Values
X
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
B2U(X)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
B2T(X)
0
1
2
3
4
5
6
7
–8
–7
–6
–5
–4
–3
–2
–1
• Equivalence
– Same encodings for nonnegative
values
• Uniqueness
– Every bit pattern represents
unique integer value
– Each representable integer has
unique bit encoding
•  Can Invert Mappings
– U2B(x) = B2U-1(x)
• Bit pattern for unsigned
integer
– T2B(x) = B2T-1(x)
• Bit pattern for two’s comp
integer
Mapping Between Signed &
Unsigned
Unsigned
Two’s Complement
T2U
x
T2B
X
B2U
ux
Maintain Same Bit Pattern
Unsigned
ux
Two’s Complement
U2T
U2B X
B2T
x
Maintain Same Bit Pattern
• Mappings between unsigned and two’s complement numbers:
keep bit representations and reinterpret
Mapping Signed  Unsigned
Bits
Signed
Unsigned
0000
0
0
0001
1
1
0010
2
2
0011
3
3
0100
4
4
0101
5
0110
6
0111
7
1000
-8
8
1001
-7
9
1010
-6
10
1011
-5
11
1100
-4
12
1101
-3
13
1110
-2
14
1111
-1
15
T2U
U2T
5
6
7
Mapping Signed  Unsigned
Bits
Signed
Unsigned
0000
0
0
0001
1
1
0010
2
2
0011
3
0100
4
0101
5
5
0110
6
6
0111
7
7
1000
-8
8
1001
-7
9
1010
-6
10
1011
-5
1100
-4
12
1101
-3
13
1110
-2
14
1111
-1
15
=
+/- 16
3
4
11
Conversion Visualized
• 2’s Comp. 
Unsigned
UMax
UMax – 1
– Ordering Inversion
– Negative  Big
TMax
Positive
2’s Complement
Range
0
–1
–2
TMin
TMax + 1
TMax
0
Unsigned
Range
Sign Extension
• Task:
– Given w-bit signed integer x
– Convert it to w+k-bit integer with same value
• Rule:
– Make k copies of sign bit:
– X  = xw–1 ,…, xw–1 , xw–1 , xw–2 ,…, x0
w
k copies of MSB
X
•••
•••
X
•••
k
•••
w
Sign Extension Example
short int x = 15213;
int
ix = (int) x;
short int y = -15213;
int
iy = (int) y;
x
ix
y
iy
Decimal
15213
15213
-15213
-15213
Hex
3B
00 00 3B
C4
FF FF C4
6D
6D
93
93
Binary
00111011
00000000 00000000 00111011
11000100
11111111 11111111 11000100
01101101
01101101
10010011
10010011
• Converting from smaller to larger integer data
type
• C automatically performs sign extension
Two’s Complement Addition
u
•••
v
•••
u+v
•••
Operands: w bits
+
True Sum: w+1 bits
Discard Carry: w bits
TAddw(u , v)
•••
• TAdd and UAdd have Identical Bit-Level Behavior
– Signed vs. unsigned addition in C:
int s, t, u, v;
s = (int) ((unsigned) u + (unsigned) v);
t = u + v
– Will give
s == t
TAdd Overflow
• Functionality
– True sum
requires w+1
bits
– Drop off MSB
– Treat remaining
bits as 2’s comp.
integer
True Sum
0 111…1
2w–1
PosOver
TAdd Result
0 100…0
2w –1
011…1
0 000…0
0
000…0
1 011…1
–2w –1–1
100…0
1 000…0
–2w
NegOver
Visualizing 2’s Complement Addition
NegOver
• Values
– 4-bit two’s comp.
– Range from -8 to +7
• Wraps Around
– If sum  2w–1
• Becomes negative
• At most once
– If sum < –2w–1
• Becomes positive
• At most once
TAdd4(u , v)
8
6
4
2
0
6
-2
4
2
-4
0
-6
-2
-8
-4
-8
-6
-4
u
v
-6
-2
0
2
4
-8
6
PosOver
Characterizing TAdd
• Functionality
– True sum requires
w+1 bits
– Drop off MSB
– Treat remaining bits
as 2’s comp. integer
Positive Overflow
TAdd(u , v)
>0
v
<0
<0
u
>0
Negative Overflow
u  v  22ww1 u  v  TMin w (NegOver)

TAdd w (u,v)  u  v
TMinw  u  v  TMax w
u  v  22ww1 TMax w  u  v (PosOver)

Mathematical Properties of TAdd
• Isomorphic Group to unsigneds with UAdd
– TAddw(u , v) = U2T(UAddw(T2U(u ), T2U(v)))
• Since both have identical bit patterns
• Two’s Complement Under TAdd Forms a Group
– Closed, Commutative, Associative, 0 is additive identity
– Every element has additive inverse
u
TCompw (u)  
TMinw
u  TMinw
u  TMinw
Representing Pointers
int B = -15213;
int *P = &B;
Sun
IA32
x86-64
EF
D4
0C
FF
F8
89
FB
FF
EC
2C
BF
FF
FF
7F
00
00
Different compilers & machines assign different locations to objects
Strings
How might we represent a string?
Representing Strings
char S[6] = "18243";
• Strings in C
– Represented by array of characters
Linux/Alpha
– Each character encoded in ASCII format
31
• Standard 7-bit encoding of character
38
set
32
• Character “0” has code 0x30
34
– Digit i has code 0x30+i
33
– String should be null-terminated
00
• Final character = 0
• Compatibility
– Byte ordering not an issue
Sun
31
38
32
34
33
00