Download EXAM 1 - Department of Computer and Information Science

Department of Computer and Information Science,
School of Science, IUPUI
CSCI 230
Information Representation:
Negative and Floating Point Representation
Dale Roberts, Lecturer
IUPUI
[email protected]
Dale Roberts
Negative Numbers in Binary
Four different representation schemes are used
for negative numbers
1. Signed Magnitude
Left most bit (LMB) is the sign bit :
0  positive (+)
1  negative (-)
Remaining bits hold absolute magnitude
Example:
210  0000 0010b
Try, 1000 0100b = -410
-210  1000 0010b
Q: 0000 0000 = ?
1000 0000 = ?
Dale Roberts
2. One’s Compliment
– Left most bit is the sign bit :
•
•
0  positive (+)
1  negative (-)
– The magnitude is complimented
Example:
210  0 000 0010b
-210  1 111 1101b
Exercise: try - 410 using 1’s compliment
Q: 0000 0000 = ?
1111 1111 = ?
Solution:
410 = 0 000 0100
-410 = 1 111 1011
b
b
Dale Roberts
Negative Numbers in Binary (cont.)
3. 2’s Compliment
• Sign bit same as above
• Magnitude is complimented first and a “1” is added to
the complimented digits
Example:
210  0 000 0010b
1’s compliment  1 111 1101b
+
1
-210  1 111 1110b
Exercise: try -710 using 2’s compliment
710  0000 0111
b
1’s compliment  1111 1000b
+
1
-710  1111 1001b
Dale Roberts
Negative Numbers in Binary (cont.)
Example: 7+(-3)
[hint]: A – B = A + (~B) +1
710 = 0000 0111b
310 = 0000 0011b
1’s complement 1111 1100
b
2’s complement 1111 1101  -3
10
b
7+(-3) 
0000 0111
+1111 1101
1 1111 111 carry
ignore 1 0000 0100  0000 0100  410
Dale Roberts
Three Representation of Signed Integer
Representation
0 0000
0 0001
0 0010
0 0011
0 0100
0 0101
0 0110
0 0111
0 1000
0 1001
0 1010
0 1011
0 1100
0 1101
0 1110
0 1111
1 0000
1 0001
1 0010
1 0011
1 0100
1 0101
1 0110
1 0111
1 1000
1 1001
1 1010
1 1011
1 1100
1 1101
1 1110
1 1111
Value Representation
Signed Magnitude
1's Complement 2's Complement
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
-0
-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
-11
-12
-13
-14
-15
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
-15
-14
-13
-12
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
-0
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
-16
-15
-14
-13
-12
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
Dale Roberts
Negative Numbers in Binary (cont.)
4. Excess Representation
– For a given fixed number of bits
the range is remapped such that
roughly half the numbers are
negative and half are positive.
Example: (as left)
Excess – 8 notation for 4 bit numbers
Binary value = 8 + excess-8
value
MSB can be used as a sign bit,
but
If MSB =1, positive (+ ve) number
If MSB =0, negative (- ve) number
Numbers
Binary
Value
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Notation
Excess – 8
Value
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
Dale Roberts
Fundamental Data Type
• With vs. without using sign bit
For a 16 bit binary pattern:
2 byte unsigned (Default type is int)
0000 0000 0000 0000 (  0D)
0000 0000 0000 0001 (  1D )
0000 0000 0000 0010 (  2D )
….
0111 1111 1111 1111 (  32767D  215 -1)
1000 0000 0000 0000 (  32768D  215)
….
1111 1111 1111 1111 ( 216 –1)
2 byte
int
1000 0000 0000 0000 (  -32768D  - 215 )
1000 0000 0000 0001 (  -32767D  - 215 +1)
….
1111 1111 1111 1110 (  - 2D )
1111 1111 1111 1111 (  - 1D )
0000 0000 0000 0000 (  0D )
0000 0000 0000 0001 (  1D )
0000 0000 0000 0010 (  2D )
….
0111 1111 1111 1111 (  32767D  215 -1)
Dale Roberts
Fundamental Data Type
Four Data Types in C (assume 2’s complement, byte machine)
Data Type
char
Abbreviation
Size (byte)
Range
char
1
-128 ~ 127
unsigned char
1
0 ~ 255
2 or 4
-215 ~ 215-1 or -231 ~ 231-1
2 or 4
0 ~ 65535 or 0 ~ 232-1
int
int
unsigned int
unsigned
short int
short
2
-32768 ~ 32767
unsigned short
int
unsigned short
2
0 ~ 65535
long int
long
4
-231 ~ 231-1
unsigned long
int
unsigned long
4
0 ~ 232-1
float
4
double
8
Note:
27 = 128, 215 =32768, 215 = 2147483648
Complex and double complex are not available
Dale Roberts
Fractional Numbers
Examples: 456.7810 = 4 x 102 + 5 x 101 + 6 x 100 + 7 x 10-1+8 x 10-2
1011.112
= 1 x 23 + 0 x 22 + 1 x 21 + 1 x 20 + 1 x 2-1 + 1 x 2-2
= 8 + 0 + 2 + 1 + 1/2 + ¼
= 11 + 0.5 + 0.25 = 11.7510
Conversion from binary number system to
decimal system
Examples: 111.112 = 1 x 22 + 1 x 21 + 1 x 20 + 1 x 2-1 + 1 x 2-2
= 4 + 2 + 1 + 1/2 + ¼ = 7.7510
Examples: 11.0112
2
1
0
22
21
20
2
1
-1
2-1
-2
2-2
-3
2-
3
4
x
x
½
¼
x
1/8
x
Dale Roberts
Conversion from decimal number system to binary
system
Examples:
7.7510 = (?)2
Conversion of the integer part: same as before – repeated division by 2
7 / 2 = 3 (Q), 1 (R)  3 / 2 = 1 (Q), 1 (R)  1 / 2 = 0 (Q), 1 (R)
710 = 1112
2. Conversion of the fractional part: perform a repeated multiplication by 2 and extract
the integer part of the result
0.75 x 2 =1.50  extract 1
0.5 x 2 = 1.0  extract 1
0.7510 = 0.112
write in the same order
0.0
 stop
1.

Combine the results from integer and fractional part, 7.7510 = 111.112
How about choose some of
4
2
1
1/2
1/4
=0.5
=0.25
1/8
=0.125
Examples: try 5.625B
Dale Roberts
Fractional Numbers (cont.)
Exercise 1: Convert (0.625)10 to its binary form
Solution:
0.625 x 2 = 1.25
 extract 1
0.25 x 2 = 0.5
 extract 0
0.5 x 2 = 1.0
 extract 1
0.0
 stop
 (0.625)10 = (0.101)2
Exercise 2: Convert (0.6)10 to its binary form
Solution:
0.6 x 2 = 1.2
 extract 1
0.2 x 2 = 0.4
 extract 0
0.4 x 2 = 0.8
 extract 0
0.8 x 2 = 1.6
 extract 1
0.6 x 2 =

 (0.6)10 = (0.1001 1001 1001 …)2
Dale Roberts
Fractional Numbers (cont.)
Errors
One source of error in the computations is due to back and
forth conversions between decimal and binary formats
Example: (0.6)10 + (0.6)10 = 1.210
Since (0.6)10 = (0.1001 1001 1001 …)2
Lets assume a 8-bit representation: (0.6)10 = (0 .1001 1001)2 , therefore
0.6
0.10011001
+ 0.6
 + 0.10011001
1.00110010
Lets reconvert to decimal system:
(1.00110010)b
= 1 x 20 + 0 x 2-1 + 0 x 2-2 + 1 x 2-3 + 1 x 2-4 + 0 x 2-5 + 0 x 2-6 + 1 x 2-7 + 0 x 2-8
= 1 + 1/8 + 1/16 + 1/128 = 1.1953125
 Error = 1.2 – 1.1953125
= 0.0046875
Dale Roberts
Floating Point Number Representation
If x is a real number then its normal form representation is:
x = f • Base E
where
f : mantissa
E: exponent
exponent
Example:
125.3210 = 0.12532 • 103
mantissa
- 125.3210 = - 0.12532 • 103
0.054610 =
0.546 • 10 –1
The mantissa is normalized, so the digit after the fractional point is non-zero.
If needed the mantissa should be shifted appropriately to make the first digit
(after the fractional point) to be non-zero & the exponent is properly adjusted.
Dale Roberts
Example:
134.1510 = 0.13415 x 103
-2
0.002110 = 0.21 x 10
101.11B =
0.011B =
AB.CDH=
0.00ACH=
Dale Roberts
Assume we use 16-bit binary pattern for normalized binary
form based on the following convention (MSB to LSB)
Sign of mantissa (±)= left most bit (where 0: +; 1: - )
Mantissa (f)= next 11 bits
Sign of exponent (±)= next bit (where 0: +; 1: - )
Exponent (E) = next three bits
x = ± f • Base
f = 0.?1?2?3?4…?11 ?12…?15
±E
E : converted to binary, b1b2b3
MSB
LSB
?1
+:0
- :1
?2
?3
?4
?5
?6
?7
?8
?9
?10 ?11
b1
b2
b3
+:0
- :1
Dale Roberts
Floating Point Number Representation
Question:
How the computer expresses the 16-bit approximation of
1110.111010111111 in normalized binary form using the
following convention
Sign of mantissa = left most bit (where 0: +; 1: - )
Mantissa = next 11 bits
Sign of exponent = next bit (where 0: +; 1: - )
Exponent = next three bits
Answer:
Step 1: Normalization
1110.111010111111 = + 0.1110111010111111 * 2
+4
Step 2: “Plant” 16 bits
sign
1 bit
mantissa
11 bits
sign exponent
1 bit
3 bits
the 16 bit floating point representation is 0 11101110101 0 100
Dale Roberts
Question:
Interpret the normalized binary number
0111 0000 0000 1010
B
using the convention mentioned
Sign of mantissa = left most bit (where 0: +; 1: - )
Mantissa = next 11 bits
Sign of exponent = next bit (where 0: +; 1: - )
Exponent = next three bits
find its decimal equivalent.
Answer:
0 11100000000 1 010
B
= 0.111B * 2-2
Dale Roberts
Floating Point Number Representation (cont.)
The 32 Bit Single Precision Floating Point Format for IBM 370
Base = 16
Exponent = Excess-64 notation (i.e., compute binary equivalent, then substrate 64)
Sign = sign of number (0: positive, 1: negative)
Mantissa = normalized fraction (i.e. first digital after ‘.’ is non-zero)
sign
1 bit
exponent
7 bits
mantissa
24 bits
Example: What is the value of the following point number?
1 100 0010 1001 0011 1101 0111 1100 0010
Sign = 1  the number is negative
Exponent (E) = 100 00102 = 6610 = 2 (substrate 64, because of Excess-64 )
Mantissa (f ) = 1001 0011 1101 0111 1100 0010 = 93D7C2H
 The above floating point number is: x = (sign) f • 16 E = - 0.93D7C2 • 16 2
x = - (9 x 16-1 + 3 x 16-2 + D x 16-3 + 7 x 16-4 + C x 16-5 + 2 x 16-6) • 16 2
= - (9 x 161 + 3 x 160 + 13 x 16-1 + 7 x 16-2 + 12 x 16-3 + 2 x 16-4)
= - (144+3 +0.8125+0.02734375 + 0.0029296875 + 0.000030517578125)
= - 147.842803955078125
Dale Roberts
Acknowledgements
These slides where originally prepared by Dr. Jeffrey Huang, updated by Dale
Roberts.
Dale Roberts