1. Binary Representation

2. Binary Representation for Numbers Assume 4-bit numbers 5 as an integer  0101 -5 as an integer  How? 5.0 as a real number  How?  What about 5.5?

3. Sign Bit Reserve the most-significant bit to indicate sign Consider integers in 4 bits  Most-significant bit is sign: 0 is positive, 1 is negative  The 3 remaining bits is magnitude  0010 = 2  1010 = -2 How many possible combinations for 4 bits? How many unique integers using this scheme?

4. Two’s Complement Advantages  # of combinations of bits = # of unique integers  Addition is “natural” Convert to two’s complement (and vice versa) 1. invert the bits 2. add one 3. ignore the extra carry bit if present Consider 4-bit numbers  0010 [2] -> 1101 -> 1110 [-2]

5. Addition 0010 [2] + 1110 [-2]  0000 [ignoring the final carry—extra bit] 0011 [3] + 1110 [-2]  0001 [1] 1110 [-2] + 1101 [-3]  1011 [-5]

6. Range of Two’s Complement 4-bit numbers  Largest positive: 0111 (binary) => 7 (decimal)  Smallest negative: 1000 (binary) => -8 (decimal)  # of unique integers = # of bit combinations = 16 n bits  ?

7. Binary Real Numbers 5.5  101.1 5.25  101.01 5.125  101.001 5.75  101.11 …23232 2121 2020. 2 -1 …

8. 8 bits only 5.5  101.1 -> 000101 10 5.25  101.01 -> 000101 01 5.125  101.001 -> ?? With only 2 places after the point, the precision is.25 What if the point is allowed to move around? 2525 2424 23232 2121 2020 2 -1 2 -2

9. Floating-point Numbers Decimal  54.3  5.43 x 10 1 [scientific notation] Binary  101.001  10.1001 x 2 1 [more correctly: 10.1001 x 10 1 ]  1.01001 x 2 2 [more correctly: 1.01001 x 10 10 ]  What can we say about the most significant bit?

10. Floating-point Numbers General form: sign 1.mantissa x 2 exponent  the most significant digit is right before the dot  Always 1 [no need to represent it]  Exponent in Two’s complement 1.01001 x 2 2  Sign: 0 (positive)  Mantissa: 0100  Exponent: 010 (decimal 2)

11. Java Floating-point Numbers Sign:  1 bit [0 is positive] Mantissa:  23 bits in float  52 bits in double Exponent:  8 bits in float  11 bits in double signexponentmantissa

12. Imprecision in Floating-Point Numbers Floating-point numbers often are only approximations since they are stored with a finite number of bits. Hence 1.0/3.0 is slightly less than 1/3. 1.0/3.0 + 1.0/3.0 + 1.0/3.0 could be less than 1. www.cs.fit.edu/~pkc/classes/iComputing/FloatEquality.java

13. Abstraction Levels Binary  Data  Numbers (unsigned, signed [Two’s complement], floating point) Text (ASCII, Unicode) HTML Color Image (JPEG) Video (MPEG) Sound (MP3)  Instructions  Machine language (CPU-dependent)  Text (ASCII) Assembly language (CPU-dependent) High-level language (CPU -independent: Java, C++, FORTRAN)