1 Lecture 3 Bit Operations Floating Point – 32 bits or 64 bits 1

DCO 2 Bit Operations AND & OR | ONE'S COMPLEMENT ~ EXCLUSIVE OR ^ SHIFT (right) >> SHIFT (left) <<

DCO 3 Operation - examples AND 1 & 1 = 1; 1& 0 = 0 OR 1 |1 = 1; 1| 0 = 1; 0|0 = 0 ~ 0 =~1; 1 =~0; ^ 0^ 0 = 0; 1^1 = 0; 1^0 =1; 0^1 = 1 >> 0x010 = 0x001 <<1 >1

DCO 4 AND Example (0xf2) (0xfe) (and) & (0xf2) char c = 0xf2; char d = 0xfe; Char e = c & d; //e is 0xf2

DCO 5 OR Example (0xf2) (0xfe) (or) | (0xfe) char c = 0xf2; char d = 0xfe; char e = c | d; //e is 0xfe

DCO 6 One’s complement (0xf2) ~ (0x0d) char c = 0xf2; char e = ~c; //e is 0x0d

DCO 7 EXCLUSIVE OR (0xf2) (0xfe) (^) (0x0c) char c = 0xf2; char d = 0xfe; char e = c ^ d; //e is 0x0c

DCO 8 SHIFT >> 1 - (right) by one bit (0xf2) >> 1 (shift right by one bit) (0x79) char c = 0xf2; char e = c >>1; //e is 0x79

DCO 9 SHIFT >> 2 - by two bits (0xf2) >> 2 (shift right by one bit) (0x3c) char c = 0xf2; char e = c >>2; //e is 0x3c

DCO 10 SHIFT << 1 - (left) by one bit (0xf2) << 1 (shift right by one bit) (0xe4) char c = 0xf2; char e = c <<1; //e is 0xe4

DCO 11 SHIFT << 2 - by two bits (0xf2) >> 2 (shift right by one bit) (0xc8) char c = 0xf2; char e = c <<2; //e is 0xc8

DCO 12 Results or bit operation (example) ( 1 | 2) == 3 (1 | 3) == 3 (1 & 2) == 0 (1 & 3) == 1 (2 & 3) == 2 (0 ^ 3) == 3 (1 ^ 3) == 2 (2 ^ 3) == 1 (3 ^ 3) == 0 ~0 == -1 (signed) or 255 (unsigned). ~23 == -24 (signed) or 232 (unsigned).

DCO 13 Integer Formats, Signs, and Precision Overflow

DCO 14 Data representation 150 (decimal) is 128(2^7) + 16(2^4) + 4(2^2) + 2(2^1). We can rewrite this as: 1* *64 + 0*32 + 1*16 + 0*8 + 1*4 + 1*2 + 0*1 or (in hex), In 8 bits, for 32 bits, 2^32

DCO 15 Format In C, the size of integer types (short, int, long, etc.) is implementation dependent, but on most machines: chars are 8 bits ints and longs are 32 bits shorts are 16 bits

DCO 16 One’s complement & two’s complement //original data //reverse 1 to 0, 0 to 1 One’s complement (add one to one’s complement) Two’s complement

DCO 17 Application of twos’ complement – can be converted into two’s complement The operation becomes addition instead of subtraction Two’s complement – is – is = Convert “-” to “+”

DCO 18 Signed integer of 100 Negative sign is represented by 1 (most significant bit) 100 (decimal) 0x0064 (hex) positive -100 (decimal) 0xff9c (hex) negative

DCO 19 Example of – 101 (decimal) 101 (decimal ) is 0x0065 (hex) ( 32 bits integer) -101 is two’s complement 0xff9b Convert 0 to 1, 1 to 0 and then add (101 in decimal) (convert 1 to 0 and 0 to 1) (add 1, result)

DCO 20 Left & right Shift 101 (dec) is 0x0065 (hex) Shift right by one bit >> (101 dec) >> 1 (shift left) (50 dec), not 50.5

DCO 21 Overflow When integers are too big to fit into a word, overflow occurs. 16-bit unsigned integer, the largest is If you add (dec), it produces (binary) (overflow)

DCO 22 Floating Fixed-Point Representations – Floating-Point Representations such as 32.45, x 10, x 100 Normalization and Hidden Bits IEEE Floating Point Details

DCO 23 Fixed-Point Representations Such as The decimal point is fixed

DCO 24 Floating point A floating-point number is really two numbers packed into one string of bits. One of the numbers is simply a fixed-point binary number, which is called the mantissa. The other number acts to shift the binary point left or right to keep the mantissa in a useful range. This number is called the exponent.

DCO 25 Example of Floating E x 10^4

DCO 26 Expression 1 bit sign bit, 8 bit exponent, and 23 bit Mantissa (total 32 bits) -1^Sign * 2^(Exponent - 127) * (1 + Mantissa * 2^-23) Zero, sign bit is 0, Negative, sign bit is 1 Exponent is unsigned, minus 127. That is if the value is 128, it means 128 – 127 = 1, if the value is 256, it means 256 – 127 = 128, or the value is zero, it means 0 – 127 = -127.

DCO 27 Example

DCO 28 Expression - Mantissa Mantissa is unsigned bit, the expression is 1 + Mantissa * 2^( -23) If Mantissa is zero, the expression becomes * 2^(-23) = 1 If Mantissa is 2^23, the expression becomes 1 + 2^23 * 2 ^(-23) = = 2 It ranges from 1 to 2

DCO 29 Example 2.5 (floating point) Sign: positive (1) Exponent : : 128 (128 – 127 = 1) Mantissa: , 1.25 Result 1 x 1.25 x 2^1 = 2.5

DCO 30 Example Determine the values of Sign = 1, negative Exponent = 122 so exponent = -5 (122 – 127) Mantissa: = 1.1 So result – 1 x 1.1 x 2^(-5) 2 = (decimal)

DCO 31 Two representation of Zero When the sign bit is zero, positive, exponent is zero and Mantissa is zero When the sign bit is 1, negative, exponent is zero and Mantissa is zero In short, there are two formats of ZERO

DCO 32 IEEE Double Precision – 64 bits 64 bits with 11 exponent bits 52 mantissa bits Plus one bit sign bit C++ uses 32 bits (4 bytes) instead of 64 bits

DCO 33 Summary Bit operation, &, |, ~, >> or << Representation of data, int, short, long Floating point – sign, magnitude, mantissa for example, -3.4 x10^4 - (negative), 3.5 (mantissa), 4 magnitude 0x can be an integer, floating point, or a statement. That is why we need to define int i; Char a;