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Lecture 3

Lecture 3. Bit Operations Floating Point – 32 bits or 64 bits. 1. Bit Operations. Operator. AND & OR | ONE'S COMPLEMENT ~ EXCLUSIVE OR ^ SHIFT (right) >> SHIFT (left) <<. Operation - examples. AND 1 & 1 = 1; 1& 0 = 0

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Lecture 3

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

  2. Bit Operations Operator • 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 • << 0x001 = 0x010 >>1 DCO

  4. AND Example • 1111 0010 (0xf2) • 1111 1110 (0xfe) • ---------------- (and) & • 1111 0010 (0xf2) • char c = 0xf2; • char d = 0xfe; • Char e = c & d; //e is 0xf2 DCO

  5. OR Example • 1111 0010 (0xf2)1111 1110 (0xfe)--------------(or) |1111 1110 (0xfe) • char c = 0xf2;char d = 0xfe;char e = c | d; //e is 0xfe DCO

  6. One’s complement • 1111 0010 (0xf2)-------------- ~0000 1101 (0x0d) • char c = 0xf2;char e = ~c; //e is 0x0d DCO

  7. EXCLUSIVE OR • 1111 0010 (0xf2)1111 1110 (0xfe)-------------- (^) 0000 1100 (0x0c)char c = 0xf2;char d = 0xfe;char e = c ^ d; //e is 0x0c 0 0 --> 0 1 1 --> 0 0 1 --> 1 1 0 --> 1 DCO

  8. SHIFT >> 1 - (right) by one bit • 1111 0010 (0xf2)>> 1 (shift right by one bit)--------------------- • 0111 10001 (0x79)char c = 0xf2;char e = c >>1; //e is 0x79 DCO

  9. SHIFT >> 2 - by two bits • 1111 0010 (0xf2)>> 2 (shift right by one bit)--------------------- • 0011 1100 (0x3c)char c = 0xf2;char e = c >>2; //e is 0x3c DCO

  10. SHIFT << 1 - (left) by one bit • 1111 0010 (0xf2)<< 1 (shift right by one bit)--------------------- • 1110 0100 (0xe4)char c = 0xf2;char e = c <<1; //e is 0xe4 DCO

  11. SHIFT << 2 - by two bits • 1111 0010 (0xf2)>> 2 (shift right by one bit)--------------------- • 1100 1000 (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*128 + 0*64 + 0*32 + 1*16 + 0*8 + 1*4 + 1*2 + 0*1 or • 1001 0110 (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 • 1001 0010 //original data • 0110 1101 //reverse 1 to 0, 0 to 1 • One’s complement • 0110 1101 • + • 0000 0001 (add one to one’s complement) • 0110 1110 • Two’s complement DCO

  17. Application of twos’ complement • 1100 0011 – 0011 0001 • -0011 0001 can be converted into two’s complement • The operation becomes addition instead of subtraction • Two’s complement –0011 0001 is 1100 1111 • 1100 0011 – 0011 0001 is • 1100 0011 + 1100 1111 =0001 0010 • Convert “-” to “+” DCO

  18. Signed integer of 100 • Negative sign is represented by 1 (most significant bit) • 100 (decimal) 0x0064 (hex) • 0000 0000 0110 0100 • positive • -100 (decimal) 0xff9c (hex) • 1111 1111 1001 1100 • negative DCO

  19. Example of – 101 (decimal) 101 (decimal ) is 0x0065 (hex) 0000 0000 0110 0101 ( 32 bits integer) -101 is two’s complement 0xff9b Convert 0 to 1, 1 to 0 and then add 1 0000 0000 0110 0101 (101 in decimal) 1111 1111 1001 1010 (convert 1 to 0 and 0 to 1) 1111 1111 1001 1011 (add 1, result) DCO

  20. Left & right Shift • 101 (dec) is 0x0065 (hex) • Shift right by one bit >>1 • 0000 0000 0110 0101 (101 dec) • >> 1 (shift left) • 0000 0000 0011 0010 (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 32767 • If you add 32767 + 1 (dec), it produces 0 1111 1111 1111 1111 (binary) + 0000 0000 0000 0001 1 0000 0000 0000 0000 (overflow) DCO

  22. Floating • Fixed-Point Representations – 32.45 • Floating-Point Representations such as 32.45, 3.245 x 10, 0.3245 x 100 • Normalization and Hidden Bits • IEEE Floating Point • Details DCO

  23. Fixed-Point Representations • Such as 123.45 • 1234.78 • 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 • 1.234 • 78.945 • 12.56 • 4.5 E10 • 4.5 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 -23 2 • Mantissa is unsigned bit, the expression is 1 + Mantissa * 2^(-23) • If Mantissa is zero, the expression becomes 1 + 0 * 2^(-23) = 1 • If Mantissa is 2^23, the expression becomes 1 + 2^23 * 2 ^(-23) = 1 + 1 = 2 • It ranges from 1 to 2 DCO

  29. Example • 2.5 (floating point) • 0100 0000 0010 0000 0000 0000 0000 0000 • Sign: positive (1) • Exponent : 1000 0000 : 128 (128 – 127 = 1) • Mantissa: 1. 010 0000 0000 0000 0000 0000, 1.25 • Result 1 x 1.25 x 2^1 = 2.5 DCO

  30. Example • Determine the values of • 1011 1101 0100 0000 0000 0000 0000 0000 • Sign = 1, negative • Exponent = 122 so exponent = -5 (122 – 127) • Mantissa: 100 0000 0000 0000 0000 0000 = 1.1 • So result –1 x 1.1 x 2^(-5) 2 = -0.046875 10 (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 • 0x00341256 can be an integer, floating point, or a statement. That is why we need to define int i; Char a; DCO

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