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Computer Architecture and Operating Systems CS 3230 :Assembly Section Lecture 10

Computer Architecture and Operating Systems CS 3230 :Assembly Section Lecture 10. Department of Computer Science and Software Engineering University of Wisconsin-Platteville. Converting 0.85 to binary. 0.85 × 2 = 1.7 0.7 × 2 = 1.4 0.4 × 2 = 0.8 0.8 × 2 = 1.6 0.6 × 2 = 1.2 0.2 × 2 = 0.4

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Computer Architecture and Operating Systems CS 3230 :Assembly Section Lecture 10

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  1. Computer Architecture and Operating SystemsCS 3230 :Assembly SectionLecture 10 Department of Computer Science and Software Engineering University of Wisconsin-Platteville

  2. Converting 0.85 to binary 0.85 × 2 = 1.7 0.7 × 2 = 1.4 0.4 × 2 = 0.8 0.8 × 2 = 1.6 0.6 × 2 = 1.2 0.2 × 2 = 0.4 0.4 × 2 = 0.8 0.8 × 2 = 1.6

  3. Decimal Fractions vs Binary Floating-Point • In decimal : 0.123 = 1 × 10−1 + 2 × 10−2 + 3 × 10−3 • In binary: • 0.1012 = 1 × 2−1 + 0 × 2−2 + 1 × 2−3 = 0.625 • 110.0112 = 4 + 2 + 0.25 + 0.125 = 6.375

  4. Floating-Point Representation • In decimal, floating numbers can be represented using standard-form scientific notation • Notation: 1.Sxbe • S: significand , e: exponent , b: system base • Example: • 152853.5047 = 1.528535047×105 • S=528535047, e=5

  5. IEEE floating point representation • Types • Single Precision • 32 bits: 1 bit for the sign, 8 bits for the exponent, and 23 bits for the fractional part of the significand • Double Precision • 64 bits: 1 bit for the sign, 11 bits for the exponent, and 52 bits for the fractional part of the significand • Double Extended Precision • 80 bits: 1 bit for the sign, 16 bits for the exponent, and 63 bits for the fractional part of the significand

  6. Single precision representation

  7. Example: • How would 23.85 be stored? • First, it is positive so the sign bit is 0. • 23.85/16 = 1.490625 ,16 = 24 • Next, the true exponent is 4, so the biased exponent is 7F+4 = 8316 = 10000011 • Finally, the fraction (0.490625) is 01111101100110011001100 (remember the leading one is hidden). • How would -23.85 be stored? • Just change the sign bit: C1 BE CC CD. • Do not take the two’s complement!

  8. Converting Single-Precision to Decimal 1. If the MSB is 1, the number is negative; otherwise, it is positive. 2. The next 8 bits represent the exponent. Subtract binary 01111111 (decimal 127), producing the unbiased exponent. Convert the unbiased exponent to decimal. 3. The next 23 bits represent the significand. Notate a “1.”, followed by the significand bits. Trailing zeros can be ignored. Create a floating-point binary number, using the significand, the sign determined in step 1, and the exponent calculated in step 2 4. Shift the binary point the number of places equal to the value of the exponent. Shift right if the exponent is positive, or left if the exponent is negative 5. From left to right, use weighted positional notation to form the decimal sum of the powers of 2 represented by the floating-point binary number.

  9. Example • Convert 0 10000010 1011000000000000000000 to Decimal • The number is positive. • The unbiased exponent is binary 00000011, or decimal 3 • Combining the sign, exponent, and significand, the binary number is +1.01011 X 23 • The unnormalized binary number is +1010.11 • The decimal value is +10 3/4, or +10.75.

  10. Floating Point Arithmetic • Addition and subtraction: • To add/sub two floating point numbers, the exponents must be equal • If they are not already equal, then they must be made equal by shifting the significand of the number with the smaller exponent • Example: 10.375 + 6.34375 = 16.71875 1.0100110 × 23 + 1.1001011 × 22 16.75

  11. Floating Point Arithmetic • Multiplication: • The significands are multiplied and the exponents are added • Example: 10.375 × 2.5 = 25.9375

  12. Numeric Coprocessor • 80x86 has a math coprocessor that performs many floating point operations much faster than using a software procedure • The numeric coprocessor has eight floating point registers. Each register holds 80-bits of data. • The registers are named ST0, ST1, ST2, . . . ST7, which are organized as a stack • There is also a status register in the numeric coprocessor. It has several flags

  13. FPU Register Stack • Three-bit field named TOP in the FPU status word identifies the register number that is currently the top of stack.

  14. Load/store Floating-Point Value • FLD mem_location • Copies floating point operand from memory into the top of the FPU stack, ST(0) • FST mem_location • Copies floating point operand from the top of the FPU stack into memory • FSTP • pops the stack after copying

  15. Arithmetic Instructions

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