CS 208 w20 lecture 7 outline

• Common mistakes: applying little endian, data type widths, pointers as data
• Average score 25.6, median score 27

• Good: practice problems, discussing with neighbors, course web page
• Comments: more C in class, book can be challenging

• How would the bits of 0x8a000000 be interpreted as an int and a float? Can express in terms of powers of two.
  • \(-2^{31} + 2^{27} + 2^{25} = -59 \times 2^{25}\)
  • \(-1 \times 2^{-107}\)
• We execute the following code in C. How many bytes are the same (value and position) between i and f?

int i = 384; // 2^8 + 2^7
float f = (float) i;

• i is 0x00 00 01 80
• 384 = 0b1 1000 0000 = \(2^8 \times 1.1\)
  • s = 0
  • exp = \(E + Bias = 8 + 127 = 135 =\) 0b1000 0111
  • frac = \(M - 1 =\) 0b100...0
• 0b0100 0011 1100 0000 0000 0000 0000 0000
• f is 0x43 c0 00 00

• Denormalized: evenly spaced numbers close to 0 (when exp bits are all zero)
• Normalized: exponentially spaced numbers outside of denormalized range (when exp bits ≠ 0 and are not all 1s)

Key trade-off vs fixed-point: IEEE achieves far greater precision near zero and an order of magnitude greater range at the expense of less precision farther away from 0. With a 64-bit double, the values at which this loss of precision becomes significant are so large as to arise very rarely in practice. Conversely, very small values occur in all sorts of practical applications, making the extra precision there very valuable in approximating real arithmetic.