Computer Arithmetic

How are the bits of an IEEE 754 single-precision (32-bit) number allocated among sign, exponent and significand?

  1. A1, 11, 20
  2. B1, 8, 23
  3. C1, 5, 26
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Answer: B. 1, 8, 23

Single precision (FP32) uses 1 sign bit, 8 exponent bits and 23 significand bits, 32 in total. For normalized numbers the leading 1 of the significand is implied, so precision is 24 bits, about seven decimal digits. Double precision (FP64) uses 1, 11 and 52 bits, with a wider exponent and higher precision. 1, 11, 20 confuses it with double precision's exponent width, and 1, 5, 26 with half precision's.

  • A. An 11-bit exponent is the double-precision width.
  • B. Correct. 1 sign, 8 exponent and 23 significand bits.
  • C. A 5-bit exponent is the half-precision (FP16) width.
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More Computer Arithmetic questions

  1. In IEEE 754 double precision (64-bit), how many bits are the exponent and significand?
  2. In IEEE 754 single precision, if the stored exponent field is 130, what is the actual exponent (power of two)?
  3. What is the effect of not storing the leading 1 of the significand (the hidden bit) in IEEE 754 normalized numbers?
  4. What is the hexadecimal bit pattern of decimal 1.0 in IEEE 754 single precision?
  5. Converting −6.25 to single precision, what is the sign–exponent–significand bit string written in hexadecimal?