IEEE single precision floating-point format

Question

Why doesn't the double precision format simply double the bits in each field, rather than only doubling the fraction bits? Also, what is the hidden bit and why is it used?

Answer 1

Your assessment isn't quite true:

• IEEE754 16-bit float: 1 sign bit, 5 exponent bits, 10(+1) significand bits, exp ∈ [-14, 15]
• IEEE754 32-bit float: 1 sign bit, 8 exponent bits, 23(+1) significand bits, exp ∈ [-126, 127]
• IEEE754 64-bit float: 1 sign bit, 11 exponent bits, 52(+1) significand bits, exp ∈ [-1022, 1023]
• IEEE754 80-bit float: 1 sign bit, 15 exponent bits, 64(+0) significand bits, exp ∈ [-16382, 16383]

So nobody is quite getting doubled. More precision is presumably more useful than a wider range. Recall that the range of representable values increases ... exponentially in the size of the exponent.