Highest (existing) number in half precision IEEE 754

问题

Why is 0 11110 1111111111 and not 0 11111 1111111111 the highest half precision number?

回答1:

Because an exponent field of 11111₂ is reserved for infinities and NaNs. Section 3.4 of the IEEE 754-2008 standard says:

The range of the encoding’s biased exponent E shall include:

• every integer between 1 and 2^w − 2, inclusive, to encode normal numbers
• the reserved value 0 to encode ±0 and subnormal numbers
• the reserved value 2^w − 1 to encode ±∞ and NaNs.

Here "w" is the width of the exponent field. This applies not just to the half-precision format, but to all the IEEE 754 binary interchange formats (including the usual single and double precision formats).

In the particular case of the binary16 format, w is 5 and the top 5 bits following the sign bit describe the biased exponent E, and can range from 00000₂ to 11111₂. The top value in this range, 11111₂, is used for infinities and NaN values, and the bottom value of 00000₂ is used for zeros and subnormals. The remaining biased exponent values are used for representing finite normal numbers. So the top biased exponent value available for a finite number is 2^w - 2 = 11110₂ rather than 11111₂.

For this format, the exponent bias is 15, so the bit pattern 0 11110 1111111111 gives an unbiased exponent of 30 - 15 = 15, and represents a value of 1.1111111111₂ × 2¹⁵, or 65504.

To give another example, the largest binary32 (single-precision) float is represented by bit pattern

0 11111110 11111111111111111111111₂

来源：https://stackoverflow.com/questions/33454045/highest-existing-number-in-half-precision-ieee-754