ieee floating point format

Question

I'm having trouble converting 19861119 into ieee floating point format (single precision). I hope someone can tell me where I've lost my way.

In binary, the value is b1:00101111:00001110:01111111 (using : to mark every 8 bits from the rhs), which is b1.00101111:00001110:01111111 * 2^24. So the float is b00101111:00001110:01111111 and the biased exponent is 24 + 127 = 151 = b10010111. The float is 24 bits long but ieee format only allows 23 bits for it, which looks problematic to me. Does the format lack sufficient precision to store a yyyymmdd date?

When I write the output from Python's struct.pack("f", 19861119) to file and take a look with a hex editor, I see x4087974b. After allowing for little-endianness this is x4b978740. So Python has written a biased exponent of b01010111 = 87 and a float of b0010111:10000111:01000000 which bear little resemblance to any of the numbers I calculated. What have I missed?

Thanks in advance,

Ian

Answer 1

The number is stored rounded: 19861120 becomes 0x4B978740. The unrounded value is 0x4B97873F.

Here's 0x4B978740 in binary:

0   10010111   [1]   0010111 10000111 01000000

+   127 + 24    1                   ~ .1838150

And 2²⁴ = 16777216.

The online float calculator is great for exploring such details.

Talking about rounding modes in IEE754 could fill a whole essay...

Answer 2

The original number is:

1 0010 1111 0000 1110 0111 1111

Normalizing we have:

1,0010 1111 0000 1110 0111 1111 x 2²⁴

Since only 23 bits of the fractional part will be stored:

  1,0010 1111 0000 1110 0111 111 x 2²⁴
                              +1 (if rounded)
= 1,0010 1111 0000 1110 1000 000 x 2²⁴

= 0|100 1011 1|001 0111 1000 0111 0100 0000
  s    exp         23 bit fractional part
= 0x4B978740

And if I understood well, that's what are you getting.

Answer 3

32 bit single-precision does not have enough precision to store YYYYMMDD dates. Let's say we go up to 20 million, which is 25 bits. So you always lose one bit.

The one thing to keep in mind with floating-point formats is that in the mantissa part, the first digit is always 1. Single and double precision take advantage of that by not storing it.

In your notation: 19861119 = +1 * b1.00101111:00001110:01111111 * 2^24

So, for the three parts of the number, we have:

1. The sign is 0.
2. The exponent part is 127 + 24 = 151 = b1001011:1
3. The mantissa part is the first 23 bits after the initial zero, rounded to even: `0010111:10000111:01000000

So the full number is:

0 1001011:1 0010111:10000111:01000000
s eeeeeee e mmmmmmm mmmmmmmm mmmmmmmm

or in hex, depending on ended-ness: 4b978740 / 4087974b.