What's the bit pattern for minimal value in double 64 bit

问题

I assumed that the minimal positive value that can be put into double floating point is this

0 0000000000 00000000000000000000000000000000000000000000000000000

And in scientific form it's this:

1 x 2^{-1023}

However, this article states that:

As mentioned above, zero is not directly representable in the straight format, due to the assumption of a leading 1 (we'd need to specify a true zero mantissa to yield a value of zero). Zero is a special value denoted with an exponent field of all zero bits, and a fraction field of all zero bits.

So what's the pattern for minival positive value?

回答1:

When the exponent part is all zero, the numbers are subnormals or denormals, where the implicit digit is 0 instead of 1. Therefore

0 00000000000 0000000000000000000000000000000000000000000000000000

is zero, and the next number

0 00000000000 0000000000000000000000000000000000000000000000000001

is the smallest positive number, being equal to

0.0000000000000000000000000000000000000000000000000001₂×2^-1022 = 2^-1074 ≈ 5.0×10^-324

UPDATE: so why is the exponent -1022, when the bias is 1023? The subnormals have the same exponent as the first binade of normal floats (so 1-1023 = 1022). This is so that the spread is continuous, i.e. the largest subnormal is

0.1111111111111111111111111111111111111111111111111111₂×2^-1022

and the next floating point number (the smallest normal):

1.0000000000000000000000000000000000000000000000000000₂×2^-1022

来源：https://stackoverflow.com/questions/39944832/whats-the-bit-pattern-for-minimal-value-in-double-64-bit