Difference between Java's `Double.MIN_NORMAL` and `Double.MIN_VALUE`?

Question

What's the difference between Double.MIN_NORMAL (introduced in Java 1.6) and Double.MIN_VALUE?

Answer 1

The answer can be found in the IEEE specification of floating point representation:

For the single format, the difference between a normal number and a subnormal number is that the leading bit of the significand (the bit to left of the binary point) of a normal number is 1, whereas the leading bit of the significand of a subnormal number is 0. Single-format subnormal numbers were called single-format denormalized numbers in IEEE Standard 754.

In other words, Double.MIN_NORMAL is the smallest possible number you can represent, provided that you have a 1 in front of the binary point (what is referred to as decimal point in a decimal system). While Double.MIN_VALUE is basically the smallest number you can represent without this constraint.

Answer 2

IEEE-754 binary64 format:

s_eee_eeee_eeee_mmmm_mmmm_mmmm_mmmm_mmmm_mmmm_mmmm_mmmm_mmmm_mmmm_mmmm_mmmm_mmmm 

(1 s; 3×4−1 =11 es; 64−3×4 =52 ms)

, and its algorithm:

• If e >000_0000_0000 and <111_1111_1111: interpret as (-1)^s ×2^{e−balancer:1023} ×(base:1 +m×2^{−sub-one-pusher:52}). (These are the normal numbers.)

• If e =000_0000_0000: do the same (as line above) except base:1 is base:0, and e is e +1. (These are the subnormal numbers, except for zero which is neither subnormal/normal.)

• If e =111_1111_1111 and m =0000...0000: interpret as (-1)^s × infinity.

• If e =111_1111_1111 and m <>0000...0000: interpret as NaN. (Btwbtw: therefore there're 2× (2⁵² −1) different bit representations for NaN, cf #Quiet NaN &doubleToRawLongBits.)

Thus:

• The smallest of its possible positive numbers is 0_000_0000_0000_0000_..._0001 (Double.MIN_VALUE (also .NET's Double.Epsilon)) (a subnormal number).

• The smallest of its possible positive normal numbers is 0_000_0000_0001_0000_..._0000 (Double.MIN_NORMAL).

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Appendix:

MIN_VALUE computation:

         (-1)^s:0 ×2^{(e:0+1)−balancer:1023} ×(base:0 +m:1 ×2^{−sub-one-pusher:52})

      = 1 ×2⁻¹⁰²² ×2⁻⁵²

      = 2⁻¹⁰⁷⁴ (~4.94 × 10⁻³²⁴)

, and MIN_NORMAL computation:

         (-1)^s:0 ×2^{e:1 −balancer:1023} ×(base:1 +m:0 ×2^{−sub-one-pusher:52})

      = 1 ×2⁻¹⁰²² ×1

      = 2⁻¹⁰²² (~2.225 × 10⁻³⁰⁸)