Understanding how these floating point numbers work?

Per IEEE 754-2008:

• NaN: If the exponent field is all ones and the significand field is not zero, the floating-point datum is a NaN, regardless of the sign field. Preferably, a QNaN has the leading bit of the significand field 1 and a signaling NaN has 0, but this is not required.
• Infinite: If the exponent field is all ones and the significand field is zero, the datum is (−1)^s • ∞, where s is the sign field. (I.e., +∞ if the sign is 0 and −∞ if the sign is 1.)
• Normal: If the exponent field is neither all zeros nor all ones, the datum is (−1)^s • (1 + f • 2^−q) • 2^{e - bias}, where s is the sign field, f is the significand field, q is the number of bits in the significand field, e is the exponent field, and bias is the exponent bias (127 for 32-bit floating-point).
• Subnormal: If the exponent field is all zeros, and the significand field is not, the datum is (−1)^s • (0 + f • 2^−q) • 2^{1 - bias}. Note the two differences from normal: 0 is added to the significand instead of 1, and 1 is used for the exponent (before subtracting bias). This means subnormals have the same exponent as the smallest normals but are decreased by reducing the significand.
• Zero: If the exponent field is all zeroes, and the significand field is also all zeros, the datum is (−1)^s • 0. (Note that IEEE 754 distinguishes +0 and −0.)

The exponent used with subnormals is 1 rather than 0 so that the numbers change from (normal) 1.000…000•2¹⁻¹²⁷ to (subnormal) 0.111…111•2¹⁻¹²⁷. If 0 were used, there would be a jump to 0.0111…1111•2¹⁻¹²⁷.

The formula for the values of subnormals works for zeros too. So zeros do not actually need to be listed separately above.