Float precision bits

Question

In this wiki article it shows 23 bits for precision, 8 for exponent, and 1 for sign

Where is the hidden 24th bit in float type that makes (23+1) for 7 significand di

Answer 1

As you write, the single-precision floating-point format has a sign bit, eight exponent bits, and 23 significand bits. Let s be the sign bit, e be the exponent bits, and f be the significand bits. Here is what various combinations of bits stand for:

If e and f are zero, the object is +0 or -0, according to whether s is 0 or 1.

If e is zero and f is not, the object is (-1)^s * 2^1-127 * 0.f. "0.f" means to write 0, period, and the 23 bits of f, then interpret that as a binary numeral. E.g., 0.011000... is 3/8. These are the "subnormal" numbers.

If 0 < e < 255, the object is (-1)^s * 2^e-127 * 1.f. "1.f" is similar to "0.f" above, except you start with 1 instead of 0. This is the implicit bit. Most of the floating-point numbers are in this format; these are the "normal" numbers.

If e is 255 and f is zero, the object is +infinity or -infinity, according to whether s is 0 or 1.

If e is 255 and f is not zero, the object is a NaN (Not a Number). The meaning of the f field of a NaN is implementation dependent; it is not fully specified by the standard. Commonly, if the first bit is zero, it is a signaling NaN; otherwise it is a quiet NaN.