Explain why the hidden bit of floating point format does not need to be represented

Question

Explain why the hidden bit of floating point format does not need to be represented.

help? I know that the hidden bit exists for more precision but why doesn't it need to be represented?

Answer 1

If you mean by the hidden bit the the one preceding the mantissa H.xxxxxxx, H=hidden, the answer is that it is implicitly 1, when exponent>0 and it's zero, when exponent==0.

Omitting the bit, when it can be calculated from the exponent, allows one more bit of precision in the mantissa.

Answer 2

@Aki's answer is correct. The implied bit is not always 1.

For 0.0, -0.0 & denormal numbers (gradual underflow) the implied bit if any would be 0.
Those numbers are all the ones having a zero biased exponent.

Non finite float (+inf -inf NaN) don't need any concept of implied bit at all.
Though, adding such a bit wouldn't hurt
Those numbers are all the ones having biased exponent set to all ones.

For every other float, implied bit would be 1.

Technically, the answer is that we don't need to store the implied bit because we put additional logic in FPU circuitry to reconstruct it from above rules ;).

For example, the implied bit could be obtained by ORing all the bits of biased exponent, so it's not that expensive after all.

Answer 3

It will be always one so we don't represent it You will only get a hidden bit when you normalize a binary number and the meaning of normalization is writing a number in the form of

1.xxxx x 2^x (eg: 110.11 becomes 1.1011x2^2)

so the first bit always becomes 1 so we don't need to represent it