what is problem there?

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closed with the note:
resolved

1

smallest fractional form in 32 bit will be 0.00000.....1(23rd bit). Weight of this bit is 2^-23. Just used fractional form for denormalized case. You will get 2^-149

0

I got

smallest fractional form in 32 bit will be 0.00000.....1(23rd bit). Weight of this bit is 2^-23.

But. I did not get

Just used fractional form for denormalized case. You will get 2^-149

I am interpreting it as you added -126 to -23 to get -149. But what is this -126?

- To get unbiased exponent in normalized numbers, we do $e-b$, where $e$ is biased exponent and $b$ is bias. $b$ in IEEE 754 is 127. So it will be $e-127$ for normalized numbers. I know exponent in denormalized numbers is zero. So if we are to calculate unbiased exponent of denormalized numbers in similar way that in normalized numbers, it would have been $0-127=-127$, not $-126$.
- I just checked this link. In the table of summary section, it says value of positive denormalized real is $0.f\times 2^{(-b+1)}$, which will correctly give $2^{-149}$ . But, I didnt get from where that $-b\color{red}{+1}$ came. Especially that $+1$.
- Also isnt denormalized numbers unbiased? Or they indeed are biased, just that their exponent is 0, more precisely a biased zero.

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bro super thanks!!!!

Maaan you seems to be enlighted one!!!! Did you just come up with that reasoning? Or you found it somewhere? Anyways thanks.

Maaan you seems to be enlighted one!!!! Did you just come up with that reasoning? Or you found it somewhere? Anyways thanks.

0

That link only I put in earlier comment and asked doubt on it:

Just tell from where $-b\color{red}{+1}$ came on that page in summary table at the end

2

Denormalized numbers are used to fill the gap between 0 and smallest positive number in normalized form.

what is smallest positive number in normalized form? S=0 M =0 and E =1 so it will become 1.0 * 2 ^ -126

Now if number is smaller than this, it will be of form 0.M * 2 ^ -126

Hence when E=0 to make it -126 than +1 is added. i.e E- bias +1 = 0-127+1= -126.

what is smallest positive number in normalized form? S=0 M =0 and E =1 so it will become 1.0 * 2 ^ -126

Now if number is smaller than this, it will be of form 0.M * 2 ^ -126

Hence when E=0 to make it -126 than +1 is added. i.e E- bias +1 = 0-127+1= -126.