1 1 vote Why floating point in de-normalized normal form has range between : $\pm1\times2^{-149}$ and $\pm(1 - 2 ^{-23})\times2^{-126}$ Digital Logic floating-point-representation number-theory digital-logic + – Durgesh Singh 1.2k views answer comment Share Follow Print See all 6 Comments 6 6 Comments reply Show 3 previous comments MiNiPanda commented Jan 13, 2018 reply Follow flag Yeah..saw it :P 0 0 replyShare Durgesh Singh commented Jan 13, 2018 reply Follow flag got the clarity but i do not understand the -126 part.. if exponent is all zeros then substracting 127 should give -127 (IEEE 754 single precision format)? 0 0 replyShare MiNiPanda commented Jan 13, 2018 reply Follow flag Yes that is what confused me also. But what i think is, suppose a binary no. is represented in 1.M*2(e-127) form as 1.0111..1 * 2-127 where e=0. But here in this denormalized form we are shifting that point towards left by one bit i.e. 0.10111...1 to make in the form of 0.M. So the exponent increases by one unit i.e. -126. Let me know what you think. I may be wrong. 0 0 replyShare Please log in or register to add a comment.
