GATE CSE 2008 | Question: 4

Question

In the IEEE floating point representation the hexadecimal value $0\text{x}00000000$ corresponds to

• A. The normalized value $2^{-127}$
• B. The normalized value $2^{-126}$
• C. The normalized value $+0$
• D. The special value $+0$

Comments

refer this:

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Also Refer :
Digital Logic: GATE CSE 2021 Set 2 | Question: 4
Digital Logic: GATE CSE 2024 | Set 2 | Question: 4

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Thanks bro

Answer 1

$$\begin{array}{|c|c|c|}\hline \textbf{S} & \textbf{BE} & \textbf{M} &  \textbf{Value} \\\hline 0/1 & { {\text{All 0’s}}} & { {\text{All 0’s}}} & 0 \\\hline 0 & { {\text{All 1’s}}} & { {\text{All 0’s}}} & +\infty \\\hline 1 & { {\text{All 1’s}}} & { {\text{All 0’s}}} & -\infty \\\hline 0/1 & { {\text{All 1’s}}} & \text{Non-zero}& \text{NaN} \\\hline\end{array}$$The answer is option D.

Reference: http://steve.hollasch.net/cgindex/coding/ieeefloat.html

Comments

Sir, what is the difference between option C and option D?

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earlier to IEEE 758 representation, floating numbers couldn't differentiate -0 and +0. So, they come up with IEEE 758 floating poing representation in 32 bit and 64 bit and they marked +0 and -0 representation special that's all I read somewhere, don't know exact history.

Answer 2

D. The special value +0.

Ref: http://steve.hollasch.net/cgindex/coding/ieeefloat.html

Comments

Answer is not A because Exponent and Mantisa both are zero which is equal to special value +0.

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Very Nice and useful link it clears all IEEE Floating Point Representation Doubts.

Thank you Sir!.

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Answer is not A because range of single precision normalized form is $\pm 2^{-126}$ to $\left (2 -2^{(-23)} \right ) \times 2 ^{127}$