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+20 votes

The value of a $\text{float}$ type variable is represented using the single-precision $\text{32-bit}$ floating point format of $IEEE-754$ standard that uses $1$ $\text{bit}$ for sign, $\text{8 bits}$ for biased exponent and $\text{23 bits}$ for the mantissa. A $\text{float}$ type variable $X$ is assigned the decimal value of $−14.25$. The representation of $X$ in hexadecimal notation is

- $C1640000H$
- $416C0000H$
- $41640000H$
- $C16C0000H$

+30 votes

Best answer

$\begin{array}{|c|c|c|c|}\hline \textbf{S}&\textbf{BE}&\textbf{M}&\textbf{Value}\\\hline 0/1&\text{All 0's}&\text{All 0's}&0\\0&\text{All 1's}&\text{All 0's}&+\infty\\1&\text{All 1's}&\text{All 0's}&-\infty\\0/1&\text{All 1's}&\text{Non zero}&\text{NaN}\\\hline\end{array}$

$\textbf{Bias} = 2^{N-1}-1$

$N-\text{Number of bits to represent exponent in binary}$

$14.25 = 1110.01000 = 1.11001000\times 2^3$

Biased exponent $= \text{actual} + \text{bias} = 3 + \text{bias}$

where; $\text{bias} = 2^{8-1}-1 = 127$

$\text{biased exponent} = 3+127 = 130 = 10000010$

Therefore, number represented as = **1 10000010 11001000000000000000000**

on converting to hexadecimal we get $(C1640000)_{16}$

0

sometimes in the question it is given that 1 in normalised form will be added implicitly in that case we would shift the decimal point before first bit or would we follow the same as shown [email protected], @Arjun

+5 votes

M=14.25=1110.01= 1.11001*2^3

M=11001 MSB 1 is for sign bit , since exponent is 8 bit biased so, 2^8 -1= 127.. E= 127 +3= 130=10000010

So , 1 foe sign bit 10000010(8 bits) for exponent and 1100100000....0(23bits)= C1640000H

M=11001 MSB 1 is for sign bit , since exponent is 8 bit biased so, 2^8 -1= 127.. E= 127 +3= 130=10000010

So , 1 foe sign bit 10000010(8 bits) for exponent and 1100100000....0(23bits)= C1640000H

0

https://www.doc.ic.ac.uk/~eedwards/compsys/float/

These might come handy with floating point numbers.

+2 votes

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