GATE CSE 2014 Set 2 | Question: 45

Question

The value of a $\text{float}$ type variable is represented using the single-precision $\text{32-bit}$ floating point format of $\text{IEEE-754}$ 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

• A. $\text{C1640000H}$
• B. $\text{416C0000H}$
• C. $\text{41640000H}$
• D. $\text{C16C0000H}$

Answer 1

Since No is negative S bit will be 1
Convert 14.25 into binary 1110.01
Normalize it : 1.11001 X 2 ^ 3
Biased Exponent (Add 127) : 3 + 127 = 130 (In binary 10000010)
Mantissa : 110010.....0 (Total 23 bits)

Num represented in IEEE 754 single precision format :

1 10000010 11001000000000000000000

In Hex (Group of Four bits) -

1100 0001 0110 0100 0000 0000 0000 0000

Num becomes : C1640000

Comments

@  Regina Phalange

very nice explanation.. :)