floating point

Question

Consider the following floating point format

31                                        30                                 23                                                                     0 

S

E

M

Bias=31

Assume only 0 are padded while shifting a field

The decimal number 0.15625×2^5 has the following hexadecimal representation without normalization.
A)24 28 00 00
B)24 10 00 00
C)21 28 00 00
D)21 10 00 00

Comments

Answer is (a)

As Bias given is 31 so we have to consider it in our solution.

Then B.E become : A.E + Bias = 5 + 31 = 36 (0100100)

Now,  data is :
                             = 0.15625 × 2^5
                             = 0.00101 × 2^5 (by converting)
                             
Now, we have to store data without Normalization
So,  
        Sign(S) = 0 (+ve)
        B.E = 0100100 (7 bit)
        Mantissa(M) = 01000......up to 23 bits (0 padding)

So stored data is :
      0 0100100 00101000000000000000000

Which is in Hexadecimal : 24 28 00 00 ( grouping 4 bits)

Answer 1

decimal number 0.15625×2^5 =5=$101\times 2^{0}$=$1.01\times 2^{2}$

Exponent will be 128+2=130

Mantissa=$010000000000000000000000$

So, 32 bit representation will be

$0$

$1000010$

$01000000000000000000000$

Hexadecimal representation will be 21200000

Comments

why did u take bias 128 when it is given bias 31

 

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Why this question is wrong?

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This answer is wrong