GATE2012-7

Question

The decimal value $0.5$ in IEEE single precision floating point representation has

• A. fraction bits of $000\dots 000$ and exponent value of $0$
• B. fraction bits of $000\dots 000$ and exponent value of $−1$
• C. fraction bits of $100\dots 000$ and exponent value of $0$
• D. no exact representation

Comments

But in case if u use explicit then option c

Please explain

Answer 1

(B) is the answer. IEEE 754 representation uses normalized representation when the exponent bits are all non zeroes and hence an implicit '1' is used before the decimal point. So, if mantissa is

0000..0

it would be treated as

1.000..0

and hence the exponent need to be -1 for us to get 0.1 which is the binary representation of 0.5.

More into IEEE floating point representation:
http://steve.hollasch.net/cgindex/coding/ieeefloat.html

Comments

why didn't we used biasing here?

I mean to say saved exponent will be $127-1 = 126$

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@thor yes, exponent bits stored will be that of "126", but question asks for the value and it is of "-1".

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Nice explaination .... Thank you

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I think Biasing has been done here because we have to add the bias value to 1 that will make it 127 +1 (128) and In IEEE floating point representation Exponent is stored in 2's complement representation (continuous run of 8 One's) and its value is -1

Answer 2

In IEEE biasing of exponent is must.

Step 1: decimal 0.5 --> binary 0.1

Step 2: normalize binary 0.1 --> 1.0 * 2^-1

Step 3: exponent -1 + 127 = 126 = binary 01111110

Step 4: remove hidden digit from 1.0 --> 0 (1 is implicit in IEEE representation)

Step 5: 0.5 is positive - the sign bit is zero: 0

The next eight bits are the exponent: 01111110

The next 23 bits are the mantissa: 000000000000000000000

Binary result (32 bits): 10111111000000000000000000000000

I think there is something wrong with the question. Arjun Sir, please explain.

Comments

Question did not ask for the "exponent bits" but its value. So, "-1" is enough.

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Thank You so much sir.

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@ Rajendra Dangwal

sign bit is 0 so firt bit(MSB) in ur ans must be 0 i.e 00111111000000000000000000000000