+1 vote
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## A 32-bit floating-point number (follows IEEE Standard)is represented by a 8-bit signed exponent, and a 23-bit fractional mantissa. The base of the scale factor is 16, The range of the exponent is ___________ .

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32 bit means single precision floating point number

Where exponent will be of 8 bits

So, range of exponent will be $-2^{8-1}-1$ to $2^{8-1}$ i.e.$-127$ to $128$
by Veteran (117k points)
+1
-$2^{8-1}-1$ why???? -$2^{8-1}$ to $2^{8-1}-1$ why not???
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no , range of 2's complement is $-2^{n-1}-1$ to $-2^{n-1}$ how far I remember
+1

@ srestha  I think for the IEEE 754 binary32 case, an exponent value of 127 represents the actual zero (i.e. for 2e − 127 to be one, e must be 127). Exponents range from −126 to +127 because exponents of −127 (all 0s) and +128 (all 1s) are reserved for special numbers.

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@iarnav , could you please explain what is meaning of "The base of the scale factor is 16" ?
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@ ankitgupta.1729 "The base of the scale factor is 16" means that exponent is raised to 16.

For example- (-1)^s * 1.M * 16^ (exponent-bias)

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i think scale factor will not effect the range of exponent
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@srestha

It is -2^[n-1] to +2^[n-1]-1