When assigning bits to the floating-point representation, a decision is to be made for the total number of space required for mantissa an...
When assigning bits to the floating-point representation, a decision is to be made for the total number of space required for mantissa and exponent. For assigning the bits, a choice is given in the floating-point program.
Usually, we are likely to choose for the higher numbers of bits for mantissa for precision and provide less space for exponent. This might not be precise as it leaves out the possible outcomes. In order to use the space optimally, full use of bits is necessary. In order to achieve precision, normalization is essential in the floating-point representation.
To normalize the number of bits, largest possible magnitude should be represented by the mantissa. To illustrate this, lets take 8 bit representation in the Table. B2 for positive number.
Denary Representation |
Floating-point binary representation |
0.125 x 2^{4} |
0 001 0100 |
0.25 x 2^{3} |
0 010 0011 |
0.5 x 2^{2} |
0 100 0010 |
Table B2. Representing floating-point in binary by using Denary 2 with using four bits for mantissa and exponent in positive.
Denary Representation |
Floating-point binary representation |
-0.25 x 2^{4} |
1 100 0100 |
-0.5 x 2^{3} |
1 100 0011 |
-1.0 x 2^{2} |
1 000 0010 |
Table B3. Representing floating-point binary by using four bit for mantissa and exponent in negative.
As seen in the table, the mantissa with the highest magnitude represents the numbers of the two most important bits. This is how we can know if a number is in normalized representation or not. The table shows how numbers are normalized. In Table B3, the mantissa portion is shifted to left where 0 comes before 1. In positive number, as the floating-point representation is increased, the exponent side is decreased by 1.
Same goes for Table B3, where numbers are in negative. In this representation, the significant bits in mantissa and exponent are of 4 bits where 1 comes before 0 in order to be normalized.
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