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What is the effect of increasing the number of bytes allocated to the mantissa?
The mantissa holds the bits which represent the number, increasing the number of bytes for the mantissa increases the number of bits for the mantissa and so increases the size of the number which can be accurately held, ie it increases the accuracy of the stored number.
How do you represent floating point number in microprocessor?
It is somewhat complicated (search for the IEEE floating-point representation for more details), but the basic idea is that you have a few bits for the base, and a few bits for the exponent. The numbers are stored in binary, not in decimal, so the base and the exponent are the numbers "a" and "b" in a x 2b.
How many bits will it take to store a 7-digit telephone number in BCD?
28-bits
How are floating point numbers handled as binary numbers?
Floating point numbers are typically stored as numbers in scientific notation, but in base 2. A certain number of bits represent the mantissa, other bits represent the exponent. - This is a highly simplified explanation; there are several complications in the IEEE floating point format (or other similar formats).Floating point numbers are typically stored as numbers in scientific notation, but in base 2. A certain number of bits represent the mantissa, other bits represent the exponent. - This is a highly simplified explanation; there are several complications in the IEEE floating point format (or other similar formats).Floating point numbers are typically stored as numbers in scientific notation, but in base 2. A certain number of bits represent the mantissa, other bits represent the exponent. - This is a highly simplified explanation; there are several complications in the IEEE floating point format (or other similar formats).Floating point numbers are typically stored as numbers in scientific notation, but in base 2. A certain number of bits represent the mantissa, other bits represent the exponent. - This is a highly simplified explanation; there are several complications in the IEEE floating point format (or other similar formats).
How is scientific notation related to the floating point representation used by computers?
Floating point numbers are stored in scientific notation using base 2 not base 10.There are a limited number of bits so they are stored to a certain number of significant binary figures.There are various number of bytes (bits) used to store the numbers - the bits being split between the mantissa (the number) and the exponent (the power of 10 (being in the base of the storage - in binary, 10 equals 2 in decimal) by which the mantissa is multiplied to get the binary/decimal point back to where it should be), examples:Single precision (IEEE) uses 4 bytes: 8 bits for the exponent (encoding ±), 1 bit for the sign of the number and 23 bits for the number itself;Double precision (IEEE) uses 8 bytes: 11 bits for the exponent, 1 bit for the sign, 52 bits for the number;The Commodore PET used 5 bytes: 8 bits for the exponent, 1 bit for the sign and 31 bits for the number;The Sinclair QL used 6 bytes: 12 bits for the exponent (stored in 2 bytes, 16 bits, 4 bits of which were unused), 1 bit for the sign and 31 bits for the number.The numbers are stored normalised:In decimal numbers the digit before the decimal point is non-zero, ie one of {1, 2, ..., 9}.In binary numbers, the only non-zero digit is 1, so *every* floating point number in binary (except 0) has a 1 before the binary point; thus the initial 1 (before the binary point) is not stored (it is implicit).The exponent is stored by adding an offset of 2^(bits of exponent - 1), eg with 8 bit exponents it is stored by adding 2^7 = 1000 0000Zero is stored by having an exponent of zero (and mantissa of zero).Example 10 (decimal):10 (decimal) = 1010 in binary → 1.010 × 10^11 (all digits binary) which is stored in single precision as:sign = 0exponent = 1000 0000 + 0000 0011 = 1000 00011mantissa = 010 0000 0000 0000 0000 0000 (the 1 before the binary point is explicit).Example -0.75 (decimal):-0.75 decimal = -0.11 in binary (0.75 = ½ + ¼) → 1.1 × 10^-1 (all digits binary) → single precision:sign = 1exponent = 1000 0000 + (-0000 0001) = 0111 1111mantissa = 100 0000 0000 0000 0000 0000Note 0.1 in decimal is a recurring binary fraction 0.1 (decimal) = 0.0001100110011... in binary which is one reason floating point numbers have rounding issues when dealing with decimal fractions.
