如何计算双精度+浮点精度 [英] How to Calculate Double + Float Precision

问题描述

I have been trying to find how to calculate the Floating/Double precision/range numbers -3.402823e38 .. 3.402823e38 and -1.79769313486232e308 .. 1.79769313486232e308.

For int32 you would do 2^32=4294967296/2 you get a range of -2147483648 to 2147483647. So how do i figure out the precision numbers for float and double. I think i am searching the wrong terms since nothing is coming up anywhere.

解决方案

Well, both types actually look like the following:

[sign] [exponent] [mantissa]

representing a number in the following form:

[sign] 1.[mantissa] × 2^[exponent]

with the size of the exponent and mantissa varying. For float the exponent is eight bits wide, while double has an eleven-bit exponent. Furthermore, the exponent is stored unsigned with a bias which is 127 for float and 1023 for double. This results in a range for the exponent of −126 through 127 for float and −1022 though 1023 for double.

The exponent is the exponent for 2^something so when calculating 2¹²⁷ you'll get 1.7 × 10³⁸ which gets you in the approximate range of the float maximum value. Similarly for double with 9 × 10³⁰⁷.

Obviously those numbers are not exactly those we expect. This is where the mantissa comes into play. The mantissa represents a normalized binary number that always begins with "1." (that's the normalized part). The rest is simply the digits after the dot. Since the maximum mantissa is then roughly 1.111111111... in binary, which is almost 2, we'll get approximately 3.4 × 10³⁸ as float's maximum value and 1.79 × 10³⁰⁸ as the maximum value for double.

[EDIT 2011-01-06] As Mark points out below (and below the question), the exact formula is the following:

where e is the number of bits in the exponent and p is the number of bits in the mantissa, including the aforementioned implicit bit (due to normalization). The formula replicates what we have seen above, only now accurate. The first factor, 2^{2e − 1}, is the maximum exponent, multiplied by two (we save the two in the second factor then). The second factor is the largest number we can represent below one. I said above that the number is almost two. Since we exaggerated the exponent by a factor of two in this formula, we need to account for that and now have a number that is almost one. I hope it's not too confusing.

In any case, for float (with e = 8 and p = 24) we get the exact value 340282346638528859811704183484516925440 or roughly 3.4 × 10³⁸. double then yields (with e = 10 and p = 53) 179769313486231570814527423731704356798070567525844996598917476803157260780028538760589558632766878171540458953514382464234321326889464182768467546703537516986049910576551282076245490090389328944075868508455133942304583236903222948165808559332123348274797826204144723168738177180919299881250404026184124858368 or roughly 1.80 × 10³⁰⁸.

[/EDIT]

Another thing: You're bringing up the term "precision" in your question but you quote the ranges of the types. Precision is a quite different thing and refers to how many significant digits the type can retain. Again, the answer here lies in the mantissa which is 23 and 52 bits for float and double, respectively. Since the numbers are stored normalized we actually have an implicit bit added to that, which puts us at 24 and 53 bits. Now, the way how digits after the decimal (or binary here) point work is the following:

 1.   1     0     1     1
 ↑    ↑     ↑     ↑     ↑
2^0  2^-1  2^-2  2^-3  2^-4
 =    =     =     =     =
 1   0.5   0.25  0.125 0.0625

So the very last digit in the double mantissa represents a value of roughly 2.2 × 10⁻¹⁶ or 2⁻⁵², so if the exponent is 1, this is the smallest value we can add to the number – placing the double precision around 16 decimal digits. Likewise for float with roughly seven digits.

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