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C++ floating point precision [duplicate]

Possible Duplicate:
Floating point inaccuracy examples

double a = 0.3;
std::cout.precision(20);
std::cout << a << std::endl;

result: 0.2999999999999999889

double a, b;
a = 0.3;
b = 0;
for (char i = 1; i <= 50; i++) {
  b = b + a;
};
std::cout.precision(20);
std::cout << b << std::endl;

result: 15.000000000000014211

So.. 'a' is smaller than it should be. But if we take 'a' 50 times - result will be bigger than it should be.

Why is this? And how to get correct result in this case?

like image 334
Davinel Avatar asked May 26 '10 00:05

Davinel


2 Answers

Computers store floating point numbers in binary, not decimal.

Many numbers that look ordinary in decimal, such as 0.3, have no exact representation of finite length in binary.
Therefore, the compiler picks the closest number that has an exact binary representation, just like you write 0.33333 for 1⁄3.

If you add many floating-point numbers, these tiny difference add up, and you get unexpected results.

like image 128
SLaks Avatar answered Oct 04 '22 16:10

SLaks


Why is this?

Because floating-point numbers are stored in binary, in which 0.3 is 0.01001100110011001... repeating just like 1/3 is 0.333333... is repeating in decimal. When you write 0.3, you actually get 0.299999999999999988897769753748434595763683319091796875 (the infinite binary representation rounded to 53 significant digits).

Keep in mind that for the applications for which floating-point is designed, it's not a problem that you can't represent 0.3 exactly. Floating-point was designed to be used with:

  • Physical measurements, which are often measured to only 4 sig figs and never to more than 15.
  • Transcendental functions like logarithms and the trig functions, which are only approximated anyway.

For which binary-decimal conversions are pretty much irrelevant compared to other sources of error.

Now, if you're writing financial software, for which $0.30 means exactly $0.30, it's different. There are decimal arithmetic classes designed for this situation.

And how to get correct result in this case?

Limiting the precision to 15 significant digits is usually enough to hide the "noise" digits. Unless you actually need an exact answer, this is usually the best approach.

like image 33
dan04 Avatar answered Oct 04 '22 14:10

dan04