Davinel - 1 year ago 110

C Question

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?

Answer Source

To get the correct results, don't set precision greater than available for this numeric type:

```
#include <iostream>
#include <limits>
int main()
{
double a = 0.3;
std::cout.precision(std::numeric_limits<double>::digits10);
std::cout << a << std::endl;
double b = 0;
for (char i = 1; i <= 50; i++) {
b = b + a;
};
std::cout.precision(std::numeric_limits<double>::digits10);
std::cout << b << std::endl;
}
```

Although if that loop runs for 5000 iterations instead of 50, the accumulated error will show up even with this approach -- it's just how floating-point numbers work.