double-precison

[Guzfraba]

I wont to store in a variable a real number,for example 0.132547698,but the double can store just 6 digits.Take a look:

#include <iostream>
using namespace std;
int main() {
	double a,b,c;
	a=1.0/3.0;
	b=1.23456789;
	c=1234.56789;
	cout<<a<<endl<<b<<endl<<c;

	cin.ignore();
	return 0;
}

the output:

0.333333
1.23457
1234.57

What should I do ? I've seen on other topics that the double data type is much larger...
Thanks :)

[Thumper]

The double is capable of storing up to 15 decimal places, and does so in your code. The problem is 'cout' truncates it to a certain precision.

#include <iostream>
#include <iomanip>
using namespace std;

int main()
{
    double a = (1.0/3.0);
    cout << setprecision(15) << a << endl;
}

you can read about setprecision here: http://cplusplus.com/reference/iomanip/setprecision/

That will allow you to change the decimal precision of 'cout'.

[Chervil]

Just because you don't see the extra digits displayed, it doesn't mean they don't exist.

	double a,b,c;
	a = 1.0/3.0;
	b = 0.333333;
	c = a - b;

	cout << "a: " << a << endl;
	cout << "b: " << b << endl;
	cout << "c: " << c << endl;

Output:

a: 0.333333
b: 0.333333
c: 3.33333e-007

Notice that c is not zero.