How can I display a higher precision value in c++ [duplicate]

Solution 1:

You can set the precision directly on std::cout and use the std::fixed format specifier.

double d = 3.14159265358979;
cout.precision(17);
cout << "Pi: " << fixed << d << endl;

You can #include <limits> to get the maximum precision of a float or double.

#include <limits>

typedef std::numeric_limits< double > dbl;

double d = 3.14159265358979;
cout.precision(dbl::max_digits10);
cout << "Pi: " << d << endl;

Solution 2:

Use std::setprecision:

#include <iomanip>
std::cout << std::setprecision (15) << 3.14159265358979 << std::endl;

Solution 3:

Here is what I would use:

std::cout << std::setprecision (std::numeric_limits<double>::digits10 + 1)
          << 3.14159265358979
          << std::endl;

Basically the limits package has traits for all the build in types.
One of the traits for floating point numbers (float/double/long double) is the digits10 attribute. This defines the accuracy (I forget the exact terminology) of a floating point number in base 10.

See: http://www.cplusplus.com/reference/std/limits/numeric_limits.html
For details about other attributes.

Solution 4:

The iostreams way is kind of clunky. I prefer using boost::lexical_cast because it calculates the right precision for me. And it's fast, too.

#include <string>
#include <boost/lexical_cast.hpp>

using boost::lexical_cast;
using std::string;

double d = 3.14159265358979;
cout << "Pi: " << lexical_cast<string>(d) << endl;

Output:

Pi: 3.14159265358979

Solution 5:

In C++20 you'll be able to use std::format to do this:

std::cout << std::format("{}", M_PI);

Output (assuming IEEE754 double):

3.141592653589793

The default floating-point format is the shortest decimal representation with a round-trip guarantee. The advantage of this method compared to the setprecision I/O manipulator is that it doesn't print unnecessary digits.

In the meantime you can use the {fmt} library, std::format is based on. {fmt} also provides the print function that makes this even easier and more efficient (godbolt):