lexical_cast strtof strtold lose accuracy?

Question

Here:

#include <iostream>
#include <cstdlib>
#include <boost/lexical_cast.hpp>

int main(void) {
    const char * str = "277499.84";

    std::cout << boost::lexical_cast<double>(str) << std::endl;
    std::cout << strtof(str, NULL) << std::endl;
    std::cout << strtold(str, NULL) << std::endl;
    std::cout << atof(str) << std::endl;

    return 0;
}

output:

Why the output are not 277499.84?

The default precision is 6 I suppose, just like for `printf`'s `%f`. Change it if you don't like it. — Kerrek SB, Apr 23 '16 at 11:29

score 3 · Accepted Answer · answered Apr 23 '16 at 11:30

It's not the operations themselves losing accuracy, but the output.

You can use the I/O manipulator std::setprecision to control the numeric precision. The following will use the full precision of a double (assuming the stream is set for decimal output).

double value = boost::lexical_cast<double>(str);
std::cout << std::setprecision( std::numeric_limits<double>::digits10 + 1 ) << value;

Or you can use std::ios_base::precision. This is useful if you want to restore the precision to the original value after.

auto old_precision = cout.precision( std::numeric_limits<double>::digits10 + 1 );
cout << value;
cout.precision( old_precision );

lexical_cast strtof strtold lose accuracy?

1 Answers1