Question

I figured out that double type on my machine corresponds to this Wikipedia article, and long double corresponds to this text: x86 Extended Precision Format.

That’s why floor(52/log2(10)) or 15 digits of double should be correct and floor(63/log2(10)) or 19 digits of long double are trusted.

The code:

int main()
{
    double d=0.1;
    long double ld=0.1;

    std::cout.precision(19);
    std::cout.setf(std::ios_base::scientific);

    std::cout << d << std::endl;
    std::cout << ld << std::endl;

    return 0;
}

gives the output:

1.0000000000000000555e-01
1.0000000000000000555e-01

If we set cout.precision to 16, the output will be:

1.0000000000000001e-01
1.0000000000000001e-01

It’s allright, that 17th digit of double in 1st output and 16th digit of double in second output is incorrect. But why is it incorrect for long double? Is there a way to get all 19 correct digits of long double variables?

If I try printf("%.19Le\n",ld); for long double, I get exactly the same result.

I use OpenSUSE 12.1 and g++ 4.6.2.

Answer 1

The 0.1 literal constant is not a long double, it is probably a double .

You may want to initialize ld with

 long double ld = (long double) 1.0 / (long double) 10.0;

So that the division involves two long double operands, so hopefully is done in long double (I am not sure of that, you have to double check the appropriate C++ standards).

And you probably want to compile with GCC supporting the latest standard. With a GCC 4.7 I would suggest to compile with g++ -Wall -std=c++11 but with your 4.6 you may need to say -std=c++0x