Convert double to String with fixed width

Answer 1

Is that what you want? Here, we calculate the amount of available precision and set ostream accordingly.

    #include <iostream>
    #include <iomanip>

    using namespace std;

    int main(int argc, char* argv[])
    {
    // Input
    double value =  102.1213239999;
    // Calculate limits
    int digits = ( (value<1) ? 1 : int(1+log10(double(abs(value)))) );
    int width = 10;
    int precision = (((width-digits-1)>=0) ? (width-digits-1):0);

    // Display
    cout.setf(ios::fixed);
    cout.precision(precision);
    cout<<setw(10)<<value<<endl;


    return 0;

    }
    OUTPUT: 102.121324

Btw, if you want a truckload of ways to compute digits, here's how.

Answer 2

How about lexical cast?

double x = 102.1213239999;
std::cout << boost::lexical_cast<std::string>(x).substr(0,10);

Its not exactly what you asked for. I was just trying to think outside the box.
You may also want to look at this question for a discussion on formatting differences between C and C++ and check out the Boost Format Library

Answer 3

You didn't specify the language, so here goes the python solution:

d=123.12345678910111213 '%.10f'%d

this will give you: '123.1234567891'

Answer 4

You can use osteram::width and ostream::precision function to achieve your goal, like this

std::ostringstream out;
out.width(10);
out.precision(10);
out << 123.12345678910111213;

Although it won't add zeros after the point in order to respect width but it will add spaces (or any character of you choise) before the number. So You'll get ' 102' or '0000000102' (if you call out.fill('0');) instead of '102.000000' if you pass 102 as a input value.

Answer 5

Use boost::lexical_cast