double to string conversion with fixed width

Question

I would like to print a double value, into a string of no more than 8 characters. The printed number should have as many digits as possible, e.g.

5.259675
48920568
8.514e-6
-9.4e-12

I tried C++ iostreams, and printf-style, and neither respects the provided size in the way I would like it to:

cout << setw(8) <<  1.0 / 17777.0 << endl;
printf( "%8g\n", 1.0 / 17777.0 );

gives:

5.62525e-005
5.62525e-005

I know I can specify a precision, but I would have to provide a very small precision here, in order to cover the worst case. Any ideas how to enforce an exact field width without sacrificing too much precision? I need this for printing matrices. Do I really have to come up with my own conversion function?

A similar question has been asked 5 years ago: Convert double to String with fixed width , without a satisfying answer. I sure hope there has been some progress in the meantime.

Answer 1

This seems not too difficult, actually, although you can't do it in a single function call. The number of character places used by the exponent is really quite easy to predict:

const char* format;
if (value > 0) {
    if (value < 10e-100) format = "%.1e";
    else if (value < 10e-10) format = "%.2e";
    else if (value < 1e-5) format = "%.3e";
}

and so on.

Only, the C standard, where the behavior of printf is defined, insists on at least two digits for the exponent, so it wastes some there. See c++ how to get "one digit exponent" with printf

Incorporating those fixes is going to make the code fairly complex, although still not as bad as doing the conversion yourself.