6

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.

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pentadecagon
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2 Answers2

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.

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Ben Voigt
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  • *and so on* is the part I'm afraid of. I would need to do the same thing for large numbers. Count the `-` for negative numbers. Check if `%e` or `%f` is to be used. But maybe you are right and this is the least cumbersome solution. Thanks. – pentadecagon May 31 '14 at 14:49
0

If you want to convert to fixed decimal numbers (e.g. drop the +/-"E" part), then it makes it a lot easier to accomplish:

#include <stdio.h>
#include <cstring>     // strcpy
#include <iostream>     // std::cout, std::fixed
#include <iomanip>      // std::setprecision
#include <new>

char *ToDecimal(double val, int maxChars) 
{
    std::ostringstream buffer;
    buffer << std::fixed << std::setprecision(maxChars-2) << val;
    std::string result = buffer.str();

    size_t i = result.find_last_not_of('\0');
    if (i > maxChars) i = maxChars;
    if (result[i] != '.') ++i;
    result.erase(i);

    char *doubleStr = new char[result.length() + 1];
    strcpy(doubleStr, (const char*)result.c_str());

    return doubleStr;
}

int main()
{
    std::cout << ToDecimal(1.26743237e+015, 8) << std::endl;
    std::cout << ToDecimal(-1.0, 8) << std::endl;
    std::cout << ToDecimal(3.40282347e+38, 8) << std::endl;
    std::cout << ToDecimal(1.17549435e-38, 8) << std::endl;
    std::cout << ToDecimal(-1E4, 8) << std::endl;
    std::cout << ToDecimal(12.78e-2, 8) << std::endl;
}

Output:

12674323
-1
34028234
0.000000
-10000
0.127800
  • Thanks, but I don't think that would be of any use, just look at the very first output line. The result is completely wrong. When I said I want an output consisting of 8 characters, I meant I want output that resembles the input to some extend. – pentadecagon May 31 '14 at 20:37
  • @pentadecagon Yes, you cannot have any larger number than 99999999 (or 9.999999) or less than -9999999 (or -0.00001) in the case of a 8 character maximum example. I'm sorry this won't work for you. –  Jun 01 '14 at 08:08