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What is the maximum length in chars needed to represent any double value?

Asked 2009-11-09T13:43:07.917
71

When I convert an unsigned 8-bit int to string then I know the result will always be at most 3 chars (for 255) and for an signed 8-bit int we need 4 chars for e.g. "-128".

Now what I'm actually wondering is the same thing for floating-point values. What is the maximum number of chars required to represent any "double" or "float" value as a string?

Assume a regular C/C++ double (IEEE 754) and normal decimal expansion (i.e. no %e printf-formatting).

I'm not even sure if the really small number (i.e. 0.234234) will be longer than the really huge numbers (doubles representing integers)?

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You can control the number of digits in the string representation when you convert the float/double to a string by setting the precision. The maximum number of digits would then be equal to the string representation of std::numeric_limits<double>::max() at the precision you specify.

#include <iostream>
#include <limits>
#include <sstream>
#include <iomanip>

int main()
{
 double x = std::numeric_limits<double>::max();

 std::stringstream ss;
 ss << std::setprecision(10) << std::fixed << x;

 std::string double_as_string = ss.str();
 std::cout << double_as_string.length() << std::endl;
}

So, the largest number of digits in a double with a precision of 10 is 320 digits.

answered 2009-11-09T14:02:13.073

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