Knowing the exact number of character needed to write float, double and integers

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1 comment, last by Aressera 11 months ago

Hi everybody,
I have a custom type-safe string format function and actually using fixed buffer of 2048 to sprintf.
How to know the exact number of character needed to write float, double and integers?
Thank you!

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For integers it's pretty easy:

// Determine log(value)/log(base)
Size logBase;
if ( nonZero )
{
	logBase = 0;
	for ( ValueType power = value; power != 0; power /= tBase )
		logBase++;
}
else
{
	logBase = 1;
}

const Size prefixLength = getNumberPrefixLengthForBase( base ); // e.g. 0x for hex
const Size stringLength = logBase + hasSign + prefixLength;

Floating point is more difficult, and will depend on how exactly you round/format the string. For example, you might have a value of 0.999 that you want to print with 2 decimal digits. This would round to 1. Do you want to display 1.00, or 1? Determining how many digits in a robust way for all values is not trivial. It takes 50+ lines of code for my implementation to do this, handing different rounding modes and exponent formats and bases. The crux of it is:

const Size prefixLength = getNumberPrefixLengthForBase( base );
Size stringLength = hasSign + prefixLength + numDigits + // sign, base prefix, significant digits.
                    (numDecimalDigits != 0) + numDecimalDigits; // decimal point and decimal digits.

if ( numExponentDigits > 0 )
    stringLength += (exponent < ValueType(0)) + numExponentDigits + 1;

This is the maximum length of the string which is used to allocate a buffer. It might be shortened once the digits are actually generated (e.g. the 0.999 → 1 case mentioned above).

Note that I don't include the null terminator in any of the above calculations.

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