C++ – the maximum length in chars needed to represent any double value

cfloating point

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)?

Best Answer

The standard header <float.h> in C, or <cfloat> in C++, contains several constants to do with the range and other metrics of the floating point types. One of these is DBL_MAX_10_EXP, the largest power-of-10 exponent needed to represent all double values. Since 1eN needs N+1 digits to represent, and there might be a negative sign as well, then the answer is

int max_digits = DBL_MAX_10_EXP + 2;

This assumes that the exponent is larger than the number of digits needed to represent the largest possible mantissa value; otherwise, there will also be a decimal point followed by more digits.

CORRECTION

The longest number is actually the smallest representable negative number: it needs enough digits to cover both the exponent and the mantissa. This value is -pow(2, DBL_MIN_EXP - DBL_MANT_DIG), where DBL_MIN_EXP is negative. It's fairly easy to see (and prove by induction) that -pow(2,-N) needs 3+N characters for a non-scientific decimal representation ("-0.", followed by N digits). So the answer is

int max_digits = 3 + DBL_MANT_DIG - DBL_MIN_EXP

For a 64-bit IEEE double, we have

DBL_MANT_DIG = 53
DBL_MIN_EXP = -1023
max_digits = 3 + 53 - (-1023) = 1079
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