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

c++floating-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 Solution

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
``````