How can variadic char template arguments from user defined literals be converted back into numeric types?
- by Pubby
This question is being asked because of this one.
C++11 allows you to define literals like this for numeric literals:
template<char...> OutputType operator "" _suffix();
Which means that 503_suffix would become <'5','0','3'>
This is nice, although it isn't very useful in the form it's in.
How can I transform this back into a numeric type? This would turn <'5','0','3'> into a constexpr 503. Additionally, it must also work on floating point literals. <'5','.','3> would turn into int 5 or float 5.3
A partial solution was found in the previous question, but it doesn't work on non-integers:
template <typename t>
constexpr t pow(t base, int exp) {
return (exp > 0) ? base * pow(base, exp-1) : 1;
};
template <char...> struct literal;
template <> struct literal<> {
static const unsigned int to_int = 0;
};
template <char c, char ...cv> struct literal<c, cv...> {
static const unsigned int to_int = (c - '0') * pow(10, sizeof...(cv)) + literal<cv...>::to_int;
};
// use: literal<...>::to_int
// literal<'1','.','5'>::to_int doesn't work
// literal<'1','.','5'>::to_float not implemented
