How do I convert a C string to a int at compile time?

问题

I want to be able to pass an integer or a double (or a string) as a template argument and in some instances convert the result to an integer and use it as a template argument for a type in the class.

Here's what I've tried:

template <typename MPLString>
class A
{
    // the following works fine
    int fun()
    {
      // this function should return the int in the boost mpl type passed to it
      // (e.g. it might be of the form "123")
      return std::stoi(boost::mpl::c_str<MPLString>::value);
    }

    // the following breaks because std::stoi is not constexpr
    std::array<int, std::stoi(boost::mpl::c_str<MPLString>::value)> arr;
};

Can I do this somehow? I've tried std::stoi and atoi but neither are constexpr... Any ideas how this could be done (I cannot change the template parameter to take an int directly, as it might be a double).

回答1:

Defining a constexpr stoi isn't too hard with regular C strings. It can be defined as follows:

constexpr bool is_digit(char c) {
    return c <= '9' && c >= '0';
}

constexpr int stoi_impl(const char* str, int value = 0) {
    return *str ?
            is_digit(*str) ?
                stoi_impl(str + 1, (*str - '0') + value * 10)
                : throw "compile-time-error: not a digit"
            : value;
}

constexpr int stoi(const char* str) {
    return stoi_impl(str);
}

int main() {
    static_assert(stoi("10") == 10, "...");
}

The throw expression is invalid when used in constant expressions so it'll trigger a compile time error rather than actually throwing.

来源：https://stackoverflow.com/questions/25195176/how-do-i-convert-a-c-string-to-a-int-at-compile-time

标签

c++

c++11

boost-mpl