How to convert a C string (containing only "0" or "1") to binary data directly according to its literal value?

For example, I have a string str, it's value is 0010010. Now I want to convert this string to a binary form variable, or a decimal variable equal to 0b0010010 (is 18).

int main() {
    string str_1 = "001";
    string str_2 = "0010";
    string str = str_1   str_2;
    cout << str << endl;

    int i = stoi(str);
    double d = stod(str);
    cout << i << " and " << d << endl;
}

I tried stoi and stod, but they all can't work. They all treated binary 0b0010010 as decimal 10010. So how can I achieve this requirement? Thanks a lot!

CodePudding user response：

std::stoi has an optional second argument to give you information about where the parsing stopped, and an optional third argument for passing the base of the conversion.

int i = stoi(str, nullptr, 2);

This should work.

Corollary: If in doubt, check the documentation. ;-)

CodePudding user response：

Or there is this fun way of doing it (at least I think it is) at compile time

#include <stdexcept>
#include <string_view>

constexpr auto from_bin(std::string_view&& str)
{
    std::size_t value{ 0ul };
    for (const char c : str)
    {
        value <<= 1;
        if ((c != '0') && (c != '1')) throw std::invalid_argument("invalid input");
        value |= (c == '1');
    }
    return value;
}


int main()
{
    static_assert(5 == from_bin("101"));
    static_assert(9 == from_bin("1001"));
    static_assert(2 == from_bin("0010"));
    return 0;
}