c – Boost.Spirit使用基数将字符串解析为数字

我有一个模板函数,它将字符串转换为数字如下：

template <typename RetType,
    typename Parser =
    typename boost::spirit::traits::create_parser<RetType>::type>
    inline std::enable_if_t<std::is_arithmetic<RetType>::value, RetType>
    cast(const std::string &input)
{
    RetType result;

    if(input.empty())
    {
        // handle this
    }
    auto itBeg = input.cbegin();
    auto itEnd = input.cend();
    if(!bsq::parse(itBeg, itEnd, Parser(), result) || itBeg != itEnd)
    {
        // handle that
    }
    return result;
}

现在我想创建一个类似于上面的函数,它将解析表示某个基数中的数字的字符串

template <typename RetType, unsigned Radix,
    typename Parser =
    typename boost::spirit::traits::create_parser<RetType>::type>
    inline std::enable_if_t<std::is_arithmetic<RetType>::value, RetType>
    cast(const std::string &input)
{
    RetType result;

    if(input.empty())
    {
        // handle this
    }
    auto itBeg = input.cbegin();
    auto itEnd = input.cend();
    if(!bsq::parse(itBeg, itEnd, Parser<RetType, Radix, 1 - 1>() /*something like this*/, result) || itBeg != itEnd)
    {
        // handle that
    }

    return result;
}

当然,解析器是无效的,但是使用基数定义“自动”算术解析器的正确方法是什么？

解决方法:

我用qi :: int_parser<>直：

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#include <boost/spirit/include/qi.hpp>
#include <type_traits>

template <typename RetType, unsigned Radix = 10, typename Parser = typename boost::spirit::qi::int_parser<RetType, Radix> >
inline typename std::enable_if<std::is_arithmetic<RetType>::value, RetType>::type
    cast(const std::string &input)
{
    RetType result;

    if(input.empty())
    {
        // handle this
    }
    auto itBeg = input.cbegin();
    auto itEnd = input.cend();
    if(!boost::spirit::qi::parse(itBeg, itEnd, Parser(), result) || itBeg != itEnd)
    {
        // handle that
        throw "oops";
    }

    return result;
}

int main() {
    std::cout << cast<int>    ("10") << "\n";
    std::cout << cast<int,  2>("10") << "\n";
    std::cout << cast<int,  8>("10") << "\n";
    std::cout << cast<int, 16>("10") << "\n";
    std::cout << cast<int, 16>("ee") << "\n";
}

打印

10
2
8
16
238

Hint: to be very accurate you might want to detect signed/unsigned types and use uint_parser<> accordingly

标签：boost-spirit-qi,c,boost-spirit
来源： https://codeday.me/bug/20190829/1762610.html