c++ - 如何使函数能够接受原始指针作为迭代器？

Question

我有两个函数可以拆分字符串并将标记添加到 vector 中:

template < typename InputIterator, typename ContainerType >
void Slice(InputIterator begin,
           InputIterator end,
           typename InputIterator::value_type delimiter,
           ContainerType& container)
{
    using CharType = InputIterator::value_type;
    InputIterator right = begin;
    while (true) {
      begin = find_if_not(right, end, [ delimiter ](CharType c) { return c == delimiter; });
      if (begin == end) {
        break;
      } else {
        right = find_if(begin + 1, end, [ delimiter ](CharType c) { return c == delimiter; });
        container.push_back(std::basic_string< CharType >(begin, right));
      }
    }
}

template < typename InputIterator, typename ContainerType >
void Slice(InputIterator begin,
           InputIterator end,
           InputIterator delimitersBegin,
           InputIterator delimitersEnd,
           ContainerType& container)
{...}

它在像这样的调用中效果很好

std::string headers;
std::vector<std::string> rawHeaders;
Slice(headers.begin(), headers.end(), '\0', rawHeaders)

并且不适用于const char*:

auto begin = headers.c_str();
auto end = begin + headers.size();
Slice(begin, end, '\0', rawHeaders);

错误是(我使用 MSVC2017 和 MSVC2013 工具链):

error C2780: 'void Slice(InputIterator,InputIterator,InputIterator,InputIterator,ContainerType &)' : expects 5 arguments - 4 provided

error C2893: Failed to specialize function template 'void Slice(InputIterator,InputIterator,InputIterator::value_type,ContainerType &)With the following template arguments:'InputIterator=const char *''ContainerType=std::vectorstd::string,std::allocator<_Ty>'

更新:添加了函数体。目前最好不要使用 std::span 等附加功能。我希望有以下字符串构造函数:

template< class InputIt >
basic_string( InputIt first, InputIt last, const Allocator& alloc = Allocator() );

更新 2:仍然不起作用，但出于不同的原因，我尝试更改分隔符类型:

 template < typename InputIterator, typename ContainerType >
  void Slice(
      InputIterator begin,
      InputIterator end,
      typename std::iterator_traits< InputIterator >::value_type delimiter,
      //typename InputIterator::value_type delimiter,
      ContainerType& container)
  {
    using CharType = typename std::iterator_traits< InputIterator >::value_type;

现在它说:

error C3861: 'find_if_not': identifier not found. see reference to function template instantiation 'void StringUtils::Slice<char*,std::vectorstd::string,std::allocator<_Ty>>(InputIterator,InputIterator,char,ContainerType &)' being compiled with [_Ty=std::string, InputIterator=char *, ContainerType=std::vectorstd::string,std::allocator<std::string>

还有:

error C3861: 'find_if': identifier not found

Answer 1

SFINAE 开始。签名

template < typename InputIterator, typename ContainerType >
void Slice(InputIterator begin,
           InputIterator end,
           typename InputIterator::value_type delimiter,
           ContainerType& container)

不是可能的候选者，因为 const char* 中没有嵌套成员 ::value_type。

您可能需要这个签名:

template < typename InputIterator, typename ContainerType >
void Slice(InputIterator begin,
           InputIterator end,
           typename std::iterator_traits<InputIterator>::value_type delimiter,
           ContainerType& container)