Question

请考虑会员功能模板。我的问题嵌入在评论表中。

template<typename T>
GetValueResult GetValue(
                          const std::string &key,
                          T &val,
                          std::ios_base &(*manipulator)(std::ios_base &) = std::dec
                       )
{
   // This member function template is intended to work for all built-in
   // numeric types and std::string. However, when T = std::string, I get only
   // the first word of the map element's value. How can I fix this?

   // m_configMap is map<string, string>
   ConfigMapIter iter = m_configMap.find(key);

   if (iter == m_configMap.end())
      return CONFIG_MAP_KEY_NOT_FOUND;

   std::stringstream ss;
   ss << iter->second;

   // Convert std::string to type T. T could be std::string.
   // No real converting is going on this case, but as stated above
   // I get only the first word. How can I fix this?
   if (ss >> manipulator >> val)
      return CONFIG_MAP_SUCCESS;
   else
      return CONFIG_MAP_VALUE_INVALID;
}

Answer 1

流上的<<和>>运算符用于处理由空格分隔的标记。因此，如果字符串看起来像“1 2”，那么您的stringstream只会在第一个1上读取<<。

如果你想要多个值，我建议你在流上使用循环。这样的事可能会......

//stringstream has a const string& constructor
std::stringstream ss(iter->second); 

while (ss >> manipulator >> value) { /* do checks here /* }

有了这个，我建议你看看Boost，尤其是lexical_cast，它可能会开箱即用。

用于将std :: string转换为任意数字类型或std :: string的成员函数模板

1 个答案: