如何使用模板和继承来纠正和改进这个 c++ 代码？

Question

我正在学习 c++，大致知道我想做什么，但我做错了，我很累。做这样的事情的最好方法是什么（首先是正确的方法）：

// Query.hpp
class Query {
public:
 Query();

 template<typename T>
 std::vector<boost::shared_ptr<BaseResult> > run();

private:
 std::string sql;
};

// Firstly, something like this:
// I would like to do the equivalent of passing in a type T that would be
// derived from BaseResult and create a new instance of it when adding
// to vector of BaseResult:
template<typename T>
std::vector<boost::shared_ptr<BaseResult> > Query::run() {
 std::vector<boost::shared_ptr<BaseResult> > results;

 // ResultSet is from the mysql c++ connector
 boost::shared_ptr<sql::ResultSet> res(stmt->executeQuery(this->sql));

 // I want to add a new T to results here
 while (res->next()) {
  results.push_back(new T(res));
 }

 // RVO prevents copy in release - yes?
 return results;
}

// Query.cpp
Query::Query() {
}


// main.cpp
void foo(const std::vector<boost::shared_ptr<BaseResult> >& results) {
 // loop through calling virtual method on each item
}

int main(int argc, char* argv[]) 
 // Determine query type
 ProgramOptions opts(argc, argv);

 // Should this indeed be a pointer because
 // of code below or is it wrong?
 std::vector<boost::shared_ptr<BaseResult> >* results;

    // Secondly, something like this:
 if (opts.getquerytype() == "type1") {

  // I'd like to get a
  // std::vector<boost::shared_ptr<BaseResult> > returned here
  // containing many instances of a
  // type derived from BaseResult

  // I'd like to be able to do something like this
  // Is this assignment correct?
        *results = getQuery().run<DerivedResult>();
 }
 else {
  // I'd like to get a
  // std::vector<boost::shared_ptr<BaseResult> > returned here
  // containing many instances of a
  // different type derived from BaseResult

  // I'd like to be able to do something like this
  // Is this assignment correct?
        *results = getQuery().run<DifferentDerivedResult>();
 }

    foo(results);
}

Answer 1

首先，你的作业不正确

*results = getQuery().run<DifferentDerivedResult>();

取消引用一个未初始化的指针！

我会对您的代码进行一些更改，首先，必须在任何地方使用 vector<...> 会导致 RSI，typedef 它

class Query {
public:

 // Query::ResultType
 typedef std::vector<boost::shared_ptr<BaseResult> > ResultType; 

 Query();

 // next, pass in the vector where the results will be stored...
 template<typename T>
   void run(ResultType& records);

private:
 std::string sql;
};

现在实现：

template<typename T>
void Query::run(ResultType& records) {

 // ResultSet is from the mysql c++ connector
 boost::shared_ptr<sql::ResultSet> res(stmt->executeQuery(this->sql));

 // I want to add a new T to results here
 while (res->next()) {
  records.push_back(new T(res));
 }

 // No need to worry about RVO
}

现在在你的主要：

Query::ResultType results; // not a pointer!

然后你可以调用每种类型，例如

getQuery().run<DifferentDerivedResult>(results);

这将对您的设计进行最小的更改，如果您想完全避免 BaseResult 类型，您可以模板化所有内容 - 但我不确定 BaseResult 为您提供什么功能以及这是否可能..