从构造函数中的成员变量中减去模板参数 [英] Deduce template parameter from member variable in constructor

问题描述

我有一个C ++库，它是C库的包装器。工厂类 Creator 创建对象，每个对象表示C库功能的一部分。
这些类构造函数是私有的，以确保所有对象都通过 Creator 类创建，因为C库需要一些内部魔术（在下面我省略，简化示例）。

I have a C++ library which is a wrapper for a C library. A factory class Creator creates objects of which each represents parts of the functionality of the C library. These class constructors are private to ensure that all objects are created through that Creator class because the C library requires some internal magic (which I omitted in the following, simplified example).

一切正常，但我有以下问题：

It all works, but I have the following "problem":

c $ c> UsesAandB 我必须指定 A<> 两次的模板参数：
一旦在成员声明code> std :: shared_ptr< A< int>> ）并且一次在构造函数的初始化列表中（ creator-> createA< int> code>）。

In class UsesAandB I have to specify the template parameter of A<> twice: Once in the member declaration (std::shared_ptr<A<int>>) and once in the initialization list of the constructor (creator->createA<int>).

由于我已经知道成员 aInt 将是 std :: shared_ptr< A< int>> ，我如何使用这些知识来调用相应的 createA< int>方法在构造函数中不重复 int 或者如何避免调用 createA （）？

Since I already know that member aInt will be of type std::shared_ptr<A<int>>, how can I use this knowledge to call the corresponding createA<int>() method in the constructor without repeating int or how can I avoid the call to createA<int>() at all?

#include <memory>

class Creator;

template<typename T>
class A
{
    friend class Creator;
private:
    A(int param) {}
    T value;
};

class B
{
    friend class Creator;
private:
    B(){}
};

class Creator
{
public:
    template<typename T>
    std::shared_ptr<A<T>> createA(int param) { return std::shared_ptr<A<T>>(new A<T>(param)); }
    std::shared_ptr<B> createB() { return std::shared_ptr<B>(new B());}
};

class UsesAandB
{
public:
    UsesAandB(std::shared_ptr<Creator> creator)
        : creator(creator),
          aInt(creator->createA<int>(0)),
          aDouble(creator->createA<double>(1)),
          b(creator->createB())
   {

   }
private:
    std::shared_ptr<Creator> creator;
    std::shared_ptr<A<int>> aInt;
    std::shared_ptr<A<double>> aDouble;
    std::shared_ptr<B> b;
};

int main()
{
    auto creator = std::make_shared<Creator>();
    UsesAandB u(creator);
    return 0;
}

推荐答案

shared_ptr ，您需要将其传递给模板函数，如下所示：

To get the type from the shared_ptr, you need to pass it to a template function like this one:

template <typename U>
shared_ptr< A<U> > CreateA( std::shared_ptr<Creator>& c,
                            const shared_ptr< A<U> >& p,
                            const U& val )
{
    return c->createA<U>(val);
}

然后只需：

aInt(CreateA(creator, aInt, 0 )),

此处的示例 - http://ideone.com/Np7f8t

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