C ++通过一个派生类shared_ptr的一个模板函数 [英] C++ passing a derived class shared_ptr to a templated function

问题描述

首先东西应该工作，那么一些不。为什么它是不是这个问题。

First something that should work, then something that doesn't. Why doesn't it is the question.

我声明了两个类：

class Base { ... };
class Derived : public Base { ... };

然后我有以下的其他功能：

I then have the following function elsewhere:

void foo(shared_ptr<Base> base);   

以下code应该工作吧？

The following code should work right?

share_ptr<Derived> derived;
foo(derived);

现在，忘了上面，我宣布三类：

Now, forget the above, I declare three classes:

class Foo { ... };
template <typename TYPE> class Base { ... };
class Derived : public Base<Foo> { ... };

在其他地方，我宣布一个模板函数：

Elsewhere, I declare a templated function:

template <typename TYPE> void foo(shared_ptr<Base<TYPE> > base); 

以下code不工作：

The following code does not work:

shared_ptr<Derived> derived;
foo(derived);

报告说，没有发现匹配函数foo（...），它接受 share_ptr＆LT;衍生GT;

首先，应该在原来的例子工作？第二，你认为什么可能是在我有一个的shared_ptr 来是从一个专门的基类派生的类。

First, should the original example work? And second, what do you think could be the issue in the second example where I have a shared_ptr to a class that is derived from a specialized base class.

推荐答案

我不认为编译器将通过间接的那样的水平。相反，你可以明确地实例化类型设置为富富。例如，通过G ++下面编译：

I don't think the compiler will go through a level of indirection in that way. Rather, you can explicitly instantiate foo with TYPE set to Foo. Eg, the following compiles via g++:

#include<boost/shared_ptr.hpp>

class Foo {};
template <typename T> class Base {};
class Derived : public Base<Foo> {};

template<typename T>
int foo(boost::shared_ptr<Base<T> > base) {return 0;}

boost::shared_ptr<Derived> derived;
int t = foo<Foo>(derived);

int main() {return 0;}

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