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我在工作时有一个复杂的代码库，并且我创建了一个模拟问题的小例子，这里是下面的代码。C++ - 智能指针 - 在模板中投射智能指针
<以下代码供参考> - 如果我们有boost库和FastDelegate.h与项目链接，此代码是可编译的。如果您需要完整的可编译示例项目，请告诉我，我可以给您发邮件。
我有两个问题，需要帮助解决它们。
- 如下面的代码所示，我有一个类作为另一个类对象的参数类型作为模板。现在，当我在UserClass的构造函数（Line 107）中初始化下面的类时，出现错误，因为mBaseAcceptor是一个类型为base Class的模板参数的类，但是我需要执行mbaseAcceptor（new derivedAcceptor_t）。铸造问题如何解决这个问题？这里
错误是
./boost/smart_ptr/shared_ptr.hpp:387:9: error: comparison between distinct pointer types ‘Acceptor<DerivedClass>*’ and ‘Acceptor<BaseClass>*’ lacks a cast
的另一个问题是线108，即使我奇迹般地通过派生类的另一种受体说，解决这个问题，这是我使用mDerivedAcceptor，在108行我做
mDerivedAcceptor->SetDelegate(fastdelegate::MakeDelegate(this, &UserClass::HandleDelegate));
然后我得到错误说
"error no matching function call for HandleDelegate(DerivedClass&, bool).
这是有意义的，因为HandleDelegate具有类型的BaseClass的，并通过存储委托（这是一个FUNC参数。 ptr），我们必须用适当的参数调用函数。但如何解决这个问题。
- 如果我将派生类中的Handler内部转换为Acceptor类，它将工作，当我只传递baseClass指针？
代码
/*
* smart_pointer_1.cpp
*
* Created on: Jul 26, 2011
* Author: balaji
*/
#include <algorithm>
#include <boost/foreach.hpp>
#include <boost/scoped_ptr.hpp>
#include <boost/shared_ptr.hpp>
#include "FastDelegate.h"
#include <iostream>
using namespace std;
template <class Handler>
class Acceptor {
public:
typedef fastdelegate::FastDelegate1<Handler &, bool> delegate_t;
Acceptor();
void Initialize(Handler *&handle);
void SetDelegate(delegate_t delegate) { mDelegate = delegate; }
private:
int mValues[2];
delegate_t mDelegate;
};
template <class Handler>
Acceptor<Handler>::Acceptor()
{
std::cout << "In Constructor: " << __FUNCTION__ << std::endl;
mValues[0] = 1;
mValues[1] = 2;
}
template <class Handler>
void Acceptor<Handler>::Initialize(Handler *&handle){
if (!handle) {
std::cout << __FUNCTION__ << " : created" << std::endl;
handle = new Handler();
} else {
std::cout << __FUNCTION__ << " : Error exception" << std::endl;
}
if (mDelegate && mDelegate(*handle)) {
std::cout << "Ok Called Handle in " << __FUNCTION__ << std::endl;
} else {
std::cout << "Not Called Handle in " << __FUNCTION__ << std::endl;
}
handle->displayComputer();
}
class BaseClass {
std::string mComputer;
public:
BaseClass() {
std::cout << "In Base Constructor: " << __FUNCTION__ << std::endl;
mComputer = "Mac";
}
virtual void displayComputer() {
std::cout << "Computer type is " << mComputer << std::endl;
}
};
class DerivedClass : public BaseClass {
std::string mLanguage;
public:
DerivedClass() {
std::cout << "In Derived Constructor: " << __FUNCTION__ << std::endl;
mLanguage = "C++";
}
void displayComputer() {
std::cout << "Language is " << mLanguage << std::endl;
}
};
class UserClass {
public:
UserClass();
UserClass(bool);
typedef Acceptor<BaseClass> baseAcceptor_t;
typedef Acceptor<DerivedClass> derivedAcceptor_t;
typedef boost::shared_ptr<BaseClass> basePtr_t;
void CallDelegate(BaseClass&);
private:
boost::shared_ptr<baseAcceptor_t> mBaseAcceptor;
boost::shared_ptr<derivedAcceptor_t> mDerivedAcceptor;
BaseClass *mConnBasePtr;
bool HandleDelegate(BaseClass& baseDelegate);
};
UserClass::UserClass() : mBaseAcceptor(new baseAcceptor_t)
{
std::cout << "In Constructor: " << __FUNCTION__ << std::endl;
mBaseAcceptor->SetDelegate(fastdelegate::MakeDelegate(this, &UserClass::HandleDelegate));
mBaseAcceptor->Initialize(mConnBasePtr);
}
UserClass::UserClass(bool value)
{
std::cout << "In Constructor: " << __FUNCTION__ << std::endl;
mBaseAcceptor.reset(new derivedAcceptor_t); // <<========== Problem Here because of improper casting
mBaseAcceptor->SetDelegate(fastdelegate::MakeDelegate(this, &UserClass::HandleDelegate)); // <<=== Also here because of improper type passed to MakeDelegate function ptr. Please note HandleDelegate has an argument of type BaseClass, but Acceptor is derived class
mBaseAcceptor->Initialize(mConnBasePtr);
}
bool UserClass::HandleDelegate(BaseClass& baseDelegate)
{
std::cout << "In " << __FUNCTION__ << std::endl;
return true;
}
int main() {
std::cout << "In function: " << __FUNCTION__ << std::endl;
typedef boost::shared_ptr<UserClass> userPtr_t;
userPtr_t user(new UserClass(true));
std::cout << "In function: " << __FUNCTION__ << " at end "<< std::endl;
return 0;
}
请在此处输入*相关的*代码。 SO具有出色的代码格式功能，因此让其他人可以轻松帮助您并在此处发布代码*。 – jalf
@jalf - 我的坏，编辑我的文章与代码。我拥有完整代码的原因是因为它容易复制粘贴来分析问题。 – devgp