我正在尝试实现 Signal C++ 中的模板。
这是我目前所拥有的:
Main.cpp :
//Developed by Trofimov Yaroslav on 02.04.18
#include <iostream>
#include "Signal.h"
void f1() {
std::cout << "here in f1" << std::endl;
}
void f2() {
std::cout << "F2 F2 F2" << std::endl;
}
typedef void (* VoidResultDelegate)();
int main(void) {
Signal<VoidResultDelegate> signalVoid;
signalVoid.addListener(f1);
signalVoid.addListener(f1);
signalVoid.invoke();
signalVoid.removeListener(f2);
signalVoid.invoke();
return 0;
}
Signal.h :
//Developed by Trofimov Yaroslav on 02.04.18
#ifndef _SIGNAL_H_TROFIMOV_
#define _SIGNAL_H_TROFIMOV_
#include "LinkedList.h"
template<typename FunctionType>
class Signal {
LinkedList<FunctionType> _delegates;
public:
Signal<FunctionType>(void)
: _delegates(LinkedList<FunctionType>()) {
}
~Signal<FunctionType>(void) {
}
void addListener(const FunctionType& delegated) {
_delegates.add(delegated);
}
void removeListener(const FunctionType& delegated) {
_delegates.remove(delegated);
}
void invoke() {
_delegates.startIteration();
while(_delegates.hasNext()) {
_delegates.next()();
}
}
};
#endif
LinkedList.h :
//Developed by Trofimov Yaroslav on 30.03.2018
#ifndef _LINKED_LIST_H_TROFIMOV_
#define _LINKED_LIST_H_TROFIMOV_
#include <string>
#include <iostream>
#include <typeinfo>
template<typename T>
class LinkedList {
template<typename T>
struct Node {
T _data;
Node* _next;
Node()
: _next(0){}
~Node<T>() {
if(_next) {
delete _next; _next = 0;
}
}
};
Node<T>* _head;
Node<T>* _tail;
Node<T>* _iterator;
public:
LinkedList<T>()
: _head(0), _tail(0), _iterator(0) {
};
~LinkedList<T>() {
delete _head; _head = 0;
}
void add(const T& element) {
if(!_head) {
_head = new Node<T>;
_head->_data = element;
_tail = _head;
return;
}
Node<T>* newNode = new Node<T>;
newNode->_data = element;
_tail->_next = newNode;
_tail = newNode;
return;
}
void remove(const T& element) {
if(!_head) {
return;
}
if(_head->_data == element) {
_head = _head->_next;
return;
}
Node<T>* previous = _head;
Node<T>* current = _head->_next;
while(current) {
if(current->_data == element) {
previous->_next = current->_next;
return;
}
previous = current;
current = current->_next;
}
}
void startIteration() {
_iterator = _head;
}
bool hasNext() {
return (_iterator)?true:false;
}
T& next() {
T& res = _iterator->_data;
_iterator = _iterator->_next;
return res;
}
};
#endif
所以，我想添加的是传递参数的通用方法。假设，现在而不是 typedef void (* VoidResultDelegate)();我有typedef void (* VoidResultDelegate)(int i);这意味着我想要 int参数以某种方式出现在 Signal::invoke 中方法参数列表并传递到这里_delegates.next()();这样_delegates.next()(i);或类似的东西。
在 C++ 中有可能吗？
我在想的是通过另一个typename Signal 的参数这将表示 Signal::invoke 接受的参数类型并在调用_delegates.next()();中传递给链表中的元素.但是这种方法的问题是我无法控制参数的数量(它只是一个参数)。没有人(当然我指的是编译器)没有人强制我将正确的参数作为 typename 传递。到 Signal模板。在上面的示例中，我可以通过 bool typename而不是 int在错误发生之前，没有人会注意到它。
以下是回答后的更新版本:
Main.cpp :
//Developed by Trofimov Yaroslav on 02.04.18
#include <iostream>
#include "Signal.h"
void f1(int i) {
std::cout << "here in f1" << std::endl;
}
void f2(int i) {
std::cout << "F2 F2 F2" << std::endl;
}
typedef void (* VoidResultDelegate)(int i);
int main(void) {
Signal<VoidResultDelegate, int> signalVoid;
signalVoid.addListener(f1);
signalVoid.addListener(f2);
signalVoid.invoke(-1);
signalVoid.removeListener(f1);
signalVoid.invoke(-1);
return 0;
}
Signal.h :
//Developed by Trofimov Yaroslav on 02.04.18
#ifndef _SIGNAL_H_TROFIMOV_
#define _SIGNAL_H_TROFIMOV_
#include "LinkedList.h"
template<typename FunctionType, typename... Args>
class Signal {
LinkedList<FunctionType> _delegates;
public:
Signal<FunctionType, parameter>(void)
: _delegates(LinkedList<FunctionType>()) {
}
~Signal<FunctionType, parameter>(void) {
}
void addListener(const FunctionType& delegated) {
_delegates.add(delegated);
}
void removeListener(const FunctionType& delegated) {
_delegates.remove(delegated);
}
void invoke(Args&& ... args) {
_delegates.startIteration();
while(_delegates.hasNext()) {
(_delegates.next())(std::forward<Args>(args)...);
}
}
};
#endif
LinkedList.h :
//Developed by Trofimov Yaroslav on 30.03.2018
#ifndef _LINKED_LIST_H_TROFIMOV_
#define _LINKED_LIST_H_TROFIMOV_
#include <string>
#include <iostream>
#include <typeinfo>
template<typename T>
class LinkedList {
template<typename T>
struct Node {
T _data;
Node* _next;
Node()
: _next(0){}
~Node<T>() {
if(_next) {
delete _next; _next = 0;
}
}
};
Node<T>* _head, _tail, _iterator;
public:
LinkedList<T>()
: _head(0), _tail(0), _iterator(0) {
};
~LinkedList<T>() {
delete _head; _head = 0;
}
void add(const T& element) {
if(!_head) {
_head = new Node<T>;
_head->_data = element;
_tail = _head;
return;
}
Node<T>* newNode = new Node<T>;
newNode->_data = element;
_tail->_next = newNode;
_tail = newNode;
return;
}
void remove(const T& element) {
if(!_head) {
return;
}
if(_head->_data == element) {
_head = _head->_next;
return;
}
Node<T>* previous = _head;
Node<T>* current = _head->_next;
while(current) {
if(current->_data == element) {
previous->_next = current->_next;
return;
}
previous = current;
current = current->_next;
}
}
void startIteration() {
_iterator = _head;
}
bool hasNext() {
return (_iterator)?true:false;
}
T& next() {
T& res = _iterator->_data;
_iterator = _iterator->_next;
return res;
}
};
#endif
最佳答案
您可以对 Signal 类进行部分特化，以在单独的模板参数中获取返回类型和参数。
// Declare the template without any definition
template<typename FunctionType>
class Signal;
// Add partial specialization
template<typename ReturnType, typename... Args>
class Signal<ReturnType(*)(Args...)> {
// Now you have access to return type and arguments
// Other things...
void invoke(Args... args) {
_delegates.startIteration();
while(_delegates.hasNext()) {
_delegates.next()(args...);
}
}
}
我遗漏了 perfect forwarding从 invoke 函数来保持简单。有了它，它看起来像这样。
void invoke(Args&&... args) {
_delegates.startIteration();
while(_delegates.hasNext()) {
_delegates.next()(std::forward<Args>(args)...);
}
}
关于c++ - 在 C++ 中实现 Signal 模板，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/49620375/