我在使用预先编写的C ++的obvserver设计模式代码时遇到了问题。在attach方法中,当我将它声明为类时,它说类型Observer无法解析。任何人都可以帮我解决这个问题吗?我在这个方法中也遇到了views.push_back的另一个问题。它抛出错误“无效的参数,候选者是无效的push_back(Observer * const&)”。我是C ++的新手,并不太了解这些错误。
#include <iostream>
#include <vector>
using namespace std;
class Subject {
// 1. "independent" functionality
vector < class Observer * > views; // 3. Coupled only to "interface"
int value;
public:
void attach(Observer *obs) {
views.push_back(obs);
}
void setVal(int val) {
value = val;
notify();
}
int getVal() {
return value;
}
void notify();
};
class Observer {
// 2. "dependent" functionality
Subject *model;
int denom;
public:
Observer(Subject *mod, int div) {
model = mod;
denom = div;
// 4. Observers register themselves with the Subject
model->attach(this);
}
virtual void update() = 0;
protected:
Subject *getSubject() {
return model;
}
int getDivisor() {
return denom;
}
};
void Subject::notify() {
// 5. Publisher broadcasts
for (int i = 0; i < views.size(); i++)
views[i]->update();
}
class DivObserver: public Observer {
public:
DivObserver(Subject *mod, int div): Observer(mod, div){}
void update() {
// 6. "Pull" information of interest
int v = getSubject()->getVal(), d = getDivisor();
cout << v << " div " << d << " is " << v / d << '\n';
}
};
class ModObserver: public Observer {
public:
ModObserver(Subject *mod, int div): Observer(mod, div){}
void update() {
int v = getSubject()->getVal(), d = getDivisor();
cout << v << " mod " << d << " is " << v % d << '\n';
}
};
问题是您声明类的顺序。类Subject
使用类Observer
,但后来声明了Observer
,所以当编译器尝试编译Subject
时,不知道Observer
是什么。然而,由于Subject
也使用Observer
,你不能只是改变声明的顺序。
这个难题的答案是转发声明类Observer
,所以至少编译器知道Observer
是一个类,即使它不太了解它。在class Observer;
定义之前添加class Subject
。您也可以从矢量声明中删除单词class
。
class Observer;
class Subject {
// 1. "independent" functionality
vector < Observer * > views; // 3. Coupled only to "interface"
...