Question

对于以下代码：

#include <iostream>
#include <string>

using namespace std;

class Foo2;
class Foo3;

template <class T>
class Foo1 {
  public:
    Foo1();
    void print() {
      cout << "My name is: " << name << endl;
    }

    T getNext(){
      return nextLink;
    }

    string name;
    T nextLink;

};

class Foo2 : public Foo1 {
  public:
    Foo2(){
      name = "Foo2";
    }
};


class Foo3 : public Foo1 {
  public:
    Foo3(){
      name = "Foo3";
    }
};

template <class T>
class LinkedList {



public:
    T curr;
    T first;

void add(T node){
  if(first == NULL){
    first = node
  }
  node->nextLink = this;
  curr = node;
}
T getNext(){
  return next;
}
void printAll(){
  T curr = first;
  cout << "Contents are: " ;
  while(curr != NULL){
    cout << curr.print() << ", ";
    curr = curr.getNext();
  }
}

};

int main() {
  LinkedList<?> list;
  list.add(new Foo2());
  list.add(new Foo3());
  list.printAll();
  return 0;
}

我正在尝试实现通用链表，我意识到我可以导入<list>但这不适合我的项目。我正在尝试拥有Foo2和Foo3个对象的链接列表 - 上面是我可以完成的最好的，因为我是C ++的新手。

错误：

generic.C: In instantiation of Foo1<Foo2>:
generic.C:26:   instantiated from here
generic.C:22: error: Foo1<T>::nextLink has incomplete type
generic.C:6: error: forward declaration of âclass Foo2
generic.C: In instantiation of Foo1<Foo3>:
generic.C:34:   instantiated from here
generic.C:22: error: Foo1<T>::nextLink has incomplete type
generic.C:7: error: forward declaration of class Foo3
generic.C: In member function void LinkedList<T>::add(T):
generic.C:50: error: expected ; before } token
generic.C: In member function T LinkedList<T>::getNext():
generic.C:55: error: ânextâ was not declared in this scope
generic.C: In function âint main()â:
generic.C:69: error: template argument 1 is invalid
generic.C:69: error: invalid type in declaration before â;â token
generic.C:70: error: request for member âaddâ in âlistâ, which is of non-class type âintâ
generic.C:71: error: request for member âaddâ in âlistâ, which is of non-class type âintâ
generic.C:72: error: request for member âprintAllâ in âlistâ, which is of non-class type âintâ

Answer 1

你需要使用T *，而不是T.看起来像你来自Java，一切都是参考。 C ++模板中没有?。我认为你需要先拿一本关于基础C ++的书，然后再回到模板。

Answer 2

我认为问题是“？”在LinkedList

中

如果是这种情况，那么您应该使用LinkedList<Foo1 *>。

为什么不能使用std :: list？也许我们可以帮助你，使用你自己的实现会好得多。

Answer 3

尽管你的断言恰恰相反，你给出的例子可以用std::list来解决：

std::list<Foo1 *> list;

list.push_back(new Foo2());
list.push_back(new Foo3());

for (std::iterator<Foo1 *> it = list.begin(); it != list.end(); ++it)
{
    (*it)->print();
}

显然，这里存在潜在的内存泄漏......

Answer 4

结合这些位，看起来应该可行：

int main() {
  std::list<boost::variant<Foo2, Foo3> > list;
  list.push_back(Foo2());
  list.push_back(Foo3());
  printAll(list); // You'd still need to write this obviously.
  return 0;
}

C ++通用链接列表

4 个答案: