c++ - 如何解决有关迭代器构造函数的一系列错误，涉及预期左值、缺少转换和可行性

Question

我有一个程序是我在 2000 年为一个类(class)编写的，我想我会尝试运行它只是为了好玩。它曾经在 2000 年编译过，但我现在遇到了一堆错误。自那门课以来，我几乎没有看过 C++ 代码，所以我有点困惑。我解决了一堆关于 iostream.h 和 list.h 的投诉(把“std::”放在一堆像 std::cerr 等东西的前面)。我不知道这是否导致了当前错误，或者我的迭代器是否存在固有错误，但唯一需要解决的问题涉及一堆迭代器构造函数。首先，这是错误示例:

SolutionList.cpp:10:22: error: no matching constructor for initialization of 'PieceList::iterator'
        PieceList::iterator i=rhs.first();
                            ^ ~~~~~~~~~~~
./PieceList.h:27:5: note: candidate constructor not viable: expects an l-value for 1st argument
                                iterator(iterator& rhs){data = rhs.data;}
                                ^
./PieceList.h:28:5: note: candidate constructor not viable: no known conversion from 'PieceList::iterator' to 'std::list<Piece>::iterator &'
      (aka '__list_iterator<value_type, __void_pointer> &') for 1st argument
                                iterator(std::list<Piece>::iterator& rhs){data = rhs;}
                                ^
./PieceList.h:26:5: note: candidate constructor not viable: requires 0 arguments, but 1 was provided
                                iterator(){}
                                ^

我认为这是代码的相关部分:

class PieceList
{
    public:
        PieceList(){}
        PieceList(PieceList& rhs){pieces = rhs.pieces;}
        PieceList(std::list<Piece>& rhs){pieces = rhs;}

        friend std::ifstream& operator>>(std::ifstream&,PieceList&);
        friend std::istream& operator>>(std::istream&,PieceList&);

        class iterator
        {
            public:
                friend class PieceList;

                iterator(){}
                iterator(iterator& rhs){data = rhs.data;}
                iterator(std::list<Piece>::iterator& rhs){data = rhs;}

                Piece operator*(){return *data;}
                iterator& operator=(iterator& rhs){data=rhs.data;return *this;}
                iterator& operator=(std::list<Piece>::iterator& rhs){data = rhs;return *this;}
                bool operator==(iterator& rhs){return data==rhs.data;}
                bool operator==(const iterator& rhs) const{return data==rhs.data;}
                bool operator!=(iterator rhs){return data != rhs.data;}
                iterator& operator++(){++data;return *this;}
                iterator& operator--(){--data;return *this;}
            private:
                std::list<Piece>::iterator data;
        };

        iterator first(){iterator i;i.data=pieces.begin();return i;}
        iterator last(){iterator i;i.data=pieces.end();return i;}

    private:
        std::list<Piece> pieces;
};

所以我知道它不喜欢第二个构造函数，但我不确定如何修复它。我无法通过谷歌搜索错误来理解我的任何谷歌结果。谁能指出我正确的方向？

Answer 1

您不能将临时变量绑定(bind)到非常量引用。

    iterator(iterator const& rhs) : data(rhs.data) {}
    iterator(std::list<Piece>::iterator rhs) : data(rhs) {}

请注意，迭代器通常按值传递

Live On Coliru

#include <iterator>
#include <list>
#include <string>
#include <iostream>

struct Piece {
    std::string name;
};

class PieceList
{
    public:
        PieceList(){}
        PieceList(PieceList const& rhs){pieces = rhs.pieces;}
        PieceList(std::list<Piece> const& rhs){pieces = rhs;}

        friend std::ifstream& operator>>(std::ifstream&,PieceList&);
        friend std::istream& operator>>(std::istream&,PieceList&);

        class iterator
        {
            public:
                friend class PieceList;

                iterator(){}
                iterator(iterator const& rhs) : data(rhs.data) {}
                iterator(std::list<Piece>::iterator rhs) : data(rhs) {}

                Piece operator*(){return *data;}
                iterator& operator=(iterator& rhs){data=rhs.data;return *this;}
                iterator& operator=(std::list<Piece>::iterator& rhs){data = rhs;return *this;}
                bool operator==(iterator& rhs){return data==rhs.data;}
                bool operator==(const iterator& rhs) const{return data==rhs.data;}
                bool operator!=(iterator rhs){return data != rhs.data;}
                iterator& operator++(){++data;return *this;}
                iterator& operator--(){--data;return *this;}
            private:
                std::list<Piece>::iterator data;
        };

        iterator first(){iterator i;i.data=pieces.begin();return i;}
        iterator last(){iterator i;i.data=pieces.end();return i;}

    private:
        std::list<Piece> pieces;
};

int main() {
    PieceList pl { {
        { "aap" },
        { "noot" },
        { "mies" },
    } };
    for(auto it=pl.first(); it!=pl.last(); ++it) {
        std::cout << (*it).name << "\n";
    }
}

打印

aap
noot
mies

PieceList 的构造函数也有类似的问题。您的迭代器类也缺少 operator->。我将其保留为 ExerciseForTheReader™