Question

我正在研究RPN表达式的计算器，我收到的链接器错误似乎无法解决。我正在尝试使用双重调度来选择正确的操作和值。我将从类层次结构的顶部开始向下移动直到到达Operand。它不是来自运营商。

integer.obj : error LNK2001: unresolved external symbol "public: virtual class std::shared_ptr<class Operand> __thiscall Operand::perform_addition(class std::vector<class std::shared_ptr<class Token>,class std::allocator<class std::shared_ptr<class Token> > > &)" (?perform_addition@Operand@@UAE?AV?$shared_ptr@VOperand@@@std@@AAV?$vector@V?$shared_ptr@VToken@@@std@@V?$allocator@V?$shared_ptr@VToken@@@std@@@2@@3@@Z)
tokenizer.obj : error LNK2001: unresolved external symbol "public: virtual class std::shared_ptr<class Operand> __thiscall Operand::perform_addition(class std::vector<class std::shared_ptr<class Token>,class std::allocator<class std::shared_ptr<class Token> > > &)" (?perform_addition@Operand@@UAE?AV?$shared_ptr@VOperand@@@std@@AAV?$vector@V?$shared_ptr@VToken@@@std@@V?$allocator@V?$shared_ptr@VToken@@@std@@@2@@3@@Z)
ut_rpn_evaluator.obj : error LNK2001: unresolved external symbol "public: virtual class std::shared_ptr<class Operand> __thiscall Operand::perform_addition(class std::vector<class std::shared_ptr<class Token>,class std::allocator<class std::shared_ptr<class Token> > > &)" (?perform_addition@Operand@@UAE?AV?$shared_ptr@VOperand@@@std@@AAV?$vector@V?$shared_ptr@VToken@@@std@@V?$allocator@V?$shared_ptr@VToken@@@std@@@2@@3@@Z)
tokenizer.obj : error LNK2001: unresolved external symbol "public: virtual class std::shared_ptr<class Operand> __thiscall Operation::perform(class std::vector<class std::shared_ptr<class Token>,class std::allocator<class std::shared_ptr<class Token> > > &)" (?perform@Operation@@UAE?AV?$shared_ptr@VOperand@@@std@@AAV?$vector@V?$shared_ptr@VToken@@@std@@V?$allocator@V?$shared_ptr@VToken@@@std@@@2@@3@@Z)
C:\Users\User\OneDrive\School\Semester 3\Course\project\Debug\ut_rpn_evaluator.exe : fatal error LNK1120: 2 unresolved externals

这些函数的定义从基类开始：

class Operation : public Token {
public:
    using pointer_type = TOKEN_PTR_TYPE(Operation);
    virtual unsigned number_of_args() const = 0;
    virtual Operand::pointer_type perform( TokenList& params );
};

下一个派生类是运算符，然后是运算符标题中的加法：

class Addition : public LAssocOperator {
    DEF_IS_CONVERTABLE_FROM(Addition)
    DEF_PRECEDENCE(ADDITIVE)
    public:
        Operand::pointer_type perform(TokenList& params);
    };

这是我在运营商的源代码文件中的实现：

Operand::pointer_type Addition::perform(TokenList& param){
    Operand::pointer_type op;
        op = op->perform_addition(params);
    return op;

}

这是操作数头文件中函数的声明：

class Operand : public Token
{
public:
    using pointer_type = TOKEN_PTR_TYPE(Operand);

    //Operation Definitions
    virtual Operand::pointer_type perform_addition(TokenList& params);
};

Integer派生自操作数，这是标题中类定义的一部分：

class Integer : public Operand {
public:
    //usings for pointer and value
private:
    //member for value
public:
    //Constructor
    value_type              get_value() const { return value_; }

        Operand::pointer_type perform_addition(TokenList& params);
};

最后，我最终在整数源文件中定义了函数：

Operand::pointer_type Integer::perform_addition(TokenList& params){
    Integer::value_type value = Integer::value_type(0);
    for (auto val : params){
        value += convert<Integer>(val)->get_value();
    }
    return make_operand<Integer>(value);
}

我一直在阅读有关C ++链接器和链接器错误的大量内容，但我无法解决这个问题。对于我正在做的愚蠢事情，我总是会收到链接器错误。有谁能发现问题？

由于

Answer 1

class Operand : public Token
{
    // ....
    virtual Operand::pointer_type perform_addition(TokenList& params);

此方法尚未在任何地方实施。如果您认为它是纯虚拟的，那么您需要在最后添加= 0。

如何找到这个令人难以置信的神秘链接器错误的原因？

1 个答案: