How to fix compiler error "class has no member named X"?

Question

I'm currently coding an expression evaluator and have run into an issue regarding polymorphism. My class hierarchy is as follows: Divide inherits from Operator which inherits from the base class Expression. When I instantiate an Object of type Divide with base class Expression and try to access the function divide(int, int) i get the following error:

"Expression.cpp: In member function ‘int Expression::evaluate()’: Expression.cpp:37:6: error: ‘class Expression’ has no member named ‘divide’"

Here is "Expression.h"

#ifndef EXPRESSION_H
#define EXPRESSION_H
#include <string>
using namespace std;
class Operator;
class Divide;
class Expression 
{
  protected:
    char **ana, *exp;

  public:
   Expression();
   ~Expression();
   Expression(char* ex);
   int evaluate();
   void tokenize();

    class EmptyException
    {
       public:
     EmptyException(string a){reason = a;};
     string getReason() const{return reason;};
       private:
     string reason;
    };

};

#endif

And here is "Expression.cpp"

#include "Expression.h"
#include "Stack.h"
#include "Queue.h"
#include "Node.h"
#include <iostream>
#include <cstring>
#include <stdio.h>
#include <ctype.h>
#include "Operator.h"
#include "Divide.h"

Expression::Expression()
{

} 

Expression::~Expression()
{
   delete [] exp;
}

Expression::Expression(char* ex)
{
   exp = ex;
   //tokenize();

}

int Expression::evaluate()
{
  Stack stack;

  Expression *tmp;
  Divide d;
  tmp = &d;
  tmp->divide(4, 2);
  stack.push(tmp);
   tmp = stack.pop();

 }

 void Expression::tokenize()
 {  
   int space = 0;

   for(int i  =0 ; i < strlen(exp); i++)
   {
     if(exp[i] == ' ')
     space++;
   } 

   char ** ana = new char*[space + 1];
   ana[0] = strtok(exp, " ");

   for(int i = 1 ; i < space + 1; i++)
   {        
   ana[i]= strtok (NULL, " ");
   }

  }

And now the "operator.h"

#ifndef OPERATOR_H
#define OPERATOR_H
#include "Expression.h"

using namespace std;

class Operator : public Expression
{
  protected:
    bool unary, binary;
  public:
    Operator();
    bool isUnary() const;
    bool isBinary() const;
};

#endif

"Operator.cpp"

#include "Operator.h"
#include <string>
#include <cmath>
#include "Expression.h"

Operator::Operator()
{

}

bool Operator::isUnary() const 
{
      if(unary)
    return true;
      else
    return false;

}

bool Operator::isBinary() const 
{
      if(binary)
    return true;
      else
    return false; 
}

"Divide.h"

#ifndef DIVIDE_H
#define DIVIDE_H

#include <string>
#include "Operator.h"
class Expression;
using namespace std;


class Divide : public Operator
{
        private:
      char id;
    public:
        Divide();
        Divide(char);
        int divide(int a, int b);
        char identity() const;
};

#endif

"Divide.cpp"

#include "Divide.h"

int Divide::divide(int a, int b)
{
  return a/b;
}

Divide::Divide(char _id)
{
 id = _id; 
}

char Divide::identity() const
{
  return id;
}

and for in case the makefile

main: Expression.o Equation.o Operator.o Divide.o Stack.o Queue.o main.o
   g++ -static main.o Equation.o Expression.o Operator.o Divide.o Stack.o Queue.o -o main

main.o : main.cpp 
        g++ -c main.cpp 

Equation.o : Equation.cpp Equation.h
    g++ -c Equation.cpp

Expression.o : Expression.cpp Expression.h 
    g++ -c Expression.cpp 

Operator.o : Operator.cpp Operator.h Expression.h 
    g++ -c Operator.cpp 

Divide.o : Divide.cpp Divide.h Operator.h
    g++ -c Divide.cpp 

Stack.o : Stack.cpp Stack.h Node.h
    g++ -c Stack.cpp

Queue.o : Queue.cpp Queue.h Node.h
    g++ -c Queue.cpp

I have left out all the other operators such as eg. Plus, minus etc because they are all identical to what Divide is and would just complicate this question even further.

score 4 · Answer 1 · answered Oct 01 '12 at 19:30

4

It's pretty clear, right? Expression doesn't have the method divide.

Expression *tmp;
Divide d;
tmp = &d;
tmp->divide(4, 2); //tmp is an Expression*

Also... why would you do this? What's wrong with:

Divide d;
d.divide(4,2);

answered Oct 01 '12 at 19:30

Luchian Grigore

253,575
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457
625

Well my stack takes an Expression* so that i dont end up using more than one stack to cater for multiple types, ie Divide, Add, Multiply, power etc. So when I parse my input equation and begin to push onto the stack, I want to be able to create a Divide or Add object that will inherit from Operator and Expression (base class). – Chris Oct 01 '12 at 19:40
@Chris that still doesn't explain why you need the local object. – Luchian Grigore Oct 01 '12 at 19:41
@Chris judging by your comments, it seems there's a deeper problem, which is why I recommend this - http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list the question (as you intend it at least, not as posted) is too broad to be answered on SO. – Luchian Grigore Oct 01 '12 at 19:45

score 1 · Answer 2 · answered Oct 01 '12 at 19:30

1

Your problem is that you are calling divide on the variable called temp which is a pointer to an Expression here: tmp->divide(4, 2); You should call divide on your d variable.

answered Oct 01 '12 at 19:30

bohney

1,187
8
12

I understand what your saying, but i dont want d to call divide, i want it to be polymorphic? – Chris Oct 01 '12 at 19:42
@Chris you have no virtual methods in your code, how can anything be polymorphic? – Luchian Grigore Oct 01 '12 at 19:43
Ok, i see what you mean, but in this case divide doesn't need to be virtual, does it? What am I not understanding about the statement Expression *tmp = new Divide(); ? Is this not possible? – Chris Oct 01 '12 at 19:49
@Chris it is, but `tmp` is still a pointer to an `Expression`... – Luchian Grigore Oct 01 '12 at 19:51

How to fix compiler error "class has no member named X"?

2 Answers2

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