Getting confused by a simple program

Ok so I think I'm having trouble with a class in my program so I created a small program to try to see what is going on. I am confused by the results. It seems that just when I think I understand C++ and OOP I find out I have been doing something wrong.
This is my BaseData class( in base.h):

Code:

---

class BaseData
{
public:
        BaseData(){
                static int ctor=0;
                printf("ctor %d\n",ctor);
                ctor++;

                name = new char[30];
                strcpy(name,"");
        }

        ~BaseData(){
                static int dtor=0;
                printf("dtor %d\n",dtor);
                dtor++;

                delete[] name;
        }

        BaseData(const BaseData& p){
                static int cctor=0;
                printf("copy ctor %d\n",cctor);
                cctor++;

                name = new char[30];
                *this = p;
        }

        BaseData& operator=(const BaseData& p){
                static int assign=0;
                printf("assign %d\n",assign);
                assign++;

                if(this != &p)
                {               
                        strcpy(name,p.name);
                        sensNum = p.sensNum;
                }
                return *this;
        }

        LPTSTR                name;
        BYTE                sensNum;
};

---

This is my main:

Code:

---

#include <windows.h>
#include <stdio.h>
#include "base.h"

int main()
{
        BaseData b, t1;
        BaseData* t2, t3;

        strcpy(b.name,"Name");
        b.sensNum = 42;
        printf("b: %s %d\n",b.name,b.sensNum);

        t1 = b;
        printf("t1: %s %d\n",t1.name,t1.sensNum);

        t2 = new BaseData(b);
        printf("t2: %s %d\n",t2->name,t2->sensNum);
        delete t2;
        /*       
        t3 = new BaseData; //this statement won't compile
        t3 = b;
        printf("t3: %s %d\n",t3->name,t3->sensNum);
        delete t3;
        */
        return 0;
}

---

The output is:
ctor 0
ctor 1
ctor 2
b: Name 42
assign 0
t1: Name 42
copy ctor 0
assign 1
t2: Name 42
dtor 0
dtor 1
dtor 2
dtor 3

First off why is the dtor called 4 times? t3 is never created so why is it begin destroyed? Second, though I think this is a result of using printfs, why is ctor 2 printed before b: Name 42? Third and most important, the commented out portion doesn't work and I do not understand why. Can someone explain? Finally does this class leak memory and/or what is wrong with it if anything? (I just started having problems in my larger program and I think it has something to do with this class).

The line in your code :

Code:

---

BaseData* t2, t3;

---

is the same as :

Code:

---

BaseData* t2;
BaseData  t3;  // not BaseData* t3;

---

Oh ok... now that you say that I remember. I used to put the * on the variables and not the type and recently changed.

It is amazing how much you can forget when you don't use it for awhile. I now have:

BaseData b, t1, *t2, *t3;

Well that answers my first and second questions, still have the third.

NEVERMIND, all I had to do now is change t3 =b to *t3=b.

you fixed all my problems and I feel stupid for having them. thanks for the help.

Quote:

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Originally posted by pim42
Finally does this class leak memory

---

Are you trying to wrap handling of char data in a class? If so, why not just use std::string?

Code:

---

#include <string>
class BaseData
{
  private:
      std::string name;

  public:
        BaseData(){
        static int ctor=0;
        printf("ctor %d\n",ctor);
        ctor++;
        }
};

---

Note that you really do not need a copy ctor, assignment operator, or destructor if you use std::string. I've cut the size of your class down to a fraction of what it was originally, does virtually the same thing as your original code does, and it is maintainable.

Regards,

Paul McKenzie

no this was actually a dumbed down version of a hardware sensor class. I'm interfacing a C dll so I'm using LPTSTR to match the function calls. (You might recall from pervious posts that I am required to use the same data types as the dll interface, thus no std::string).

Quote:

---

Originally posted by pim42
no this was actually a dumbed down version of a hardware sensor class. I'm interfacing a C dll so I'm using LPTSTR to match the function calls. (You might recall from pervious posts that I am required to use the same data types as the dll interface, thus no std::string).

---

Use vector<TCHAR>, or if the DLL doesn't modify your buffer, std::string::c_str(). Same thing, less headaches.

For example, if your DLL function is prototyped like this:

Code:

---

void func(LPTSTR s);

---

The way you would call it using a vector is as follows:

Code:

---

vector<TCHAR> vt( 20 ); // or whatever the number of characters
//...
func(&vt[0]);

---

The std::vector is the gateway to interfacing to 'C' functions. The vector stores its data in contiguous memory.

To specify a char *, int *, long*, whatever *, the way to do it is to create a vector of the type desired, and just specify the address of the first element when calling the function. Therefore, your entire class interface can be simplified to this:

Code:

---

#include <string>
class BaseData
{
  private:
      std::vector<TCHAR> name;

  public:
        BaseData() : name(31, 0) {
        static int ctor=0;
        printf("ctor %d\n",ctor);
        ctor++;
        }
};

---

Again, no new, delete, copy constructor, op=, or destructor is necessary. This trick of using the std::vector instead of the "new[] / delete[]" stuff that a lot of code is littered with makes the program less problematic and easier to maintain. Also, if I want to resize the name field, all I have to do is call name.resize(). I don't have to do the "new / copy / delete " coding.

Scott Meyer's book, "Effective STL", has an entire chapter dedicated to interfacing to 'C' functions using the standard containers.

Regards,

Paul McKenzie