a string class problem

Following code gave wrong output, can you explain why. Any comments for the "correct" part of the code also wellcome :-)

#include <iostream.h>
#include <string.h>

class Mystring
{
private:
char *pstring;
int length;
public:
Mystring()
{  length = 0;
   pstring = NULL;
}
Mystring(char *pchar)
{  if (pstring!=NULL) delete pstring;
   length = strlen(pchar);
   pstring = new char[length + 1];
   strcpy(pstring, pchar);
}
Mystring(Mystring &mystring1)
{  if (pstring!=NULL) delete pstring;
   length = mystring1.length;
   pstring = new char [length + 1];
   strcpy (pstring, mystring1.pstring);
}
~Mystring()
{  delete pstring;
}
int Len (void)
{  return length;
}
void Show (void)
{  cout<<pstring<<endl;
}
};

int main ()
{  Mystring Mystring1;
   Mystring Mystring2("a string");
   Mystring Mystring3(Mystring2);     //Remove these 2 lines  
   Mystring2.Show(); Mystring3.Show();//the output become OK    
   Mystring1="another string";
   Mystring1.Show();
   return 0;
}

Without operator overloading, Do you think Mystring1="another string" is OK?
learnAsked:
Who is Participating?
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

learnAuthor Commented:
Edited text of question
0
q2guoCommented:
I don't see anything wrong except you should make the
following changes

~Mystring()
{
      if (pstring !=NULL) delete pstring;
}  

also, you should move your implementation of the member functions
out of the class definition.
e.g.
void MyString::show(void)
{
      .....
}

I don't think Mystring1="another string" is ok
since what this line actually does is
creates a temperory Mystring Object that contains "another string" and assigns this temmperory object to MyString1
because your MyString class contains pointer type data members
so, =operator will have to be overloaded.

Tell me if this helps

Terry
0

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
q2guoCommented:
Forget to ask you, what output are you getting anyway?
0
Cloud Class® Course: Certified Penetration Testing

This CPTE Certified Penetration Testing Engineer course covers everything you need to know about becoming a Certified Penetration Testing Engineer. Career Path: Professional roles include Ethical Hackers, Security Consultants, System Administrators, and Chief Security Officers.

learnAuthor Commented:
Hi Q2guo,

1. The output from my code is something like "ano..er string" including some strange characters!

2. There is no operator (=) overloading in the code, how the string "another string" can go into the class and gave its value to the pointer pstring for Mystring1?

Cheers
0
q2guoCommented:
C++ compiler automatically creates the =operator function
for every object.  But, in this case, you don't want to use
the =operator provided by the compiler 'cause it will generate a bug in your program.

DOes mystring2.show() output ok in your program above?
0
learnAuthor Commented:
Yes, both Mystring2.Show() and Mystring3.Show() are OK.
0
q2guoCommented:
You are saying both Mystring2.show() and Mystring3.show()
work ok or are you saying they both give the correct output?
0
learnAuthor Commented:
Both outputs for Mystring2 and Mystring3 were correct.
0
q2guoCommented:
I tried out the program on my computer

the only problem I have is the
Mystring1="another string";
line, other than this everything works out fine


0
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
C++

From novice to tech pro — start learning today.