Overloaded Operator Function Failing

Question

Here I wrote 2 overloaded operators:

stringb operator+(const stringb &a,const stringb &b)
{
    stringb temp;
    temp.len = a.len + b.len;
    temp.p=new char[(temp.len)+1];
    strcpy(temp.p, a.p);
    strcat(temp.p, b.p);
    return (temp);
}
stringb operator-(const stringb &a,const stringb &b)
{
    stringb temp;
    temp.len=a.len-b.len;
    temp.p=new char[temp.len+1];
    strcpy(temp.p,a.p);
    return (temp);
}

However, when I compile the actual code, the whole code works except the part when I call these operator, and I get garbage out put. What's wrong with my functions?

EDIT: Declaration of stringb class:

class stringb
{
    public:
        char *p;
int len;
public:
    stringb()
    {
    len=0;
    p=0;
    }
stringb(const char *s)
{
    len=strlen(s);
    p=new char[len+1];
    strcpy(p,s);

}
stringb(const stringb &s)
{
    len=s.len;//strlen(s);
    p=new char[len+1];
    strcpy(p,s.p);
}
~stringb()
{
    delete p;
}

friend int operator==(const stringb &a,const stringb &b);
friend stringb operator+(const stringb &a,const stringb &b);
friend stringb operator-(const stringb &a,const stringb &b);
friend void putstring(const stringb a);

};

Just one thing to throw out their, what happens when you call strlen on a pointer that points to null? You have a default contractor that defaults your pointer to NULL. However, if the copy constructor is called it calls strlen on a pointer that points to null. This is going to crash your code as well. However, the correct answer is g-makulik below. — Freddy, Nov 23 '13 at 17:18

score 2 · Answer 1 · edited May 23 '17 at 12:19

2

Your problem is here:

~stringb()
{
    delete p; // <<<<<
}

This will be executed if temp goes out of scope in your operator definition.

To get your code working you need a proper implementation of the Rule of Three in your stringb class. You may also have a look at this nice IDE one sample, P0W has setup from your code.

edited May 23 '17 at 12:19

Community

1
1

answered Nov 23 '13 at 16:15

πάντα ῥεῖ

1
13
116
190

No! You'll need to implement proper rule of 3 in `stringb` – πάντα ῥεῖ Nov 23 '13 at 16:18
Here's a good explanation and example: [Wikipedia 'Rule of three'](http://en.wikipedia.org/wiki/Rule_of_three_%28C++_programming%29) – πάντα ῥεῖ Nov 23 '13 at 16:33
1

SO's contribution http://stackoverflow.com/questions/4172722/what-is-the-rule-of-three – john Nov 23 '13 at 16:35
@Anurupa_Dey See it [your](http://ideone.com/0EjAoQ) implementation and **Rule of Three** – P0W Nov 23 '13 at 16:48
@g-makulik You should add "rule of three" statement into your answer – P0W Nov 23 '13 at 16:50
1

@g-makulik I don't think the OP understands your answer. Maybe walking him through what would cause his code to crash and why? I would respond, but you're answer is correct. I just don't think he is familiar enough with the Rule of Three. Just a thought, great answer by the way. – Freddy Nov 23 '13 at 17:13

score 0 · Answer 2 · answered Nov 23 '13 at 16:16

0

Try returning a reference:

stringb& operator+(const stringb &a, const stringb &b);
{
   stringb *temp = new stringb();
   temp->len = a.len + b.len;
   [...]
   return *(temp);
}

answered Nov 23 '13 at 16:16

Eutherpy

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7
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Did This: `stringb *temp = new stringb; temp->len = a.len - b.len; temp->p=new char[temp->len+1]; strcpy(temp->p,a.p); return *(temp);` Still crashes. – Anurupa_Dey Nov 23 '13 at 16:27
Seems the error is in the -operator implementation. + works just fine. – Anurupa_Dey Nov 23 '13 at 16:39
Well, what if b is longer? How about abs(a.len-b.len)? – Eutherpy Nov 23 '13 at 16:42
still crashes even after using abs, while the + one works superfine! – Anurupa_Dey Nov 23 '13 at 16:48
Don't return a reference to memory allocated on the free store (heap). After this function returns, there are no pointers to the resource it allocates. – Yakk - Adam Nevraumont Nov 23 '13 at 23:31

Overloaded Operator Function Failing

2 Answers2