Operator = override

I am trying to duplicate the string library (to learn it better... and I won't lie, it's for an assignment).

But I am having trouble trying to override the = operator.

Here is what I have:

Code:

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mystring& mystring::operator=(const mystring& str){
   
    char *copiedStr;
   
    if (*this == str) return *this;
   
    if (this->size() != str.size())
    {
        delete this;
        copiedStr = new char[str.size()];
        strcpy (copiedStr, str.c_str());
        this = copiedStr;
    }
   
    return *this;
}

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I will admit that I do realize that this code is all jacked up. I have been playing with it so much that I don't even know where I originally started, so it's prolly best to just restart. But I'm stubborn.

Here's the idea... (at least, how I think it should work)
We'll call passing string as str1 and the destination (in this case "this") as str2.

If str1 is the same as str2, do nothing
Otherwise copy str1 to str2.

I am assuming that you would have to make sure they are the same size, and I am under the impression that if they are not the same size, the easiest thing to do would be create a new array. Once the array is created, just copy str1 to str2 and return str2.

However, I should be returning *this and I don't know how to change "this" to point to the new array.

The error I am getting now is

Code:

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error: lvalue required as left operand of assignment

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(referring to "this = copiedStr;")

Can someone please point me in the right direction or show me where I can look to see how the string class is coded? I imagine it's got to be posted somewhere, but I can't find it.

>>I am assuming that you would have to make sure they are the same size...
If you think about it, the only point where you have to make a new array is if str1 is bigger than str2. Otherwise there is room.

>>copiedStr = new char[str.size()];
Must be
copiedStr = new char[str.size() + 1];
because each C-style string must end with a '\0' which is not counted towards the length.

Now, as for this
this = copiedStr;
What you need to do is get rid of the old string you have (which should some member that's a pointer), and substitute it with the new.

So,
delete [] m_str;
m_str = CopiedStr;

And then you return *this, a reference to the object itself, and not the actual string (that's an implementation details that the user should not know about!).

There are a few ways to do this. One way is to use the copy and swap idiom:

Code:

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mystring& mystring::operator=(const mystring& str) {
    mystring temp(str);
    swap(temp);
    return *this;
}

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Now all you have to do is make sure your copy constructor and destructor is correct, and write a swap member function.

Quote:

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Originally Posted by Elysia

delete [] m_str;
m_str = CopiedStr;

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Thank you for this. The only question I have is, what is the "m_str"? Is that the current string?

laserlight... thanks for your advice, but if I did that, wouldn't I still have to make a similar code that I am trying to do now with a swap member function? Or is that easier? Sorry, I'm kind of a noob.

Thank you so much to both of you for your help.

Quote:

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Originally Posted by Amoxaphobic

Thank you for this. The only question I have is, what is the "m_str"? Is that the current string?

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It is whatever member variable that holds the string in your class. So I guess you could refer to it as "the current string," yes.

Quote:

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laserlight... thanks for your advice, but if I did that, wouldn't I still have to make a similar code that I am trying to do now with a swap member function? Or is that easier? Sorry, I'm kind of a noob.

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No, swap simply swaps two variables. Take the contents of A and put it into B, and the take contents of B and put it into A. A swap function would be very trivial, as you see.
But you will have to make sure your destructor and copy constructor works properly. But then, they should always be working.
When creating a new class, you should implement a proper constructor, copy constructor and destructor. Then you can simply write the assignment operator (if it makes sense) by using this swap idom, and it works.

Okay, so if that's the case, how do I access it (the m_str)? The function only passes the second string (the string in the which the "current" string will be). So, I guess that's where I am having all the issues. I have a feeling that's it's simple, but I just don't know what it is.

Access what? Just so you know, the current instance your code is executing in can be acquired by *this.

So... I guess I'm a bit confused. Isn't that what I had?

Something to the extent of:
delete [] this;
*this = copiedStr;


If I do that, I get an error? Sorry...

Back to basics...

If I have

mystring str;
str = "Hello World";

Then the compiler has no idea how to assign a const char[] to a mystring. Therefore, we have to tell the compiler how to do it.
We do this by creating an assignment operator, operator = (const char*). Note that the assignment operator is implemented inside mystring.
So what happens when we execute that code?

The compiler calls str.operator = (const char*).
Inside this function, *this refers to str. That is, *this is of type mystring. But we can't just say *this = CopiedStr, because that is exactly what we are trying to define here.
You know how you built this string, so you know what must be done to assign a new string into it.

At the end, we return *this to enable chaining, such as
mystr = a = b = c = d, ... ;

*this will simply return the object itself, so it is not something you should be concerned with. You just need to make sure to successfully complete the copy operation, and then we end it with that return statement.

Right, back to the basics. Sorry for that.

I got it... I have to refer to how it's made. In this particular case I have a ptr_buffer (this->ptr_buffer) which is where the pointer actually is. So I use that.

Thank you so much for your help and I apologize for the noob questioning.
I think your signature is correct, and I'd like to know as well.... how the hell do you know everything? :)

Quote:

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Originally Posted by Amoxaphobic

I got it... I have to refer to how it's made. In this particular case I have a ptr_buffer (this->ptr_buffer) which is where the pointer actually is. So I use that.

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Right. That's how you do it.

Quote:

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Thank you so much for your help and I apologize for the noob questioning.

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No need to apologize. We learn by asking what we don't understand and experiment.

Quote:

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I think your signature is correct, and I'd like to know as well.... how the hell do you know everything? :)

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OK, that's not to brag or anything, but after you use a language for a certain amount of years, you get good at it.
If you keep at it, you'll come back here one day and laugh at such a silly question and answer them for other newbies ;)

You cannot use delete this; or delete[] this; from inside an assignment operator. What you can do, but really really should not do, is to call the destructor on this, and then the constructor on this, directly from the assignment operator.
Have a read of this GOTW article: GotW #23: Object Lifetimes - Part II

By using the copy and swap idiom, your data type becomes nicely exception safe as a bonus.

Also, do not compare the strings before deciding to do the assignment. That does not save any work. Firstly if they are unequal, which is more likely, then it obviously is just extra overhead. Secondly, if they are actually equal then the only way you know this is when the string compare has walked the entire length of both strings and compared them character for character all the way along. That is unfortunately the same amount of work as doing a copy except without the allocation and deallocation. The longer the string is, the smaller the portion of time is taken by the memory management itself.