Thread: String Class

Originally Posted by Daved

I'm sure you'll find out as soon as you use them. It's rather obvious.

You should make the first one return a const char& or a char since the function is const and you shouldn't be providing non-const access to the member data. But that's not the obvious thing that is wrong with them.

I changed them and I think they are okay now:

Code:

char& String::operator [](int i) const 
{
	if(i >= 0 && i < stringlength)
		return sink;
	else
		return buf[i];
}

char& String::operator [](int i)
{
	if(i >= 0 && i < stringlength)
		return sink;
	else
		return buf[i];
}

Originally Posted by vart

if i is -1 you will return buf[-1] - out of bounds access

so, should it be this:

Code:

char& String::operator [](int i) const 
{
	if(i <= 0 && i < stringlength)
		return sink;
	else
		return buf[i];
}

char& String::operator [](int i)
{
	if(i <= 0 && i < stringlength)
		return sink;
	else
		return buf[i];
}

Originally Posted by vart

in this case using i == stringlength will give you the out of bound access

I don't understand.

Originally Posted by vart

What exactly you do not understand?

if i is stringlength than
i<=0 is false, the if is false and you go to the else getting buf[stringlength]
this is out of bounds access

oh, so you are saying replace what I have in the if() with i == stringlength.

How do I make the print function use printf() to output the name of the string, the string literal, and the size of the string literal with the three all displayed on one line per string object?

Originally Posted by vart

What is string name?

Like if I make a string:

String s1;

and then I do this:

s1.setName("s1");

Then I should be able to show the name of the object which in this case is s1.

Originally Posted by vart

like

Code:

printf("%s: %s(%d)\n", name, buf, stringlength);

?

But why do you need a printf if you trying to write in C++?

isn't printf used in C++?

Going back to your operator[], you fixed the wrong thing. I'm guessing you tried it and an exception was thrown? You have to try to understand why that happens. When I said it was obvious, I didn't mean it would necessarily be obvious when you first look at it (otherwise you wouldn't have coded it that way), I just meant that once you realized what it was it would make sense. It sounds like you still don't know what the flaw was.

So your fix was to return sink instead of throwing an exception in the first if statement. That wasn't where the problem was. The problem was the if. Pick a number and run it through the if and see where you get. You should always do this in your mind or on paper to test the if to see if it does what you want. Let's say the stringlength is 8 and the user wants the character at position 3. Follow your if in the earlier code and see what gets executed.

>> isn't printf used in C++?
It can be, but cout is generally used instead.

I made some changes:

Code:

char& String::operator [](int i) const 
{
	if(i < 0 || i> stringlength)
		return sink;
	else
		return buf[i];
}

char& String::operator [](int i)
{
	if(i < 0 || i>stringlength)
		return sink;
	else
		return buf[i];
}

Also, how do I make a +operator() that acts as a unary operator for the string class.