My own regex...............class?

i was thinking of designs, and i came across this one which led me to question, would this be an gross misuse of a class?

i'm fairly certain that this won't compile, but just pretend it will for now

Code:

---

string operator=(const Regex &r)
{
        return r.return_string;
}

class Regex
{
        friend string operator=(const Regex &);
       
        public:
        //ONE public function
        Regex(string character_string, string expression);
       
        protected:
        string return_string;             
        //....a bunch of private funtions
}

int main()
{
        string to_match = "a2bcd4";
        string result = "";       

        //pretend like there is no need to escape any \'s here
        string expression = "[a-z][0-9]\L\w+\E\d";
       
        result = Regex(to_match, expression);       
}

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You can't return a value from out of a constructor. Thats effectively what you're trying to do in main, with the line

Code:

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result = Regex(to_match, expression);

---

Maybe the line you're looking for is

Code:

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  Regex r(result, to_match, expression)

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You can change the compiler to take three arguments, the first a pointer or reference. That would work, I think.

but i'm not returning a value from the constructor -

i'm returning it from operator=(const Regex &)


however, in my "foo-bar testing" , i cannot seem to get ::operator=() to be able to access my private class elements. i know i have done the same thing with cout and <<, but it's not quite working as i had thought.

I think I see, yes...

You want the compiler to mangle your code (on the right-hand-side of result=Regex(...); ) into sth like this?

Code:

---

operator=(const Regex r(to_match, expression));

---

I'm not sure if that can work. It would be quite neat if it did, though.

Methinks you're going to have to find a syntax for C++ somehwere and use that to check if what you want to do is possible.

I think what you want to do is define the conversion from a Regex to a string, instead of overloading the assignment operator.

It must be a member function of the Regex class with a signiture something like:

operator string() const;

and would look something like:

Code:

---

Regex::operator string() const
{
  return return_string;
}

---

Quote:

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Originally Posted by Koedoe

I think I see, yes...

You want the compiler to mangle your code (on the right-hand-side of result=Regex(...); ) into sth like this?

Code:

---

operator=(const Regex r(to_match, expression));

---

I'm not sure if that can work. It would be quite neat if it did, though.

Methinks you're going to have to find a syntax for C++ somehwere and use that to check if what you want to do is possible.

---

Sure it will work, basically it creates a temporary object on the right and then using the operater= assigns the member string to the string variable on left. Syntax-wise it's the same as this:

Code:

---

string result;
result = string("This is a string");

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EDIT: I know what you are doing wrong, the global operator= overload requires 2 parameter, not one as in the member function overload. So it needs to be:

Code:

---

string operator=(string & result, const Regex &r)
{
        result = r.return_string;
        return &result;
}

---

you could return r.return_string but since it's const you would have to return it to a const string otherwise you'd create a backdoor to a private member variable.