Conversion operators

Hi all

I need some explanation on understanding the concept of conversion operators.

I read about conversion operators and found that they automatically convert one object into another one.

Here is some code.

Code:

---


class A {
ifstream file;
string fileName;

public:

A(const string& fname)
: fileName(fname), file(fname.c_str()) {}

string name() const { return fileName; }

void name(const string& newName) {
fileName = newName;
}

operator ifstream&() { return file; }
};

int main() {

A a("A.cpp");
cout << a.name() << endl;

a.close(); // Compile time error. close() not member of A

}

---

I am little bit confused in this.
Since automatic type conversion, converts object "a" to type of ifstream then it should be able to call close() function of ifstream. So why does it giving error.


I will very thankful if some one answered my following question:

1) HOW conversion operator work's, I mean to say how calls are gone to conversion operator?

2) What is the criteria to convert one object to another, is this rule is universal?

I have read some where about conversion operator:

Quote:

---

Automatic type conversion happens only in function calls, not during member selection

---

What does it mean????

Thanks for your time.

Thanks
Nickman

Can

Your big misconception is that is somehow magically transforms the object from one type to another. That is simply not true.
It creates a new object, then transforms this into the specified type and returns it.
The cout statement calls the ifstream conversion operator which simply returns a reference to its member file. So the transformed object is then returned from the conversion operator and passed to cout.

There are several things you should be aware of, though:
Conversion operators should be avoided. They can cause a lot of implicit conversion that causes headaches.
Returning a non-reference to a private member variable is usually considered bad because it breaks encapsulation.

Thanks for reply....

So that's mean that now "a" is object of type ifstream. So why it's now able to call close() function of ifstream?

Thanks once again.

Nickman

Quote:

---

Originally Posted by nickman

So that's mean that now "a" is object of type ifstream. So why it's now able to call close() function of ifstream?

---

Like I said, no, a is NOT an ifstream object. It is and always will be an A. Conversion operators does not transform the instance. It returns a transformed object.

Hi

I am still not getting.

Quote:

---

It returns a transformed object.

---

What is the transformed object?

Could you please eloborate more on this????


Thanks
Nickman

Taking your original code,

Code:

---

#include <string>
#include <fstream>

using namespace std;

class A
{
private:
        ifstream file;
        string fileName;

public:
        A(const string& fname)
                : fileName(fname), file(fname.c_str()) {}

        string name() const { return fileName; }

        void name(const string& newName)
        {
                fileName = newName;
        }

        operator ifstream&() { return file; }
};

int main()
{
        A a("A.cpp");
        ifstream & transformed_obj = a;
        cout << a.name() << endl;
        transformed_obj.close();
}

---

Hi

So what is object "a" is here?

Quote:

---

ifstream & transformed_obj = a;

---

An instance of class A or transformed object as you are using it as initializer for transformed_obj .

Thanks
Nickman

There is only one a, and it's an instance of class A.
The compiler then calls the conversion operator in A which returns an ifstream object which is assigned to transformed_obj.

I got it...... Elysia thanks for your time....

I had played with your code and found out that if comment out line:

When I comment out line:

Code:

---

operator ifstream&() { return file; }

---

Then compiler is giving me error that cannot convert instance of A to ifstream.

It means that conversion operator is called only when the compiler see that(Suppose) in this expression there is requirement of other object type and THERE IS CONVERSION OPERATOR FUNCTION WHICH CAN RETURN THE DESIRABLE OBJECT.

Also when I comment out the same line in my original code it was not giving any error( because no conversion was required at that time).

Please correct me if wrong.

Thanks a lot for your help.!!!!!!!

You are correct. Now keep in mind what I mentioned earlier about conversion operators and references.

Thanks Elysia.... I'll keep in my mind :)

Let me put it differently.

Code:

---

struct convertible {
  int get() const { return 0; }
  operator int() const { return 0; }
};

---

There isn't much difference between get() and the conversion operator. The only difference, really, is that the latter has a weird name and is called implicitly under some circumstances, while the former must be called explicitly. Other than that, they're the same.