Question about operator overloading and hard-coded strings.

Hello. I'm trying to review operator overloading in C++, and I've run into a problem that involves both it and typecasting. This line of code is giving me issues:

Code:

---

Integer integer = "34";

---

The message is:

Code:

---

conversion from 'const char [3]' to non-scalar type 'Integer' requested|

---

At first this was coming up with the following operator overload:

Code:

---

Integer Integer::operator=(const string &str)
{
    value = str;
    return *this;
}

---

so I tried changing it to this to no avail:

Code:

---

Integer Integer::operator=(const char *str)
{
    value = *str;
    return *this;
}

---

What's the deal? There's almost definitely a simple fix to this, but I'm at a loss at the moment as to what it is. Thanks!

The problem is that, while your code looks like assignation, it really uses copy construction. You need to provide a copy constructor that takes a string literal.

P.S. Also, operator = is supposed to return a REFERENCE not an actual copy.


EDIT: As Elysia pointed out, this is not quite correct. It uses the constructor, not the copy constructor!

Integer integer = "34";
This is an initialization, the same as
Integer integer("34");
Initialization utilizes the constructor (not the copy constructor).

Integer integer;
integer = "34";
This is assignment. It uses the assignment operator.

Quote:

---

Originally Posted by Elysia

Initialization utilizes the constructor (not the copy constructor).

---

Meh, yeah you're right. Brainfart.

@OP: Sorry about that. :)

Thanks. I'm kind of surprised it wasn't a more minor, technical detail about overloading syntax or typecasting.

You're probably better off making a constructor for it instead anyway rather than doing:
Integer integer;
integer = "34";
Then thanks to the copy-constructor, it will allow you to do both.
Of course you may also want to consider whether the constructor should be marked explicit.