error: passing [...] discards qualifiers

[carlorfeo]

hello
I'm learning how to use operators:
this code compiles and works fine.

struct Item {
	int data;
	
	Item  (const int value = 0) { data = value; }

	Item operator+ (const Item &item) {
		Item result (data + item.data);
		return result;
	}
	Item operator* (const Item &item) {
		Item result (data * item.data);
		return result;
	}
};

But if I want to make data private...

class Item {
private:
	int data;
public:
	Item  (const int value = 0) { data = value; }
	int & Data () { return data; }

	Item operator+ (const Item &item) {
		Item result (data + item.Data ());
		return result;
	}
	Item operator* (const Item &item) {
		Item result (data * item.Data ());
		return result;
	}
};

int main () {
	Item a (3);
	Item b (4);
	Item c;
	c = a + b * a;

	return 0;
}

I get:
error: passing ‘const Item’ as ‘this’ argument of ‘int& Item::Data()’ discards qualifiers

what does it mean?
thanks

[laserlight]

Provide a const version of Data(), e.g.,

int Data () const { return data; }

Also, member operator+ and operator* should be const. You can make use of return value optimisation with return Item(data + item.Data()), though perhaps named return value optimisation might still be applied.

EDIT:
Also, use the constructor initialisation list:

Item(int value = 0) : data(value) {}

[carlorfeo]

thanks laserlight, it works fine now.
Although I missed the fact that I don't really need to use the function Data() inside the class, I just could read data directly.

May you explain what "discard qualifiers" would mean exactly? I couldn't get a good idea from stuff found in the web.

[laserlight]

May you explain what "discard qualifiers" would mean exactly?

item was passed by const reference. But item.Data() returns a reference (i.e., the const qualifier is discarded). Suppose this was allowed. Then, using the reference returned by item.Data(), you could modify item's contents, even though item is const. So discarding the const qualifier here is treated as an error.

[carlorfeo]

ok, great
thank you