Question

If i have the following:

class Array{
public:
Array (int n=2) : _n(n), buf(new int[n]) {}

int & operator[](int i) const
{
return _buf[i];
}

int _n;
int* _buf;
};



int main()
{
Array arr1;
const Array arr2;

arr1[0]=1;
arr1[1]=2;

arr2[0]=arr1[0];
++arr2[0];

std:: << arr2[0]<<std::endl;
}

the program will compile and will print ‘2’, but im a bit confused.

what does the const declaration in arr2 protects exactly?

obviously not the content of the object?

please help me to understand.

thank you!

Answer 1

The C++ compiler only enforces that const will protect the members of this object, not other objects referenced indirectly (e.g. via pointer). _buf is a pointer member and what it points to is not protected by the compiler.

However, many classes override based on const in order to also protect access to associated objects. To do that, you’d write:

int & operator[](int i) { return _buf[i]; }
const int & operator[](int i) const { return _buf[i]; }

This propagates const on the object to const on the subscripted element.