Question

Given the following code:

class TestA
{
    private:
        char Temp;

    public:
        char *Ptr;

        TestA(){Ptr = NULL; Temp = 'A'; Ptr = &Temp;}
        void Function(){Ptr = &Temp; Temp = 'B';}

        void operator=(const TestA &ItemCopy)
        {
            //ItemCopy.Temp = 'N'; //Not permitted
            printf("%c!\n",ItemCopy.Temp);
            Ptr = ItemCopy.Ptr; //This is okay
            *Ptr = 'M'; //This is okay, but it re-assigns ItemCopy.Temp. What?
            printf("%c!\n",ItemCopy.Temp);
        }
};

int main()
{
    TestA Temp1,Temp2;

    Temp1.Function();
    Temp2 = Temp1;
}

Produces the following:

B
M

Even though ItemCopy is const. Why am I permitted to indirectly modify it or even take a non-const copy of the pointer?

Answer 1

Because ItemCopy is const, ItemCopy.Ptr has and effective type of char * const. The pointer is const but the item pointed to can be modified. This means that the assignment:

*ItemCopy.Ptr = 'M';

is meaningful and allowed (the underlying object is not itself const), it is also legal to copy the pointer and assign through it as you have done. A direct assignment ItemCopy.Temp = 'M' would not be legal but that doesn’t meant that you can’t modify the variable ItemCopy.Temp if there is another non-const access path as you have.