Pointer confusion c++

Question

I have the following practical work to complete to learn c++, I've been spending a long time looking for an answer and reading through all my work but I'm a bit confused could I get clarification of where I am going wrong here. The code looks like below:

SafeArray& SafeArray::operator =(const SafeArray& other) {
//Add code here
return *this;
}

We have to implement code that assigns one array to another. Now if I understand that code correctly the method takes a formal parameter of a constant "SafeArray" called other. The & after the SafeArray means that the actual array itself is passed and not a copy while const means that it can not be changed. Similarly the method returns the actual array itself and not a copy of it.

As such I thought that this would be a simple case of creating a pointer, referencing the memory location of "other" and then returning the de-referenced result. In my attempts to actually code this thought I've had no luck.

I tried doing something like the following:

SafeArray* ptr = &other; //This should have created a SafeArray pointer to the memory location of the array "other".

The problem here is that I get the error:

main.cpp:31:23: error: invalid conversion from ‘const SafeArray*’ to ‘SafeArray*’ [-fpermissive]

I guess the reason for this is that I am trying to convert something I'm not allowed to alter into something I can.

I can write the code like this:

const SafeArray* = &other;

But then I can't return the value properly either. I'm clearly misunderstanding something here theoretically I presume, could someone please explain what I am missing. I'm more than happy to work the coding out myself but I can't quite grasp this method call.

Answer 1

To correctly define this operator, you simply have to copy the contents of your SafeArray object other passed as parameter into this .

The only way to know how to complete this function is to understand the contents, ie data members, of the SafeArray class.