Question

As I read here: http://www.cplusplus.com/reference/stl/array/operator%5B%5D/

it appears that saying a[2] would return the memory address (a reference) of the second element of a.

So how is

a[2]=5

a valid assignment, as that would mean I change the memory address of a[2] to location 5 (that might be possible, but usually you want to change the value, not the address) . Unless the = operator knows how to deal with this situation.

I know that it doesn’t change the memory address, so what’s actually going on here?

Answer 1

The difference between a reference and a pointer is that a reference is dereferenced automagically. Hence you don’t need such things as *(a[2]) = 5.

The following code shows this:

int baseVar = 42;            // This
int &sameVar = baseVar;      //   and this are the same memory
                             //   with two different names.
int *pBaseVar = &baseVar;    // This is separate memory that happens
                             //   to point to the baseVar memory.

Changing either of sameVar or *pBaseVar will change baseVar itself. Changing pBasevar itself will not affect basevar, it will simply cause to former to point to a different location.

Under the covers (though this is, of course, implementation dependent, basevar is probably considered (by the compiler/code) to be the int at a specific address (let’s say 0x12345678), sameVar is considered that, too.

pBaseVar is considered a pointer at (for example) 0x11112222 which happens to contain the value 0x12345678:

                      +------------+
pBaseVar (0x11112222) | 0x12345678 |--+
                      +------------+  |
   +----------------------------------+
   |
   V                  +----+
baseVar (0x12345678)  | 42 |
sameVar (same)        |    |
                      +----+