Question

If I want a pointer to point to another variable I make it do so by giving the address to the variable with &.

int foo = 10;
int *bar = &foo;

Now If I follow the same logic as above and instead creates a reference type.

int foo = 10;
int &bar = &foo; 

I would think this should work, but it doesn’t. Why?

Answer 1

Because a pointer and a reference are not the same thing.
You can think of a reference as just meaning “another name for” or “alias”.

In other words, bar is just another name for foo in your example.
When you do int &bar = foo;, you want bar to be another name for foo, you don’t want to assign the address of foo to bar. However, pointers store the address of the object it points at, hence with a pointer you need the address-of operator to get the address of foo and assign that to the bar pointer.