Trying to get my bearings with *pointers and &references, and how they work together

[SlowCoder]

New member, and new to C++!

1. Do pointers always point to a reference?
int b = 1; int *a = &b; vs int b = 1; int *a = b;

2. If any, what is the functional difference between int* a and int *a

I'm sure I'll have a lot more questions ...

[helios]

1. Don't confuse the address-of operator with a reference declaration. This:

int b = 1;
int *a = &b;

and this:

int b = 1;
int &a = b;

use the same character (ampersand), but they mean very different things. The fact that the same character is used is completely incidental. It's just how the syntax is. The syntax could also have been

int b = 1;
pointer<int> a = addressof(b);
reference<int> c = b;

2. There's no difference.

[George P]

1. No. int* a = new int(); assigns a with the address of a memory location on the heap/free store.

FYI, the & in your code snippet is the address-of operator, not a reference.
https://www.learncpp.com/cpp-tutorial/introduction-to-pointers/
https://www.learncpp.com/cpp-tutorial/references/

Yes, the same symbol has two different uses. The context determines which is which.

2. There is no difference. The location doesn't matter to the compiler, the whitespace is for us "hoo-mans." (See my first link, Introduction to pointers).

The online tutorial I linked is a good place to learn C++. Poke around, chances are reading the lessons will answer a lot of questions and generate more for ya.

[jonnin]

int b = 1; int *a = b;
this is a subtle error. if it works (it should at least throw a warning, but it may depend on compiler flags, I forget, I have not tried to do anything like that in a very long time!) ... anyway if it works it says that a is a pointer that will touch location 1. location 1 is almost certainly not valid memory for your program (most computers, the operating system claims the 'low' order memory areas simply because they are typically allocated first, -- you need memory? 100% of it is available, so I give you location 0..) and if you tried to access it (*a = 42;) it should crash. Modern OS prevent accessing memory that your program is not marked as owning.

pointers and references are very similar. Consider this..
- a variable is a 'name' (eg int x, x is the name) in your code for a block of 1 or more bytes somewhere (eg 4 bytes on the stack for a typical int x).
- a reference can put a new name on a variable. Now you have 2 names that can refer to the same block of bytes.
- a pointer gives you the location of a block of bytes. It can GET to that block and modify it, so it is 'indirectly' a new name for the same data, but you have to 'dereference' it to get there. One way to think of a pointer is that if your computer's memory is a giant array of bytes, a pointer holds the INDEX into that array, just like you can say int i = 3; arr[i] = 11; //i indirectly gets you to a useful location... its not exactly the same, but conceptually, its very close.
- a pointer can refer to a block of bytes that never existed before (the new keyword). A reference MUST refer to an existing thing.