implicit pointers?

[DonnaPin]

Could someone please explain what an implicit pointer is, The book I am learning from (currently learning about returning references) says "When a function returns a reference, it returns a implicit pointer to its return value". I've tried looking this up but all I can find is stuff that either I dont think has anything to do with it, or it seems like it might be too advanced so I do not understand what it is talking about.

[helios]

It sounds like the author misspoke. There's no such thing as an "implicit pointer". What the author meant was that references are more or less equivalent to pointers:

int &f(bool x){
    static int a = 1;
    static int b = 2;
    if (x)
        return a;
    return b;
}

which is functionally equivalent to

int *f(bool x){
    static int a = 1;
    static int b = 2;
    if (x)
        return &a;
    return &b;
}

[mbozzi]

"implicitly de-referenced pointers" are the basis of an awful analogy used to (badly) teach references.

[againtry]

"When a function returns a reference, it returns a (sic) implicit pointer to its return value".

It appears this quote comes from
https://tutorialink.com/cpp/return-by-reference.cpp

The statement and indeed the example are almost meaningless, definitely useless.

If there is an implicit pointer, it is 'this', hardly a sufficiently important revelation for anyone learning c++ to get side-tracked on.

[dhayden]

It's misleading to think of a reference as an implicit pointer. References are sometimes implemented with a pointer, but a reference is not a pointer.

A reference is a synonym for an existing object. It lets you have two names for the same variable.

int x;
int &y{x);   // x and y refer to the same int 

You can't make a reference refer to a different object, just like you can't make a variable x refer to a different object either.