Is C++ Array passed by reference or by pointer?

Question

In school, our lecturer taught us that the entire array was passed by reference when we pass it to a function,.

However, recently I read a book. It

Answer 1

Pass-by-pointer is a bit of a misnomer. It doesn't happen in C++. There is only pass-by-value and pass-by-reference. Pointers in particular are passed by value.

The answer to your question is: it depends.

Consider the following signatures:

void foo(int *arr);
void bar(int *&arr);
void baz(int * const &arr);
void quux(int (&arr)[42]);

Assuming you are passing an array to each of these functions:

• In foo(arr), your array is decayed to a pointer, which is then passed by value.
• In bar(arr), this is a compiler error, because your array would decay to a (temporary) pointer, and this would be passed by reference. This is nearly always a bug, since the reason you would want a mutable reference is to change the value of the referent, and that would not be what would happen (you would change the value of the temporary instead). I add this since this actually does work on some compilers (MSVC++) with a particular extension enabled. If you instead decay the pointer manually, then you can pass that instead (e.g. int *p = arr; bar(p);)
• In baz(arr), your array decays to a temporary pointer, which is passed by (const) reference.
• In quux(arr), your array is passed by reference.

What your book means by them being similar is that passing a pointer by value and passing a reference are usually implemented identically. The difference is purely at the C++ level: with a reference, you do not have the value of the pointer (and hence cannot change it), and it is guaranteed to refer to an actual object (unless you broke your program earlier).