Is array variable a reference in C++?

Question

I am just learning C++ and a little confused about arrays and references. I wrote the following program:

void printArray(int arr[]) {
    cout << arr[0] << arr[1] << arr[2] << endl;
}

int main() {
    int arr[3] = {5, 7, 9};
    int *aPtr = &arr[0];
    cout << aPtr[0] << aPtr[1] << aPtr[2] << endl;
    int *bPtr = arr;
    cout << bPtr[0] << bPtr[1] << bPtr[2] << endl;
    printArray(arr);
}

The output is :

579
579
579

And I have two questions:

Is array in C++ a reference? I mean, can I state that i = &i[0]?
If answer to the first question is yes, can I say that as array is reference we don't need to use & in arguments of the function printArray. I mean, we don't declare the function this way printArray(int &arr[])?

Possible duplicate of [What is array decaying?](https://stackoverflow.com/questions/1461432/what-is-array-decaying) — UnholySheep, May 13 '18 at 16:55
You actually don't have a single reference in that code. You only have pointers and the addressof-operator — UnholySheep, May 13 '18 at 16:56
It's of course good to know about C-style arrays and how/when they decay to pointers, but in the long run I'd advise to *not* use them in your own code. Instead use [std::array](http://en.cppreference.com/w/cpp/container/array) when the size is fixed at compile time or [std::vector](http://en.cppreference.com/w/cpp/container/vector) when you need a dynamically sized array. — Jesper Juhl, May 13 '18 at 17:00
`&i[0]` is a pointer. Pointers and references are different things. References are aliases; new names for an already existing object. — axiac, May 13 '18 at 17:00
@axiac: Actually a reference binds to a memory location, just like a pointer does. There certainly are syntactic differences, but the only semantic difference between a reference and a const pointer is the lifetime extension that occurs when a temporary object (prvalue) is directly bound to a reference. — Ben Voigt, May 13 '18 at 17:25

score 6 · Accepted Answer · answered May 13 '18 at 16:57

No, an array is not a reference in C++. It is an array, the length of which forms part of the type (so for example the type of int[3] is not the same as that of int[42]).

What can be confusing is that C++ inherits from C the strange features that

array function parameters have their type "adjusted" to pointer
array names can "decay" to pointers very easily. That makes it possible to assign an array to a pointer.

Point 1 above means that these two function declarations are completely equivalent:

// two ways do declaring the same function
void foo(int a[42]);
void foo(int* a);

and point 2 means you can call the function passing it an array:

int a[3] = {};
int b[100] = {};

foo(a);
foo(b);

and other funny stuff, for example, the type of expression +a being int*.

Is array variable a reference in C++?

1 Answers1