[c++] C++ pass an array by reference

is this allowed to pass an array by reference ?

 void foo(double& *bar) 

Seems that my compiler says no. Why? What is the proper way to pass an array by reference? Or a work around? I have an array argument that my method should modify and that I should retrieve afterwards. Alternatively, I could make this array a class member, which works fine, but it has many drawbacks for other part of my code (that I would like to avoid).

Thanks and regards.

As you are using C++, the obligatory suggestion that's still missing here, is to use std::vector<double>.

You can easily pass it by reference:

void foo(std::vector<double>& bar) {}

And if you have C++11 support, also have a look at std::array.

For reference:

• http://de.cppreference.com/w/cpp/container/vector
• http://de.cppreference.com/w/cpp/container/array

Like the other answer says, put the & after the *.

This brings up an interesting point that can be confusing sometimes: types should be read from right to left. For example, this is (starting from the rightmost *) a pointer to a constant pointer to an int.

int * const *x;

What you wrote would therefore be a pointer to a reference, which is not possible.

8.3.5.8 If the type of a parameter includes a type of the form “pointer to array of unknown bound of T” or “reference to array of unknown bound of T,” the program is ill-formed

Yes, but when argument matching for a reference, the implicit array to pointer isn't automatic, so you need something like:

void foo( double (&array)[42] );

or

void foo( double (&array)[] );

Be aware, however, that when matching, double [42] and double [] are distinct types. If you have an array of an unknown dimension, it will match the second, but not the first, and if you have an array with 42 elements, it will match the first but not the second. (The latter is, IMHO, very counter-intuitive.)

In the second case, you'll also have to pass the dimension, since there's no way to recover it once you're inside the function.

If you want to modify just the elements:

void foo(double *bar);

is enough.

If you want to modify the address to (e.g.: realloc), but it doesn't work for arrays:

void foo(double *&bar);

is the correct form.

Here, Erik explains every way pass an array by reference: https://stackoverflow.com/a/5724184/5090928.

Similarly, you can create an array reference variable like so:

int arr1[] = {1, 2, 3, 4, 5};
int(&arr2)[5] = arr1;