Passing by reference issue?

Question

The problem i am trying to pass array char arr[6] = {"1","2",etc.} to a function that takes parameter like this void foo(char* &arr,...) and it does not work. Can anyone explain it to me please ?

Because an array is not a pointer. And a pointer is not an array — StoryTeller - Unslander Monica, Sep 12 '17 at 14:56
Those questions are answered [here](https://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list/388282). — nwp, Sep 12 '17 at 14:56
Possible duplicate of [Passing an array by reference](https://stackoverflow.com/questions/5724171/passing-an-array-by-reference) — AMA, Sep 12 '17 at 14:56

eerorika · Accepted Answer · 2017-09-12T15:10:42.880

char arr[6] is an array.

char* &arr is a(n lvalue) reference to a pointer.

Now, since the argument is not of correct type, it has to be converted. An array implicitly decays (decaying is a kind of conversion) into a pointer to first element.

But this decayed pointer is a temporary (an rvalue). Non-const lvalue references can not refer to rvalues, so it would be ill-formed to call foo with an array argument.

You can create a pointer variable; that can be passed to foo:

char* ptr = arr;
foo(ptr, ...);

The function may then modify that pointer (i.e. make it point to some other char object), since the reference is non-const.

PS. There is something very wrong with the initialization of your array. "1" and "2" are not char objects.

Passing by reference issue?

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