Question is make an array of 10 integers that's fine
int array[10];
Now question is
how to make a reference to an array which I have declared above ?
I tried this
int &ra = a;
But it's giving me error...
Please provide me details about this error and how to make reference of an array.
int (&ra)[10] = a;
Alternatively, you can use a typedef to separate this into the type for "array of 10 ints" and having a reference there-to, as in:
typedef int int10[10];
int10& my_ref = a;
The problem with your int &ra = a; is that it tells the compiler to create a reference of type int that refers to an array of 10 ints... they're just not the same thing. Consider that sizeof(int) is a tenth of the size of an array of ten ints - they occupy different amounts of memory. What you've asked for with the reference's type could be satisfied by a particular integer, as in int& ra = a[0];.
I appreciate it's a bit confusing that int* p = a; is allowed - for compatibility with the less type-safe C, pointers can be used to access single elements or arrays, despite not preserving any information about the array size. That's one reason to prefer references - they add a little safety and functionality over pointers.
For examples of increased functionality, you can take sizeof my_ref and get the number of bytes in the int array (10 * sizeof(int)), whereas sizeof p would give you the size of the pointer (sizeof(int*)) and sizeof *p == sizeof(int). And you can have code like this that "captures" the array dimension for use within a function:
template <int N>
void f(int (&x)[N])
{
std::cout << "I know this array has " << N << " elements\n";
}
int& not const int& and it knows you cannot bind a temporary to a non-const reference. So that is the error you get. Try changing int& to const int& and you'll see you get another error, perhaps a more helpful one (but perhaps not).
const reference to a const variable without using a const_cast<>. That would defeat the protection const is designed to provide. You can bind a const reference to a non-const variable though.The reference to array will have type int (&a)[10].
int array[10];
int (&a)[10] = array;
Sometimes it might be useful to simplify things a little bit using typedef
typedef int (&ArrayRef)[10];
...
ArrayRef a = array;
This is a reference to an array of of ints of size 10:
int (&ra)[10];
so
int (&ra)[10] = a;
int &ra is a reference to an int for starters. Even after many years I still find some C declaration syntax confusing. I would use std::array<int> or std::tr1::array<int> instead.You can typedef the array type (the type should not be incomplete) as follow:
#define LEN 10
typedef int (array)[LEN];
int main(void)
{
array arr = {1, 2, 3, 4, 5}; //define int arr[10] = {1, 2, 3, 4, 5};
array arr2; //declare int arr[10];
array *arrptr = &arr; // pointer to array: int (*arrptr)[10] = &arr;
array &&arrvef; // declare rvalue reference to array of 10 ints
array &ref = arr; // define reference to arr: int (&ref)[10] = arr;
}
