Thread: simple class

If I remember correctly the name of the array is a constant reference to the array, aka a constant pointer. Thus to return an array just return a pointer of the same type, aka

Code:

#include <iostream>

using std::cin;
using std::cout;
using std::endl;

int* getArray(void);

int myArr[3] = {0, 1, 2};

int main() {

	int* myPointer;

	myPointer = getArray();

	for(int i = 0; i < 3; i++) {
		cout << myPointer[i]<< endl;
	}

	cin.get();
	return 0;
}
int* getArray(void) {

	return myArr;
}

To be nitpicky:

>>the name of the array is a constant reference to the array, aka a constant pointer.
Arrays can often be represented as pointers, but an array is NOT a pointer; and an array name is NOT a 'constant reference to the array'.

Example:

Code:

int array[256];  //Array
int* pointer = array;  //Pointer that's pretending to be an array

typedef int t[256];  //Typedef for an array of 256 int
const t& x = array;  //Constant reference to an array of 256 int

In most cases you can substitute a pointer for an array or vice versa, but there are subtle differences such as use of sizeof() to determine the length of the array - it'll work with an array but not with a pointer. And, of course, you can't use an array to point at another object

So the solution is, as andyhunter did, you can return the string (character array) as a char* since pointers can act like an arrays anyway. Just be careful, because making any changes to the 'array' that you get from the accessor will also be changed in the actual array; and don't try using sizeof() to get the length of the array either.

You are right -- my bad. What I should have said to be more specific is that the name of the array can be used as the constant reference to the first value in the array.

>>the name of the array can be used as the constant reference to the first value in the array.
The name of the array can be interpreted as a pointer to the 0'th element in the array.