passingArraytoFunction.cpp

/*

	The snippet demos the method of passing an array to a function and 
	return the array from the function.
	Array is passed by reference into the function, because a pointer is
	passed instead of a clone copy. If the array is modified inside the 
	function, the modifications are applied to the caller's copy. You could declare the
	array parameter as const to prevent the array from being modified inside the function.
	The size of the array is not part of the array parameter, and needs to be passed in 
	another int parameter. Compiler is not able to deduce the array size from the array 
	pointer, and does not perform array bound check.

*/

#include <iostream>

using namespace std;

// Function prototypes
int max(const int arr[], int size);
void replaceByMax(int arr[], int size);
void print(const int arr[], int size);

int main() {
	const int SIZE = 4;
	int numbers[SIZE] = { 11, 22, 33, 22 };
	print(numbers, SIZE);
	cout << max(numbers, SIZE) << endl;
	replaceByMax(numbers, SIZE);
	print(numbers, SIZE);

	system("pause");
}

// Return the maximum value of the given array.
// The array is declared const, and cannot be modified inside the function.
int max(const int arr[], int size) {
	int max = arr[0];
	for (int i = 1; i < size; ++i) {
		if (max < arr[i]) 
				max = arr[i];
	}
	return max;
}

// Replace all elements of the given array by its maximum value
// Array is passed by reference. Modify the caller's copy.
void replaceByMax(int arr[], int size) {
	int maxValue = max(arr, size);
	for (int i = 0; i < size; ++i) {
		arr[i] = maxValue;
	}
}

// Print the array's content
void print(const int arr[], int size) {
	cout << "{";
	for (int i = 0; i < size; ++i) {
		cout << arr[i];
		if (i < size - 1) cout << ",";
	}
	cout << "}" << endl;
}