Why Does C Treat Array Parameters as Pointers?

In C, array parameters are treated as pointers mainly to,

• To increase the efficiency of code
• To save time

It is inefficient to copy the array data in terms of both memory and time; and most of the time, when we pass an array our intention is to just refer to the array we are interested in, not to create a copy of the array.

The following two definitions of fun() look different, but to the compiler, they mean exactly the same thing. 

void fun(int arr[]) { 
    // body
}
// This is valid


void fun(int *arr) { 
    // body
}
// This is valid too

It's preferable to use whichever syntax is more accurate for readability.

Note: If the pointer coming in really is the base address of a whole array, then we should use [ ].

Example: In this example, the array parameters are being used as pointers.

// C Program to demonstrate that C treats array parameters
// as pointers
#include <stdio.h>

void findSum1(int arr[])
{
    int sum = 0;
    for (int i = 0; i < 5; i++)
        sum = sum + arr[i];
    printf("The sum of the array is: %d\n", sum);
}

void findSum2(int* arr)
{
    int sum = 0;
    for (int i = 0; i < 5; i++)
        sum = sum + arr[i];
    printf("\nThe sum of the array is: %d \n", sum);
}

// Driver code
int main()
{
    int arr[5] = { 1, 2, 3, 4, 5 };

    findSum1(arr);
    findSum2(arr);
    return 0;
}

// C++ Program to demonstrate that C treats array parameters
// as pointers
#include <iostream>
using namespace std;

void findSum1(int arr[])
{
    int sum = 0;
    for (int i = 0; i < 5; i++)
        sum = sum + arr[i];
    cout <<"The sum of the array is: " << sum;
}

void findSum2(int* arr)
{
    int sum = 0;
    for (int i = 0; i < 5; i++)
        sum = sum + arr[i];
    cout <<"\nThe sum of the array is: "<< sum;
}

// Driver code
int main()
{
    int arr[5] = { 1, 2, 3, 4, 5 };

    findSum1(arr);
    findSum2(arr);
    return 0;
}

// this code is contributed by shivanisinghss2110

The sum of the array is: 15

The sum of the array is: 15