Why Does C Treat Array Parameters as Pointers?
Last Updated :
12 Apr, 2023
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
#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);
}
int main()
{
int arr[5] = { 1, 2, 3, 4, 5 };
findSum1(arr);
findSum2(arr);
return 0;
}
|
C++
#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;
}
int main()
{
int arr[5] = { 1, 2, 3, 4, 5 };
findSum1(arr);
findSum2(arr);
return 0;
}
|
Output
The sum of the array is: 15
The sum of the array is: 15
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