Access Array Elements Using Pointers

#include <stdio.h>
int main() {
    int data[5];

    printf("Enter elements: ");
    for (int i = 0; i < 5; ++i)
        scanf("%d", data + i);

    printf("You entered: 
");
    for (int i = 0; i < 5; ++i)
        printf("%d
", *(data + i));
    return 0;
}

Output

Enter elements: 1
2
3
5
4
You entered: 
1
2
3
5
4

In this program, the elements are stored in the integer array data[].

Then, the elements of the array are accessed using the pointer notation. By the way,

data[0] is equivalent to *data and &data[0] is equivalent to data
data[1] is equivalent to *(data + 1) and &data[1] is equivalent to data + 1
data[2] is equivalent to *(data + 2) and &data[2] is equivalent to data + 1
...
data[i] is equivalent to *(data + i) and &data[i] is equivalent to data + i

Visit this page to learn about the relationship between pointers and arrays.

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Returning a Pointer from a Function in C

Last updated on July 27, 2020

We have already seen a function can return data of types int , float, char etc. Similarly, a function can return a pointer to data. The syntax of a function returning a pointer is as follows.

Syntax: type *function_name(type1, type2, ...);

Some examples:

int *func(int, int); // this function returns a pointer to int

double *func(int, int); // this function returns a pointer to double

The following program demonstrates how to return a pointer from a function.

#include<stdio.h>
int *return_pointer(int *, int); // this function returns a pointer of type int

int main()
{
    int i, *ptr;
    int arr[] = {11, 22, 33, 44, 55};
    i = 4;

    printf("Address of arr = %u
", arr);

    ptr = return_pointer(arr, i);

    printf("
After incrementing arr by 4 

");

    printf("Address of ptr = %u

" , ptr);
    printf("Value at %u is %d
", ptr, *ptr);

    // signal to operating system program ran fine
    return 0;
}

int *return_pointer(int *p, int n)
{
    p = p + n;
    return p;
}

Expected Output:

Address of arr = 2686736

After incrementing arr by 4

Address of ptr = 2686752

Value at 2686752 is 55

How it works:

Since the name of an array is a pointer to the 0th element of the array. Here we are passing two arguments to the function return_pointer(). The arr is passed using call by reference (notice that name of the array is not preceded by & operator because the name of the array is a constant pointer to the 0th element of the 1-D array) and i is passed using call by value. Inside the function pointer p is incremented by n and reassigned to p. Finally, the pointer p is returned to the main() function and reassigned to ptr.

Never return a pointer to local variable from a function.

Consider the following code.

#include<stdio.h>
int *abc(); // this function returns a pointer of type int

int main()
{
    int *ptr;
    ptr = abc();
    return 0;
}

int *abc()
{
    int x = 100, *p;
    p = &x;
    return p;
}

Can you point out the problem with above code?

In the function abc() we are returning a pointer to the local variable. Recall that a local variable exists only inside the function and as soon as function ends the variable x cease to exists, so the pointer to it is only valid inside the function abc().

Even though the address returned by the abc() is assigned to ptr inside main(), the variable to which ptr points is no longer available. On dereference the ptr you will get some garbage value.

c语言学习（4）

Returning more than one value from function in C

C Program to Access Array Elements Using Pointer

In this example, you will learn to access elements of an array using a pointer.

Access Array Elements Using Pointers

Returning a Pointer from a Function in C