Advantages and Disadvantages of Array in C

An array is a collection of similar types of elements. For example, an integer array holds the elements of int types while a character array holds the elements of char types. Below is the representation of the array:

Though, array got its own set of advantages and disadvantages.

Advantages of Arrays

Below are some advantages of the array:

  • In an array, accessing an element is very easy by using the index number.
  • The search process can be applied to an array easily.
  • 2D Array is used to represent matrices.
  • For any reason a user wishes to store multiple values of similar type then the Array can be used and utilized efficiently.

Disadvantages of Arrays

Now let’s see some disadvantages of the array and how to overcome it:



Array size is fixed: The array is static, which means its size is always fixed. The memory which is allocated to it cannot be increased or decreased. Below is the program for the same:

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// C program to illustrate that the
// array size is fixed
#include <stdio.h>
  
// Driver Code
int main()
{
    int arr[10];
  
    // Assign values to array
    arr[0] = 5;
    arr[5] = 6;
    arr[7] = -9;
  
    // Print array element at index 0
    printf("Element at index 0"
           " is %d\n",
           arr[0]);
  
    // Print array element at index 11
    printf("Element at index 11"
           " is %d",
           arr[11]);
  
    return 0;
}

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Output:

Element at index 0 is 5
Element at index 11 is -1176897384

Explanation: In the above program the array of size 10 is declared and the value is assigned at a particular index. But when the value at index 11 is printed then it prints the garbage value because the array was accessed out of the bound index. In some compiler, it gives error as “Array Index Out Of Bound.”.

How to overcome: To overcome that problem use Dynamic Memory Allocation like malloc(), calloc(). It also helps us to deallocates the memory using the free() method which helps to reduce wastage of memory by releasing it. Below is the program for the same:

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// C program to illustrate the use of
// array using Dynamic Memory Allocation
#include <stdio.h>
#include <stdlib.h>
  
// Driver Code
int main()
{
    // Pointer will hold the base address
    int* ptr;
    int n = 10;
  
    // Dynamically allocates memory
    // using malloc() function
    ptr = (int*)malloc(n * sizeof(int));
  
    // Assign values to the array
    for (int i = 0; i < n; i++) {
        ptr[i] = i + 1;
    }
  
    // Print the array
    printf("The elements are: ");
  
    for (int i = 0; i < n; i++) {
        printf("%d ", ptr[i]);
    }
  
    // Free the dynamically
    // allocated memory
    free(ptr);
  
    return 0;
}

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Output:

The elements are: 1 2 3 4 5 6 7 8 9 10

Array is homogeneous:The array is homogeneous, i.e., only one type of value can be store in the array. For example, if an array type “int“, can only store integer elements and cannot allow the elements of other types such as double, float, char so on. Below is the program for the same:

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// C++ program to illustrate that
// the array is homogeneous
#include <stdio.h>
  
// Driver Code
int main()
{
    // Below declaration will give
    // Compilation Error
    int a[5] = { 0, 1, 2, "string", 9, 4.85 };
  
    return 0;
}

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Output:



100
547.000000
Ram

Output:

Explanation: The above code gives “Compilation Error” as an integer type array is assigned value to a string and float type.

How to overcome: To overcome that problem, the idea is to structure, where it can store non-homogeneous (heterogeneous) value. Below is the program for the same:

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// C program to illustrate the use of
// structure to store heterogeneous
// variables
#include <stdio.h>
  
// Structure students
struct student {
  
    int student_id;
    float marks;
    char name[30];
};
  
// Driver Code
int main()
{
    // Structure variable s1
    struct student s1 = { 100, 547, "Ram" };
  
    // Accessing structure members
    // using structure pointer
    printf("%d\n", s1.student_id);
    printf("%f\n", s1.marks);
  
    for (int i = 0;
         s1.name[i] != '\0'; i++) {
        printf("%c", s1.name[i]);
    }
  
    return 0;
}

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Output:

100
547.000000
Ram

Array is Contiguous blocks of memory: The array stores data in contiguous(one by one) memory location. Below is the representation of the same:

How to overcome: To overcome the sequential access to the array, the idea is to use the Linked list. In a Linked list, the elements are not stored in contiguous memory locations. Below is the representation of the same:

Insertion and deletion are not easy in Array: The operation insertion and deletion over an array are problematic as to insert or delete anywhere in the array, it is necessary to traverse the array and then shift the remaining elements as per the operation. This operation cost is more.



Example: For inserting 22 in 3rd position of the array then below are the steps:

Below is the program to illustrate the same:

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// C Program to insert an element at
// a specific position in an array
#include <stdio.h>
  
// Driver Code
int main()
{
    int arr[100] = { 0 };
    int i, x, pos, n = 10;
  
    // Initial array of size 10
    for (i = 0; i < 10; i++) {
        arr[i] = i + 1;
    }
  
    // Print the original array
    for (i = 0; i < n; i++) {
        printf("%d ", arr[i]);
    }
    printf("\n");
  
    // Element to be inserted
    x = 50;
  
    // Position at which element
    // is to be inserted
    pos = 5;
  
    printf("Array after inserting %d"
           " at position %d\n",
           x, pos);
  
    // Increase the size by 1
    n++;
  
    // Shift elements forward
    for (i = n - 1; i >= pos; i--) {
  
        arr[i] = arr[i - 1];
    }
  
    // Insert x at pos
    arr[pos - 1] = x;
  
    // Print the updated array
    for (i = 0; i < n; i++) {
        printf("%d ", arr[i]);
    }
  
    return 0;
}

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Output:

1 2 3 4 5 6 7 8 9 10 
Array after inserting 50 at position 5
1 2 3 4 50 5 6 7 8 9 10

How to overcome: To overcome the above problem using a Linked List. Below is the representation of the same:

Below is the program to implement the same:

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// C program to insert an element at
// a position using linked list
#include <stdio.h>
#include <stdlib.h>
  
// Structure for the linked list
struct node {
    int data;
    struct node* next;
};
  
// head Node
struct node* head;
  
// Function to insert any element
// at the end of the linked list
int insert_last(int k)
{
    struct node *ptr, *s;
    ptr = (struct node*)
        malloc(sizeof(struct node));
    ptr->data = k;
    ptr->next = NULL;
  
    // If head is NULL
    if (head == NULL) {
        head = ptr;
    }
  
    // Else
    else {
  
        s = head;
  
        // Traverse the LL to last
        while (s->next != NULL) {
            s = s->next;
        }
        s->next = ptr;
    }
}
  
// Function to display linked list
void display()
{
    struct node* s;
  
    // Store the head
    s = head;
  
    // Traverse till s is NULL
    while (s != NULL) {
  
        // Print the data
        printf("%d ", s->data);
        s = s->next;
    }
    printf("\n");
}
  
// Function to insert any element at
// the given position of linked list
int insert_position(int a, int b)
{
    int f = 0, i;
    struct node *ptr, *s;
  
    // Allocate new Node
    ptr = (struct node*)
        malloc(sizeof(struct node));
    ptr->data = a;
  
    // If first position
    if (b == 1) {
        ptr->next = head;
        head = ptr;
    }
  
    // Otherwise
    else {
        s = head;
  
        // Move to (b - 1) position
        for (i = 0; i < b - 2; i++) {
            s = s->next;
        }
  
        // Assign node
        ptr->next = s->next;
        s->next = ptr;
    }
  
    return 0;
}
  
// Driver Code
int main()
{
    // Given Linked List
    insert_last(3);
    insert_last(1);
    insert_last(5);
    insert_last(7);
  
    printf("Current Linked List is: ");
  
    // Display the LL
    display();
  
    // Insert 6 at position 4
    insert_position(6, 4);
    printf("\n Linked List after insert"
           " 6 in 4th position: ");
  
    // Display the LL
    display();
  
    return 0;
}

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Output:

Current Linked List is: 3 1 5 7 

 Linked List after insert 6 in 4th position: 3 1 5 6 7

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