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Pointer Arithmetics in C with Examples

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  • Difficulty Level : Medium
  • Last Updated : 14 Nov, 2022
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Pointers variables are also known as address data types because they are used to store the address of another variable. The address is the memory location that is assigned to the variable. It doesn’t store any value. 

Hence, there are only a few operations that are allowed to perform on Pointers in C language. The operations are slightly different from the ones that we generally use for mathematical calculations. The operations are: 

  1. Increment/Decrement of a Pointer
  2. Addition of integer to a pointer
  3. Subtraction of integer to a pointer
  4. Subtracting two pointers of the same type
  5. Comparison of pointers of the same type.

Increment/Decrement of a Pointer

Increment: It is a condition that also comes under addition. When a pointer is incremented, it actually increments by the number equal to the size of the data type for which it is a pointer. 

For Example: 
If an integer pointer that stores address 1000 is incremented, then it will increment by 4(size of an int) and the new address it will points to 1004. While if a float type pointer is incremented then it will increment by 4(size of a float) and the new address will be 1004.

Decrement: It is a condition that also comes under subtraction. When a pointer is decremented, it actually decrements by the number equal to the size of the data type for which it is a pointer. 

For Example: 
If an integer pointer that stores address 1000 is decremented, then it will decrement by 4(size of an int) and the new address it will points to 996. While if a float type pointer is decremented then it will decrement by 4(size of a float) and the new address will be 996.

Below is the program to illustrate pointer increment/decrement: 

Pointers can be outputted using %p, since, most of the computers store the address value in hexadecimal form using %p gives the value in that form. But for simplicity and understanding we can also use %u to get the value in Unsigned int form.

C




#include <stdio.h>
// pointer increment and decrement
//pointers are incremented and decremented by the size of the data type they point to
int main()
{
    int a = 22;
    int *p = &a;
    printf("p = %u\n", p); // p = 6422288
    p++;
    printf("p++ = %u\n", p); //p++ = 6422292    +4   // 4 bytes
    p--;
    printf("p-- = %u\n", p); //p-- = 6422288     -4   // restored to original value
 
    float b = 22.22;
    float *q = &b;
    printf("q = %u\n", q);  //q = 6422284
    q++;
    printf("q++ = %u\n", q); //q++ = 6422288      +4   // 4 bytes
    q--;
    printf("q-- = %u\n", q); //q-- = 6422284       -4  // restored to original value
 
    char c = 'a';
    char *r = &c;
    printf("r = %u\n", r);   //r = 6422283
    r++;
    printf("r++ = %u\n", r);   //r++ = 6422284     +1   // 1 byte
    r--;
    printf("r-- = %u\n", r);   //r-- = 6422283     -1  // restored to original value
 
    return 0;
}

Output

p = 1441900792
p++ = 1441900796
p-- = 1441900792
q = 1441900796
q++ = 1441900800
q-- = 1441900796
r = 1441900791
r++ = 1441900792
r-- = 1441900791

Addition

When a pointer is added with a value, the value is first multiplied by the size of data type and then added to the pointer.

C




// C program to illustrate pointer Addition
#include <stdio.h>
 
// Driver Code
int main()
{
    // Integer variable
    int N = 4;
 
    // Pointer to an integer
    int *ptr1, *ptr2;
 
    // Pointer stores the address of N
    ptr1 = &N;
    ptr2 = &N;
 
    printf("Pointer ptr2 before Addition: ");
    printf("%p \n", ptr2);
 
    // Addition of 3 to ptr2
    ptr2 = ptr2 + 3;
    printf("Pointer ptr2 after Addition: ");
    printf("%p \n", ptr2);
 
    return 0;
}

Output

Pointer ptr2 before Addition: 0x7ffca373da9c 
Pointer ptr2 after Addition: 0x7ffca373daa8 

Subtraction

When a pointer is subtracted with a value, the value is first multiplied by the size of the data type and then subtracted from the pointer.

Below is the program to illustrate pointer Subtraction:

C




// C program to illustrate pointer Subtraction
#include <stdio.h>
 
// Driver Code
int main()
{
    // Integer variable
    int N = 4;
 
    // Pointer to an integer
    int *ptr1, *ptr2;
 
    // Pointer stores the address of N
    ptr1 = &N;
    ptr2 = &N;
 
    printf("Pointer ptr2 before Subtraction: ");
    printf("%p \n", ptr2);
 
    // Subtraction of 3 to ptr2
    ptr2 = ptr2 - 3;
    printf("Pointer ptr2 after Subtraction: ");
    printf("%p \n", ptr2);
 
    return 0;
}

Output

Pointer ptr2 before Subtraction: 0x7ffd718ffebc 
Pointer ptr2 after Subtraction: 0x7ffd718ffeb0 

Subtraction of Two Pointers

The subtraction of two pointers is possible only when they have the same data type. The result is generated by calculating the difference between the addresses of the two pointers and calculating how many bits of data it is according to the pointer data type. The subtraction of two pointers gives the increments between the two pointers. 

For Example: 
Two integer pointers say ptr1(address:1000) and ptr2(address:1004) are subtracted. The difference between address is 4 bytes. Since the size of int is 4 bytes, therefore the increment between ptr1 and ptr2 is given by (4/4) = 1.

Below is the implementation to illustrate the Subtraction of Two Pointers:

C




// C program to illustrate Subtraction
// of two pointers
#include <stdio.h>
 
// Driver Code
int main()
{
    int x = 6;   // Integer variable declaration
    int N = 4;
 
    // Pointer declaration
    int *ptr1, *ptr2;
 
    ptr1 = &N;  // stores address of N
    ptr2 = &x;   // stores address of x
   
    printf(" ptr1 = %u, ptr2 = %u\n", ptr1, ptr2); 
  // %p gives an hexa-decimal value,
  //We convert it into an unsigned int value by using %u
 
    // Subtraction of ptr2 and ptr1
    x = ptr1 - ptr2;
 
    // Print x to get the Increment
    // between ptr1 and ptr2
    printf("Subtraction of ptr1 "
           "& ptr2 is %d\n",
           x);
 
    return 0;
}

Output

 ptr1 = 2715594428, ptr2 = 2715594424
Subtraction of ptr1 & ptr2 is 1

Pointer Arithmetic on Arrays: 
Pointers contain addresses. Adding two addresses makes no sense because there is no idea what it would point to. Subtracting two addresses lets you compute the offset between the two addresses. An array name acts like a pointer constant. The value of this pointer constant is the address of the first element. For Example: if an array named arr then arr and &arr[0] can be used to reference array as a pointer.

Below is the program to illustrate the Pointer Arithmetic on arrays:

Program 1: 

C




// C program to illustrate the array
// traversal using pointers
#include <stdio.h>
 
// Driver Code
int main()
{
 
    int N = 5;
 
    // An array
    int arr[] = { 1, 2, 3, 4, 5 };
 
    // Declare pointer variable
    int* ptr;
 
    // Point the pointer to first
    // element in array arr[]
    ptr = arr;
 
    // Traverse array using ptr
    for (int i = 0; i < N; i++) {
 
        // Print element at which
        // ptr points
        printf("%d ", ptr[0]);
        ptr++;
    }
}

Output

1 2 3 4 5 

Program 2: 

C




// C program to illustrate the array
// traversal using pointers in 2D array
#include <stdio.h>
 
// Function to traverse 2D array
// using pointers
void traverseArr(int* arr,
                 int N, int M)
{
 
    int i, j;
 
    // Traverse rows of 2D matrix
    for (i = 0; i < N; i++) {
 
        // Traverse columns of 2D matrix
        for (j = 0; j < M; j++) {
 
            // Print the element
            printf("%d ", *((arr + i * M) + j));
        }
        printf("\n");
    }
}
 
// Driver Code
int main()
{
 
    int N = 3, M = 2;
 
    // A 2D array
    int arr[][2] = { { 1, 2 },
                     { 3, 4 },
                     { 5, 6 } };
 
    // Function Call
    traverseArr((int*)arr, N, M);
    return 0;
}

Output

1 2 
3 4 
5 6 

Comparison of pointers of the same type:

We can compare the two pointers by using the comparison operators in C. We can implement this by using all operators in C >, >=, <, <=, ==, !=.  It returns true for the valid condition and returns false for the unsatisfied condition. 

Step 1 : Initialize the integer values and point these integer values to the pointer.

Step 2 : Now, check the condition by using comparison or relational operators on pointer variables.

Step 3 : Display the output.

C




#include <stdio.h>
 
int main() {
 
    // code
    int num1=5,num2=6,num3=5; //integer input
    int *p1=&num1;// addressing the integer input to pointer
    int *p2=&num2;
    int *p3=&num3;
    //comparing the pointer variables.
   if(*p1<*p2)
   {
       printf("\n%d less than %d",*p1,*p2);
   }
   if(*p2>*p1)
   {
     printf("\n%d greater than %d",*p2,*p1);
   }
   if(*p3==*p1)
   {
      printf("\nBoth the values are equal");
   }
   if(*p3!=*p2)
   {
      printf("\nBoth the values are not equal");
   }
   
 
    return 0;
}

Output

5 less than 6
6 greater than 5
Both the values are equal
Both the values are not equal

Comparison operators on Pointers using array :

In the below approach, it results the count of odd numbers and even numbers in an array. We are going to implement this by using pointer.

Step 1 :First, declare the length of an array and array elements. 

Step 2 :Declare the pointer variable and point it to the first element of an array. 

Step 3:Initialize the count_even and count_odd. Iterate the for loop and check the conditions for number of odd elements and even elements in an array,

Step 4: Increment the pointer location ptr++ to the next element in an array for further iteration.

Step 5 : Print the result.

C




#include <stdio.h>
 
int main()
{
    int n = 10; // length of an array
 
    int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
    int* ptr; // Declaration of pointer variable
 
    ptr = arr; // Pointer points the first (0th index)
               // element in an array
    int count_even = 0;
    int count_odd = 0;
 
    for (int i = 0; i < n; i++) {
 
        if (*ptr % 2 == 0) {
            count_even++;
        }
        if (*ptr % 2 != 0) {
            count_odd++;
        }
        ptr++; // Pointing to the next element in an array
    }
    printf("\n No of even elements in an array is : %d",
           count_even);
    printf("\n No of odd elements in an array is : %d",
           count_odd);
}

Output

 No of even elements in an array is : 5
 No of odd elements in an array is : 5

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