Pointer arithmetic in C programming
Let’s suppose we created a pointer *ptr is pointing towards an int variable x which address is 2000. Which means ptr = 2000 because it is storing the address of int variable x.
Now if we shift pointer forward then pointer will start pointing 2004 because from 2000 to 2003, 4 bytes are used to store the value of int variable x.
Hence pointer will not point from 2001 to 2003 memory locations in this scenario, it will shift to 2004 memory location.
Same goes with char and others.
Now the question arises “Which operators are used to increment and decrement pointers?”
There are few arithmetic operators like ++, ——, – and + which can be use to shift pointers forward and backward.
We can also use comparison operators with pointers.
Increment a pointer
We can shift pointer forward using two operators those are ++, += and + operators.
In below example we can also use ptr+=1 or ptr = ptr + 1 instade of ptr++.
Both will give same output.
#include <stdio.h>
int main ()
{
int i,nums[]={1,2,3,4,5};
int *ptr = nums;
for(i = 0; i < 5; i++)
{
printf("n nums[%d] = %d -> value : %d",i,ptr,*ptr);
ptr++;
}
return 0;
}
Output :
nums[0] = 6422292 -> value : 1
nums[1] = 6422296 -> value : 2
nums[2] = 6422300 -> value : 3
nums[3] = 6422304 -> value : 4
nums[4] = 6422308 -> value : 5
Decrement a pointer
We can shift pointer backward using two operators those are ——, -= and – operators.
In below example we can also use ptr-=1 or ptr = ptr – 1 instade of ptr——.
Both will give same output.
#include <stdio.h>
int main ()
{
int i,nums[]={1,2,3,4,5};
int *ptr = &nums[4];
for(i = 4; i >= 0; i--)
{
printf("n nums[%d] = %d -> value : %d",i,ptr,*ptr);
ptr--;
}
return 0;
}
Output :
nums[4] = 6422308 -> value : 5
nums[3] = 6422304 -> value : 4
nums[2] = 6422300 -> value : 3
nums[1] = 6422296 -> value : 2
nums[0] = 6422292 -> value : 1
Comparison operator with pointer
We can also use comparison operators to compare value pointer with other variable’s address or with value of another pointer.
#include <stdio.h>
int main ()
{
int i=0,nums[]={1,2,3,4,5};
int *ptr = nums;
while(ptr < &nums[5])
{
printf("n nums[%d] = %d -> value : %d",nums[i],ptr,*ptr);
ptr++;
i++;
}
return 0;
}
Output :
nums[1] = 6422292 -> value : 1
nums[2] = 6422296 -> value : 2
nums[3] = 6422300 -> value : 3
nums[4] = 6422304 -> value : 4
nums[5] = 6422308 -> value : 5
CHECKOUT OTHER RELATED TOPICS
CHECKOUT OTHER QUIZZES
| Quizzes on Operator |
| Easy Quiz on Operators in C part 1 |
| Easy quiz on operators in C part 2 |
| Easy quiz on operators in C part 3 |
| Quizzes on Loops |
| Easy quiz on loops in C part 1 |
| Easy quiz on loops in C part 2 |
| Easy quiz on loops in C part 3 |
| Quizzes on Decision Making Statement |
| Easy Quiz On Decision Making Statements Part 1 |
| Easy Quiz On Decision Making Statements Part 2 |
| Quizzes on String |
| Easy quiz on String in C programming part 1 |
| Quizzes on Array |
| Easy quiz on Array in C programming – part 1 |
| 5 – Best Java program to convert decimal to binary |
| New Quizzes Coming soon.. |
PRACTICALS IN C PROGRAM
