Pointer and Array in C
Pointer and Array in C
Pointer and Array in C are closely related in C. In fact, an array declared as
int A[10];
It can be accessed using its pointer representation. The array name A is a constant pointer to the first element of the array. So A can be considered a const int*. Since A is a constant pointer, A = NULL will be an illegal statement. Arrays and pointers are the same in terms of how they are used to access memory. And, the important difference between them is that a pointer variable can take different addresses as values whereas, in the case of the array it is fixed.
Diagrammatic Representation
From the above example, it is clear that &x[0] is equivalent to x. And, x[0] is equivalent to *.
Similarly,
- Firstly &x[1] is equivalent to x+1 and x[1] is equivalent to *(x+1).
- Then &x[2] is equivalent to x+2 and x[2] is equivalent to *(x+2).
- &x[3] is equivalent to x+3 and x[3] is equivalent to *(x+3).
- &x[i] is equivalent to x+i and x[i] is equivalent to *(x+i).
Program
#include <stdio.h>
int main(){
int i,class[6],sum=0;
printf("Enter 6 numbers:\n");
for(i=0;i<6;++i){
scanf("%d",(class+i)); // (class+i) is equivalent to &class[i]
sum += *(class+i); // *(class+i) is equivalent to class[i]
}
printf("Sum=%d",sum);
return 0;
}
Output
Enter 6 numbers: 2 3 4 5 3 4 Sum=21
Pointers and 2D Array
Create pointer for the two-dimensional array
Our pointer will also be of type int because we have created the two-dimensional integer array num.
We assign the address of the first element of the array num to the pointer ptr using the address of & operator.
int *ptr = &num[0][0];
Accessing the elements of the two-dimensional array via a pointer
The two-dimensional array num will save as a continuous block in the memory. So, if we increment the value of ptr by 1 we will move to the next block in the allocated memory.
Accessing the value of the two-dimensional array via pointer using row and column of the array
If we want to get the value at any given row, a column of the array then we can use the value at the address of the * operator and the following formula.
arr[m][n] = *(ptr + (m x no_of_colmns + n))
Here, arr is a two-dimensional array, and m and n denote the mth row and nth column of the array.
ptr holds the address of the first element of the array
no_of_colmns denotes the total number of columns in the row of the array.
Program
#include <stdio.h>
int main(void) {
// 2d array
int num[3][4] = {
{1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12}
};
int
ROWS = 3,
COLS = 4,
i, j;
// pointer
int *ptr = &num[0][0];
// print the element of the array via pointer ptr
for (i = 0; i < ROWS; i++) {
for (j = 0; j < COLS; j++) {
printf("%d ", *(ptr + i * COLS + j));
}
printf("\n");
}
return 0;
}
Output
1 2 3 4 5 6 7 8 9 10 11 12
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