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How to dynamically allocate a 2D array in C?
A 2D array can be dynamically allocated in C using a single pointer. This means that a memory block of size row*column*dataTypeSize is allocated using malloc and pointer arithmetic can be used to access the matrix elements.
A program that demonstrates this is given as follows.
Example
#include <stdio.h>
#include <stdlib.h>
int main() {
int row = 2, col = 3;
int *arr = (int *)malloc(row * col * sizeof(int));
int i, j;
for (i = 0; i < row; i++)
for (j = 0; j < col; j++)
*(arr + i*col + j) = i + j;
printf("The matrix elements are:
");
for (i = 0; i < row; i++) {
for (j = 0; j < col; j++) {
printf("%d ", *(arr + i*col + j));
}
printf("
");
}
free(arr);
return 0;
}
The output of the above program is as follows.
The matrix elements are: 0 1 2 1 2 3
Now let us understand the above program.
The 2-D array arr is dynamically allocated using malloc. Then the 2-D array is initialized using a nested for loop and pointer arithmetic. The code snippet that shows this is as follows.
int row = 2, col = 3;
int *arr = (int *)malloc(row * col * sizeof(int));
int i, j;
for (i = 0; i < row; i++)
for (j = 0; j < col; j++)
*(arr + i*col + j) = i + j;
Then the values of the 2-D array are displayed. Finally the dynamically allocated memory is freed using free. The code snippet that shows this is as follows.
printf("The matrix elements are:
");
for (i = 0; i < row; i++) {
for (j = 0; j < col; j++) {
printf("%d ", *(arr + i*col + j));
}
printf("
");
}
free(arr);
Updated on: 30-Jul-2019
15K+ Views
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