I'm practicing pointers and want to substitute pointer operations in place of the arrays to traverse through the elements of the array. I have read numerous articles and cannot grasp this concept. Can someone explain?
Here I made a 2D array and iterated through it using a basic nested for-loop, but want to use pointers;
int test[3][2] = {1,4,2,5,2,8}; for (int i = 0 ; i < 3; i++) { for (int j = 0; j < 2; j++) { printf("%d\n", test[i][j]); } } 5 Answers
int test[3][2] = {{1,4},{2,5},{2,8}}; // Define a pointer to walk the rows of the 2D array. int (*p1)[2] = test; // Define a pointer to walk the columns of each row of the 2D array. int *p2 = NULL; // There are three rows in the 2D array. // p1 has been initialized to point to the first row of the 2D array. // Make sure the iteration stops after the third row of the 2D array. for (; p1 != test+3; ++p1) { // Iterate over each column of the arrays. // p2 is initialized to *p1, which points to the first column. // Iteration must stop after two columns. Hence, the breaking // condition of the loop is when p2 == *p1+2 for (p2 = *p1; p2 != *p1+2; ++p2 ) { printf("%d\n", *p2); } } In some compilers you can also use a single loop, treating a multidimensional array as a one-dimensional array read in row-major order.
This is mentioned in King's C Programming: A Modern Approach (2nd ed., p268).
#include <stdio.h> int main(void) { int test[3][2] = {{1,4},{2,5},{2,8}}, *p; for(p = &test[0][0]; p <= &test[2][1]; p++) { printf("%d\n", *p); } return 0; } Try the following and investigate
#include <stdio.h> int main(void) { int test[3][2] = { { 1,4 }, { 2,5 }, { 2,8 } }; for ( int ( *p )[2] = test ; p != test + 3; ++p ) { for ( int *q = *p; q != *p + 2; ++q ) printf( "%d ", *q ); puts( "" ); } return 0; } The putput is
1 4 2 5 2 8 The first pointer is a pointer to an object of type int[2] that is it points to the first "row" of the array and then due to increments it points to other rows.. The second pointer is a pointer to an object of type int. It points to the first element of each row in the inner loop.
Treating a 2d array as 1d array is very easy using pointer arithmetic to iterate.
void print_2d_array(int *num, size) { int counter = 0; while (counter++ < size) { printf("%i ", *num); num++; } printf("\n"); } int main() { int matrix[2][3] = {{2, 11, 33}, {9, 8, 77}}; int matrix_size = sizeof(matrix) / sizeof(int); // 24 bytes / 4 (int size) = 6 itens print_2d_array(matrix, matrix_size); return 0; } If pointer declaration is the goal of your practice, use the following initialization:
int (*pTest)[rmax][cmax] = test; Once you do that, the syntax of using pointer indexing mirrors that of array indexing, except that you have to use the * de-referencing operator.
for (int i= 0; i < 3; i++) { for (int j= 0; j < 2; j++) { printf ("%d ", *(pTest[i][j])); } printf ("\n"); } However, if pointer arithmetic is the goal of your practice, then the following will work too:
int *res = &test; for (int i = 0; i < 3; i++) { for (int j = 0; j < 2; j++) { printf ("%d ", *(res + i*2 + j)); } printf ("\n"); } OUTPUT
1 4 2 5 2 8 
