C : Processing the columns of a 2D array - c

Question

#include<stdio.h>
#define NUM_ROWS 3
#define NUM_COLS 5
int main(void){
int a[NUM_ROWS][NUM_COLS],(*p)[NUM_COLS], i;
for (p = &a[0],i=0; p < &a[NUM_ROWS]; p++,i++){
(*p)[i]=i;
}
printf("The value of a[0][0] is %d\n",a[0][0]); // I want 0
printf("The value of a[0][1] is %d\n",a[0][1]); // 1
printf("The value of a[0][2] is %d\n",a[0][2]); // 2
printf("The value of a[0][3] is %d\n",a[0][3]); // 3
printf("The value of a[0][4] is %d\n",a[0][4]); // 4
return 0;
}
Hi guys I'm a C novice, and I am trying to understand processing the columns of a 2D array.
I wanted output of 0,1,2,3,4 from row 0's columns but I had these results
The value of a[0][0] is 0
The value of a[0][1] is 0
The value of a[0][2] is 1
The value of a[0][3] is 0
The value of a[0][4] is -1
I tried to find what was wrong, but I failed to....
I will be grateful if someone explains what is wrong with my codes..

Answer 1

Your assignment in the loop is initializing the leading diagonal:
(*p)[i] = i;
To illustrate, here's an adaptation of your code that prints the whole matrix (and initializes it):
#include <stdio.h>
#include <string.h>
#define NUM_ROWS 3
#define NUM_COLS 5
int main(void)
{
int a[NUM_ROWS][NUM_COLS], (*p)[NUM_COLS], i;
/* Set all elements to -1 assuming 2's complement */
memset(a, 0xFF, sizeof(a));
for (p = &a[0], i = 0; p < &a[NUM_ROWS]; p++, i++)
{
(*p)[i] = i;
}
for (i = 0; i < NUM_ROWS; i++)
{
for (int j = 0; j < NUM_COLS; j++)
printf("%3d", a[i][j]);
putchar('\n');
}
return 0;
}
The output is:
0 -1 -1 -1 -1
-1 1 -1 -1 -1
-1 -1 2 -1 -1
Notice that the three elements on the leading diagonal are set to 0, 1, 2 and the rest are -1 as set by memset().
If you want to initialize the first row, then you simply use:
int (*p)[NUM_COLS] = &a[0];
for (int i = 0; i < NUM_COLS; i++)
(*p)[i] = i;
Or, more simply still, forget about p and use:
for (int i = 0; i < NUM_COLS; i++)
a[0][i] = i;
If you want to initialize column 0, you need:
(*p)[0] = i;
Or, again, more simply, forget about p and use:
for (int i = 0; i < NUM_ROWS; i++)
a[0][i] = i;

Answer 2

I think you want to write your for loop like this (based on you captions)
for the first row.
for (p = &a[0], i=0; i < NUM_COLS; i++){
(*p)[i] = i;
}
Here increasing the pointer p doesn't help for it will be incremented by a value of NUM_COLS every time due to pointer arithmetic.
Here is a second possible answer if you want to write and read the whole matrix within a single for loop
for (p = &a[0], i = 0; i < NUM_COLS * NUM_ROWS; i++) {
(*p)[i] = i;
}
Or the better solution would be to use two for loops as
for (p = &a[0], i = 0; p < &a[NUM_ROWS] ; p ++) {
int k = 0; //for the column count
for(;k < NUM_COLS; i++, k++) {
(*p)[k] = i;
}
}
Thanks for asking :)

Answer 3

for (p = &a[0]; p < &a[NUM_ROWS]; p++){
for(i=0;i<NUM_COLS;i++){
(*p)[i]=i;
printf("%d\n",(*p)[i]);
// I got 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4
}
}
Should I write nested loop to print out all elements of 2D?
Is there a way to print out using 1 loop with processing columns of 2D array?