Display specific row of an array using pointer and addressing techniques - c

Question

Hi I am working on a C program that declares a 2 dimensional integer array with N rows and M columns and loads it with consecutive numbers using conventional array element access methods. N and M are equal to 10. I need display a specific row of the array using this function
void showRow(int *arrayName, int rowNumber, int nColsInRow)
Then I need to use that function to display the entire array.
I have to do the same thing with the showCol function in my code below.
I have both functions in my code, but I am getting confused on how to implement them. I appreciate any help that y'all can give.
int main(void){
int array[10][10];
int i,j;
int *row0, *row1, *row2, *row3, *row4, *row5, *temp;
int add = 0;
for(i = 0; i < 10; i++)
{
for(j = 0; j < 10; j++)
{
array[i][j]= add;
add++;
}
}
for(i = 0; i < 10; i++)
{
for(j = 0; j < 10; j++)
{
printf("%d ", array[i][j]);
}
printf(" \n");
}
void showRow(int *array, int rowNumber, int nColsInRow)
{
row0 = &(array[0][0]);
row1 = &(array[1][0]);
row2 = &(array[2][0]);
row3 = &(array[3][0]);
row4 = &(array[4][0]);
row5 = &(array[5][0]);
row6 = &(array[6][0]);
row7 = &(array[7][0]);
row8 = &(array[8][0]);
row9 = &(array[9][0]);
row10 = &(array[10][0]);
for(j = 0; j < 10; j++)
{
printf(" %d", *(rowNumber + j);
}
//}
void showCol(int *array, int colNumber, int nRowsInCol)
{
col0 = &(array[0][0]);
col1 = &(array[0][1]);
col2 = &(array[0][2]);
col3 = &(array[0][3]);
col4 = &(array[0][4]);
col5 = &(array[0][5]);
col6 = &(array[0][6]);
col7 = &(array[0][7]);
col8 = &(array[0][8]);
col9 = &(array[0][9]);
col10 = &(array[0][10]);
for(j = 0; j < 10; j++)
{
printf(" %d", *(col + j);
}
}
}

Answer 1

I made a change to the showRow function to accept a void pointer. Then inside of the showRow function I cast it back to 2 dim array. After that, you can go back to treating just like you would any other 2 dim array.
void showRow(void *array, int rowNumber, int nColsInRow)
{
int (*a)[10] = (int (*)[10]) array;
int i;
for (i=0; i < nColsInRow; i++) {
printf("%d\n", a[rowNumber][i]);
}
}
int main()
{
int array[10][10];
int i,j;
int add = 0;
for(i = 0; i < 10; i++)
{
for(j = 0; j < 10; j++)
{
array[i][j]= add;
add++;
}
}
showRow (&array, 1, 10);
return 0;
}

Answer 2

Without changing the signature of the functions, try this code:
#include <stdio.h>
#include <stdlib.h>
void showRow(int *array, int rowNumber, int nColsInRow);
void showCol(int *array, int colNumber, int nRowsInCol);
int
main(void)
{
int array[10][10];
int i,j;
int *p;
int add = 0;
p = (int *)array; //Set a pointer point to this array.
for(i = 0; i < 10; i++)
{
for(j = 0; j < 10; j++)
{
array[i][j]= add;
add++;
}
}
for(i = 0; i < 10; i++)
{
for(j = 0; j < 10; j++)
{
printf("%d ", array[i][j]);
}
printf(" \n");
}
showRow( p, 2, 3);
showCol( p, 2, 3);
}
void showRow(int *array, int rowNumber, int nColsInRow)
{
int *row = &array[rowNumber*10];//row points at the particular row.
int j;
for(j = 0; j < 10; j++)
{
printf(" %d", row[j]);
}
printf("\n%d\n", row[nColsInRow]);
}
void showCol(int *array, int colNumber, int nRowsInCol)
{
int *col = &array[colNumber]; //col points at the particular col of first row.
int j, n;
for(j = 0; j < 100; j = j + 10) //Increment 10 since row size is 10.
{
printf(" %d", col[j]);
}
n = 10 * nRowsInCol;
printf( "\n%d\n", col[n]);
}
Explanation: An integer array is stored in a continuous memory.
In this case, although it is a 2 dimensional array,
it stored in 100 continuous unit of 4 bytes. (assuming
size of int is 4 in your computer.)
With int pointer, it can point to any index.
e.g. int *p = &array[58];
p points at 58th unit of this array.
Using this concept, now you can do math to get
particular row or column.