How to print first and last row in an array using C - c

Question

I have m x m array and i am trying to find out first and last row of a given matrix (using C) which is
e.g mat[3][3]={1,2,3,
4,5,6,
7,8,9};
But i could not figure out how to get first and last row using c. If some have the algorithm then please share with me.

Answer 1

Since you have m*m matrix, so first row will be mat[0][1] to mat[0][m-1] and last row will be mat[m-1][0] to mat[m-1][m-1].
for(int i = 0; i < m; i++) {
printf("%d ", mat[0][i]); //first row
}
for(int i = 0; i < m; i++) {
printf("%d ", mat[m-1][i]); //Last row
}

Answer 2

First you should understand how the matrix is organised in memory.
(Taken from http://www.fredosaurus.com/notes-cpp/arrayptr/23two-dim-array-memory-layout.html)
2 Ways you can use - As a 2D array or as a pointer accessed location. Note that everytime the row index moves, the value being accessed has to move by the number of columns in the matrix. Hence accessing the value using a pointer is more intuitive, but is really subjective.
Case1: If 2D array, and ROWS and COLUMNS are known, use the below to print any row.
//row is the row# to print, ex, 0 or ROWS-1
for i from 0 to COLUMNS-1
output = mat[row][i]
Case2: If you have input mat as a pointer to the first value in the array organised as a 2D matrix, use as below
//row is the row# to print, ex, 0 or ROWS-1
for i from 0 to COLUMNS-1
output = mat[row*COLUMNS + i]

Answer 3

As long as you are printing the matrix in the scope where it is declared, you can use the sizeof operator to print the first and last rows (or you pass that information to a function):
#include <stdio.h>
int main (void) {
int mat[3][3]={{1,2,3}, {4,5,6}, {7,8,9}};
unsigned int i;
/* printf first row */
printf ("array[0] : ");
for (i = 0; i < sizeof *mat/sizeof **mat; i++)
printf (" %d", mat[0][i]);
putchar ('\n');
/* print last row */
printf ("array[%lu] : ", sizeof mat/sizeof *mat - 1);
for (i = 0; i < sizeof *mat/sizeof **mat; i++)
printf (" %d", mat[sizeof mat/sizeof *mat - 1][i]);
putchar ('\n');
return 0;
}
Output
$ ./bin/array_sizeof
array[0] : 1 2 3
array[2] : 7 8 9
4X4 Same Code
#include <stdio.h>
int main (void) {
int mat[][4]={{1,2,3,4}, {3,4,5,6}, {6,7,8,9}, {9,8,7,6}};
unsigned int i;
/* printf first row */
printf ("array[0] : ");
for (i = 0; i < sizeof *mat/sizeof **mat; i++)
printf (" %d", mat[0][i]);
putchar ('\n');
/* print last row */
printf ("array[%lu] : ", sizeof mat/sizeof *mat - 1);
for (i = 0; i < sizeof *mat/sizeof **mat; i++)
printf (" %d", mat[sizeof mat/sizeof *mat - 1][i]);
putchar ('\n');
return 0;
}
Output
$ ./bin/array_sizeof
array[0] : 1 2 3 4
array[3] : 9 8 7 6
Passing to a Function
To print the first and last from a function, you must pass the same information. At a minimum, you will always have to pass (1) a pointer to the array, (2) the type and/or sizeof type, (3) the number of rows, and (4) the number of columns. You may be able to specify the type, size and number of columns by passing a pointer to type (combining information for 1, 2 & 4), but that must be passed in some fashion. If your array is filled dynamically, then you will have your count of rows x columns from counting your input. If you declare a static array, you have the constant. Either way, that is a required part of the information. E.g.:
#define NCOL 4
...
int main (void) {
int mat[][NCOL]={{1,2,3,4}, {3,4,5,6}, {6,7,8,9}, {9,8,7,6}};
...
print_1st_last (mat, sizeof mat/sizeof *mat);
...
}
void print_1st_last (int (*mat)[NCOL], size_t m)
{
size_t i;
printf ("array[0] : ");
for (i = 0; i < NCOL; i++)
printf (" %d", mat[0][i]);
putchar ('\n');
/* print last row */
printf ("array[%zu] : ", m - 1);
for (i = 0; i < NCOL; i++)
printf (" %d", mat[m - 1][i]);
putchar ('\n');
}

Answer 4

Create a function that does this for you
void print_first_last(int **array, int height, int width) {
unsigned v;
for (v = 0; v < width; v++) {
printf("%d ", array[0][v]);
}
puts("");
for (v = 0; v < width; v++) {
printf("%d ", array[height - 1][v]);
}
}
//main
print_first_last(mat, 3, 3);

Answer 5

for(int i = 0; i < 2; i++) {
printf("%d ", mat[0][i]); //first row
}
for(int i = 0; i < 2; i++) {
printf("%d ", mat[2][i]); //Last row
}

Answer 6

for (int i = 0; i < m; ++i)
{
// mat[0][i] first row
// mat[m - 1][i] last row
}