Functions in c language format transposing 2 arrays - arrays

Question

So, my functions in my code don't talk to each other and I cannot for the life of me figure it out. What do you put in the parenthesis of the function headings. the variable names go in them but how are they formatted? Do I need pointers? If so can someone explain them. Thanks!
int FindTranspose(????)
int main(????)
FindTranspose (????)
#include <stdio.h>
int FindTranspose(int before[3][3], int after[3][3], int i, int j){
int rows, columns;
printf("\nThe orginial matrix: \n");
for(i=0;i<3;i++){
printf("\n");
for(j=0;j<3;j++){
printf("%d ", before[i][j]);
}
}
for(rows = 0; rows < i; rows++){
for(columns = 0;columns < j; columns++){
after[columns][rows] = before[rows][columns];
}
}
printf("\n");
printf("\nThe Transpose of the matrix: \n");
printf("\n");
for(rows = 0; rows < j; rows++){
for(columns = 0; columns < i; columns++){
printf("%d ", after[rows][columns]);
}
printf("\n");
}
}
int main(){
int before[3][3], after[3][3], i, j;
for(i=0;i<3;i++){
for(j=0;j<3;j++){
printf("Enter number for array - [%d],[%d] : ",i,j);
scanf("%d", &before[i][j]);
}
}
FindTranspose (before[3][3], after[3][3], i, j);
return 0;
}
Picture of code
Picture of warnings

Answer 1

Dealing with multi-dimensional arrays as pointers is really tricky in C.
For example the multi-dimensional array before[3][3] defined in main:
The array name before is not "as usual" a pointer to before[0][0] although it's a pointer to before[0] so before actually has the type int (*) [3] which read as 'Pointer to a one-dimensional array of 3 elements'
Anyway, you should learn more about "pointer and multi-dimensional arrays".
For the moment here is the solution to your issue:
#include <stdio.h>
void FindTranspose(int (*before)[3], int (*after)[3], int i, int j){
int rows, columns;
printf("\nThe orginial matrix: \n");
for(i=0;i<3;i++){
printf("\n");
for(j=0;j<3;j++){
printf("%d ", before[i][j]);
}
}
for(rows = 0; rows < i; rows++){
for(columns = 0;columns < j; columns++){
after[columns][rows] = before[rows][columns];
}
}
printf("\n");
printf("\nThe Transpose of the matrix: \n");
printf("\n");
for(rows = 0; rows < j; rows++){
for(columns = 0; columns < i; columns++){
printf("%d ", after[rows][columns]);
}
printf("\n");
}
}
int main(){
int before[3][3], after[3][3], i, j;
for(i=0;i<3;i++){
for(j=0;j<3;j++){
printf("Enter number for array - [%d],[%d] : ",i,j);
scanf("%d", &before[i][j]);
}
}
FindTranspose (before, after, i, j);
return 0;
}
Of course, there is an easier but complex to understand solution which is dealing with the multi-dimensional array as a one-dimensional array.
Here is an example of that:
#include <stdio.h>
#define NUM_OF_ROWS 3
#define NUM_OF_COLS 3
void FindTranspose(int *before){
int rows, columns, i;
int after[NUM_OF_ROWS][NUM_OF_COLS];
int *p, (*q)[NUM_OF_COLS];
printf("\nThe orginial matrix: \n");
i = 0;
for(p = before; p < before + NUM_OF_COLS * NUM_OF_ROWS; p++){
printf("%d ", *p);
if(++i >= NUM_OF_COLS){
i = 0;
printf("\n");
}
}
columns = 0;
for(p = before; p < before + NUM_OF_COLS * NUM_OF_ROWS;){
for(q = &after[0]; q < &after[NUM_OF_ROWS]; q++){
(*q)[columns] = *p++;
}
columns++;
}
printf("\n");
printf("\nThe Transpose of the matrix: \n");
for(rows = 0; rows < NUM_OF_ROWS; rows++){
for(columns = 0; columns < NUM_OF_COLS; columns++){
printf("%d ", after[rows][columns]);
}
printf("\n");
}
}
int main(){
int before[NUM_OF_ROWS][NUM_OF_COLS], i, j;
for(i=0; i<NUM_OF_ROWS; i++){
for(j=0; j<NUM_OF_COLS; j++){
printf("Enter number for array - [%d][%d]: ", i, j);
scanf("%d", &before[i][j]);
}
}
FindTranspose ( &before[0][0] );
return 0;
}
Recommended reading: C Programming A modern approach 2nd edition for K.N. King