How to pass 2-D char array to a function? - c++

Question

I just create a simple function which takes the 2-D char array as an argument and then just print the values of that 2-D char array.
The function is given below
void final_output(char arr[][200]){
int i, j;
for(i = 0; i < r; i++){
for(j = 0; j < c; j++){
cout << arr[i][j];
}
}
}
And i call this function as
final_output(grid_2);
But i face an error
error: cannot convert 'char (*)[c]' to 'char (*)[200]' for argument '1' to 'void final_output(char (*)[200])'
final_output(grid_2);
I search for different methods to pass the 2-D array as an argument to a function but i try all of them and they give the same type of error. Please anyone help me out to remove this error. Actually i use this function in this problem of hackerrank.

Answer 1

It is a strange function but it works for me assuming you define r and c, and the grid2 is an char[][200].
void final_output(char arr[][200]) {
int i, j;
int r = 10;
int c = 200;
for (i = 0; i < r; i++) {
for (j = 0; j < c; j++) {
cout << arr[i][j];
}
}
}
int main(int argc, char* argv[])
{
char arr[10][200];
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 200; j++) {
arr[i][j] = 'a';
}
}
final_output(arr);
//it prints aaaaaaaa
}

Answer 2

I you want a method to be valid for any matrix since you need to use char**. I is more cumbersome because if you are using pointers, but it is more flexible.
void final_output(char** arr, int nrow, int ncol) {
for (int i = 0; i < nrow; i++) {
for (int j = 0; j < ncol; j++) {
cout << arr[i][j];
}
}
}
int main(int argc, char* argv[])
{
//construct the matrix (any size would work)
int nrow = 10;
int ncol = 20;
char** arr = new char* [nrow];
for (int i = 0; i < nrow; i++){arr[i] = new char[ncol];}
//assign values
for (int i = 0; i < nrow; i++) {
for (int j = 0; j < ncol; j++) {
arr[i][j] = 'a';
}
}
final_output(arr,nrow,ncol);
}