printing the alphabet in order in a matrix - c++

Question

by now I have this code
int main() {
int cols;
cout << "cols";
cin >> cols;
int rows;
cout << "rows";
cin >> rows;
char** charArray = new char*[rows];
for (int i = 0; i < rows; ++i) {
charArray[i] = new char[cols];
}
// Fill the array
for (int i = 0; i < rows; ++i) {
for (int j = 0; j < cols; ++j) {
charArray[i][j] = char('1' + i + j );
}
}
// Output the array
for (int i = 0; i < rows; ++i) {
for (int j = i; j < cols ; ++j) {
cout << charArray[i][j];
}
cout << endl;
}
// Deallocate memory by deleting
for (int i = 0; i < rows; ++i) {
delete[] charArray[i];
}
and the output is like
1 2 3
2 3 4
How can I make it to be
1 2 3
4 5 6
I tried a lot, im a newbie in programming so can you please explain me what's the matter with this problem! Thanks a lot!!

Answer 1

Just change this:
charArray[i][j] = char('1' + i + j );
by:
charArray[i][j] = char('1' + i*cols + j);
BTW:
You have a typo in the output array loop:
for (int j = i; j < cols ; ++j) {
Should be:
for (int j = 0; j < cols ; ++j) {

Answer 2

Array filling loop should be
int count = 0;
for (int i = 0; i < rows; ++i) {
for (int j = 0; j < cols; ++j) {
charArray[i][j] = char('1' +count );
count++;
}
}
try with above code

Answer 3

You are building a matrix, here right. So it's 2-dimensional.
What the output means is that your matrix looks like:
1 2 3
2 3 4
Which is only natural, since, while filling it, you wrote for each element at coordinate i,j to be value 1+i+j. That means if i and j both equal 1 (second line, second column, since index start at 0), you put value the 1+1+1=3.
If you want to get the matrix:
1 2 3
4 5 6
You'll have to fill it with
charArray[i][j] = char('1' + i*ncols + j );
This way when you fill the first line, i = 0, so it's the same.
But then on the second line, i=1, so you get 1 + 3 + 0 = 4. And then 1+3+1=5, etc.
Edit: people are fast here. By the time I write my answer, already 2 others posted theirs :D