C++ Multidimensional Array - c++

Question

i'm trying to figure out how to fill a multidimentional array in this way:
Input: rows = 3 , cols = 3 :
1 4 7
2 5 8
3 6 9
can somebody give me an idea?
P.S My task is to find how many nubers stay in the same position in both arrangements. Ex:
1 4 7 1 2 3
2 5 8 4 5 6
3 6 9 7 8 9
so the numbers that are in the same position are : 1 5 9.
i've tryied :
//n = 3 , m = 3
for(int i = 0; i <n; i++) {
for(int j = 0; j <m; j++){
if(array[i][j] == array2[i][j]) {
lol++;
}
}
}
cout<<lol;
/*
1 2 3
4 5 6
7 8 9
1 4 7
2 5 8
3 8 9
*/
it must show me 3 , but it shows 0, where is the problem?

Answer 1

Populate at initialization:
int a[3][3] = { { 1, 4, 7},
{ 2, 5, 8},
{ 3, 6, 9}
};
EDIT (unsure if resolved):
After update to question here is an example application that (with modification to accept input from user) will diff two arrays and construct an array indicating the elements that were the same and a count of the number of identical elements:
#include <iostream>
int** make_array(const size_t a_rows, const size_t a_columns)
{
int** result = new int*[a_rows];
for (size_t i = 0; i < a_rows; i++)
{
*(result + i) = new int[a_columns];
}
return result;
}
void print_array(int** a_array, const size_t a_rows, const size_t a_columns)
{
for (size_t r = 0; r < a_rows; r++)
{
for (size_t c = 0; c < a_columns; c++)
{
std::cout << *(*(a_array + r) + c) << " ";
}
std::cout << "\n";
}
std::cout << "\n";
}
int main()
{
// Example data.
int a[3][3] = { { 1, 4, 7},
{ 2, 5, 8},
{ 3, 6, 9}
};
int b[3][3] = { { 1, 2, 3},
{ 4, 5, 6},
{ 7, 8, 9}
};
size_t rows = 3;
size_t columns = 3;
// Create three arrays:
// - two input arrays
// - array that represents which elements are the same
int** in_1 = make_array(rows, columns);
int** in_2 = make_array(rows, columns);
int** diff = make_array(rows, columns);
// Populate with example data.
for (size_t r = 0; r < rows; r++)
{
for (size_t c = 0; c < columns; c++)
{
*(*(in_1 + r) + c) = a[r][c];
*(*(in_2 + r) + c) = b[r][c];
}
}
// Diff.
// The 'diff' array will hold '1' for elements that
// were the same and '0' for elements that were not.
size_t same_count = 0;
for (size_t r = 0; r < rows; r++)
{
for (size_t c = 0; c < columns; c++)
{
*(*(diff + r) + c) = *(*(in_1 + r) + c) == *(*(in_2 + r) + c);
same_count += *(*(diff + r) + c);
}
}
std::cout << "\n";
// Results.
print_array(in_1, rows, columns);
print_array(in_2, rows, columns);
print_array(diff, rows, columns);
std::cout << "Same element count: " << same_count << "\n";
// Free...
return 0;
}
Output:
$ ./cpp/main.exe
1 4 7
2 5 8
3 6 9
1 2 3
4 5 6
7 8 9
1 0 0
0 1 0
0 0 1
Same element count: 3

Answer 2

Create a dynamically allocated array if you only know the dimensions at runtime:
int** x = new int*[rows];
for ( int i = 0 ; i < rows ; i++ )
x[i] = new int[cols];
And then populate it:
for ( int i = 0 ; i < rows ; i++ )
for ( int j = 0 ; i < cols ; j++ )
x[i][j] = y;
Or better yet, use a vector of vectors, this will give you more flexibility:
std::vector<std::vector<int> > x;

Answer 3

A simple solution would be this:-
int k = 1;
for(int i = 0; i < row; i++){
for(int j = 0; j < col; j++){
a[j][i] = k; // filling it in the column first order
++k;
}
}