使用for循环创建带有星号的形状

Question

I'm studying for a programming test and one of the questions will involve looking at code which prints a shape using asterisks. The code below is similar to what will be on the test, although, the test question will output a different shape. I'm at somewhat of a loss as to how this code works. I understand the concept of a for loop, but not the role each of these for loops plays in the program. Below, is the code and it's output.

#include <iostream>
using namespace std;
int main()
{
  int m, n;
  for (m = 0; m<10; m++)
    {
      for (n = 0; n<m; n++) cout << " ";
      for (n = 0; n<(19-2*m); n++) cout << "*";
      for (n = 0; n<m; n++) cout << " ";
      cout << endl;
    }
  return 0;
}

Output:

*******************
 *****************
  ***************
   *************
    ***********
     *********
      *******
       *****
        ***
         *

Answer 1

for (m = 0; m<10; m++)                      // this one loops over the rows of the shape
{
  for (n = 0; n<m; n++) cout << " ";        // to leave spaces before the shape
  for (n = 0; n<(19-2*m); n++) cout << "*"; // to fill the shape with *
  for (n = 0; n<m; n++) cout << " ";        // to leave spaces after the shape
  cout << endl;                             // change line
}

As the guys stated, the last loop is not required to get this particular shape, but since this is for your test study, make sure to understand that too, since in the test, any similar shape could pop-up, which may require all the loops (otherwise, why the teacher put it there?.

Answer 2

for (m = 0; m<10; m++) //This is the main loop to print the 10 rows
{
  for (n = 0; n<m; n++) cout << " "; //This loops provide all the spaces before the first element of each row
  for (n = 0; n<(19-2*m); n++) cout << "*"; //prints the *s
  for (n = 0; n<m; n++) cout << " "; //Not required here....you can get the same output without this.
  cout << endl;
}