Nested loops, did not manage to do the exercise

C++
1

Hello,
I am not sure where to find support. I don’t understand Nested Loops on C++. I watched the video and did not manage to do the exercise. Can anyone explain me, please?

2

Which exercise? Do you have some code which is not working? As written your question basically asks “complete the exercise for me” so if you can ask for clarification on a more specific point it is easier for us to help.

3

Yes it was this exercise:
#include
#include
using namespace std;
int main() {
cout << left;
int rows;
cout << "Enter number of rows: ";
cin >> rows;
for (int i = 1; i <= rows ; i++){
for (int j = 0 ; j < i ; j++)
cout << “*”;
cout << endl;
}
return 0;}

I don’t understand how the code is able to print
*
**

It was in the first video of C++ for the basics

4

Hi Magnon, thank you for asking this question. I have just got up to this lesson and it stumped me for a while. Perhaps the following will help.

Inner for loop: -

for (int j = 0; j < i; j++ )

cout << “*”;

This just keeps outputting an * for as long as j is less than i.

So, when j equals 0 then the cout is *

and when j increments to 1 then the cout puts another * next to the one already there so the display is **. All this is on one line.

Obviously, this happens a lot quicker in real time and you do not see it in slow motion.

The problem is that you need to have one * on the first line, followed by two ** on another new line so that the output looks like: -
*
**

This is where the Outer Loop is used to advance the line by using endl.

The outer loop code is: -

For (int i = 1; i <= rows; i++){
Run the Inner loop code
cout << endl;
}

The outer loop runs as follows, assume rows = 2: -

Step 1: - The outer for loop value for i = 1. This is passed to the inner loop. In the first iteration of the inner loop, j = 0, i = 1 and the cout is *

Step 2: - Then the inner loop j increments to j = 1, i stays the same at i = 1 and the inner loop stops because j < i (1 < 1 is false).

Step 3: - Control goes back to the outer loop and this is where the endl is used to move to the next line.

Step 4: - Then the value of i increases to 2.

Step 5: - This value is passed to the inner loop where j has been reset to 0.

Step 6: - The inner loop cout’s a * because j = 0 and i = 2. But wait, when j increments to 1 the conditional (j < i) 1 < 2 is still true.

So another * is placed next to the one already there. We now have **.

Step 7: - The inner loop value for j increments to a value of 2 and the conditional fails (2<2) is false so the inner loop stops and the cout is **

Step 8: - The last part of the outer loop code is endl so this sets up for the next line (if there was one).

Step 9: - Control is passed back to the outer loop and i increments to 3. However, the conditional i <= rows fails because 3 <= 2 is false.

The output is: -

**

Hope that helps,
Paul

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