Drawing shapes using for loops

[R N]

Hi guys. I am working on a program that asks the user for an input on how large the shape should be, validates the input, and then displays the shape using asterisks. I figured out how to do the filled triangle but I don't know how to make inverted filled triangle, unfilled triangle, inverted unfilled triangle, filled rectangle, and unfilled rectangle. Could you help me out? Here is my code.

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// INCLUDE FILES
#include <iostream>
using namespace std;
int main()
{
    int no_lines,   // height of triangle or rectangle.
        width;      // width of rectangle.
    short choice;
    // ADDITIONAL VARIABLES
    
    // ADD A do-while LOOP TO THE PROGRAM SO THAT THE USER CAN REPEATEDLY
    // DISPLAY THE MENU, MAKE A CHOICE, AND HAVE THE APPROPRIATE STEPS FOR
    // THAT CHOICE CARRIED OUT.
    // THE LOOP SHOULD CONTINUE ITERATING UNTIL THE USER ENTERS 7 FOR THE MENU CHOICE.
    // HAVE THE PROGRAM PRINT SEVERAL BLANK LINES AFTER EACH SHAPE IS DRAWN.
        do
        {
        // Display menu.    
        cout << "THIS PROGRAM DRAWS VARIOUS SHAPES.\n" << endl;

        cout << "1. Filled Triangle" << endl;
        cout << "2. Filled Inverted Trianlge" << endl;
        cout << "3. Unfilled Triangle" << endl;
        cout << "4. Unfilled Inverted Trianlge" << endl;
        cout << "5. Filled Rectangle" << endl;
        cout << "6. Unfilled Rectangle" << endl;
        cout << "7. Exit\n" << endl;
        cin >> choice; 
        // ASK, READ AND VALIDATE CHOICE WITH while LOOP.
        
        // Process choice.
        switch (choice)
        {
               case 1: // Filled Triangle
                    // Ask, read and validate number of lines.
                    cout << "Enter the height for the filled triangle (2 - 75): ";
                    cin >> no_lines;
                    while (no_lines < 2 || no_lines > 75)
                    {
                          cout << "Number of lines has to be between 2 and 75.\n";
                          cout << "Enter the height for the filled triangle (2 - 75): ";
                          cin >> no_lines;
                    }
                    cout << endl;
                    // Draw shape.
                    // An example of a filled triangle with height 5
//                        *
//                        **
//                        ***
//                        ****
//                        *****
                    int i, j; 
                    for (i = 1; i <= no_lines; i++)
                    {
                        for (j = 1; j <= i; j++)
                            cout << "*";
                        cout << endl;
                    }
                    break;
               case 2: // Inverted filled triangle
                    // ASK, READ AND VALIDATE HEIGHT.
                    // DRAW INVERTED TRIANGLE
                    // AN EXAMPLE OF AN INVERTED TRIANGLE OF HEIGHT 10
//                        **********
//                        *********
//                        ********
//                        *******
//                        ******
//                        *****
//                        ****
//                        ***
//                        **
//                        *
                    break;
               case 3: // Unfilled triangle
                    // ASK, READ AND VALIDATE HEIGHT.
                    // DRAW UNFILLED TRIANGLE
                    // AN EXAMPLE OF A TRIANGLE OF HEIGHT 8
//                        *
//                        **
//                        * *
//                        *  *
//                        *   *
//                        *    *
//                        *     *
//                        ********
                    break;
               case 4: // Unfilled inverted triangle
                    // ASK, READ AND VALIDATE HEIGHT.
                    // DRAW UNFILLED INVERTED TRIANGLE
                    // AN EXAMPLE OF AN UNFILLED INVERTED TRIANGLE OF HEIGHT 7
//                        *******
//                        *    *
//                        *   *
//                        *  *
//                        * *
//                        **
//                        *

                    break;
               case 5:
                    // ASK, READ AND VALIDATE HEIGHT.
                    // ASKK, READ AND VALIDATE WIDTH.
                    // DRAW FILLED RECTANGLE
                    // AN EXAMPLE OF A 5 (HEIGHT) X 10 (WIDTH) FILLED RECTANGLE.
//                        **********
//                        **********
//                        **********
//                        **********
//                        **********
                    break;
               case 6:
                    // ASK, READ AND VALIDATE HEIGHT.
                    // ASKK, READ AND VALIDATE WIDTH.
                    // DRAW UNFILLED RECTANGLE
                    // AN EXAMPLE OF A 4 (HEIGHT) X 8 (WIDTH) UNFILLED RECTANGLE
//                        ********
//                        *      *
//                        *      *
//                        ********
                    break;
               case 7:
                    cout << "Exiting ...\n";
                    return 0; 
                    break;
        }
        } while (choice >= 1 && choice <= 7); 
    system("pause"); 
    return 0;
}
    

    

[R N]

Bump.

[R N]

Any ideas guys? I tried but it seems harder than it looks.

[Azagaros]

It is? It looks like loops, tracking variables and other things.

you have a variable named width, which I don't see you get from the user. For a simple triangle for case 1 would take two loops. To me the width variable is the only one I would need, since it would dictate how many lines I would create too.

for(int nWcount = 0; nWcount < width; nWcount++)
{
      for(int nIndex = 0; nIndex < nWcount; nIndex++)
      {
              cout << "*";
      }
      cout << endl;
}

all of the other cases would follow similar code and use a few if statements.

[Stewbond]

Hint for the upside down triangle: Initialize your outer for loop at the number of lines and count down.

Here's case 4:

cout << "give the height: ";
cin >> no_lines;

// first line:
for (int i = 0; i < no_lines; i++)
	cout << '*';                       // A whole bunch of stars for the first line
cout << endl;

// next lines:
for (int j = no_lines-1; j > 1; j--)       // counts from lines-1 to 1, note I'm counting down.
{
	cout << '*';                       // First star in a line
	for (int i = 0; i < j-2; i++)      // Add j-2 spaces,
	cout << ' ';                       // Writing the spaces
	cout << '*' << endl;               // Last star in a line
}

// last line
cout << '*' << endl;

You should be able to infer the others based on this type of logic.