Member Avatar for mani_singh

hello all,
i am trying to make a program using nested for loops that outputs a triangle made of *. so far i have part of the triangle, but am having diffuculty finishing it. my code so far:

#include<iostream> 
#include<conio.h>
using namespace std
void main()
{
      int n = 5;
      char usr = '*'; 
      int x;
      int y;    

       for(y=1;y<=n;y++)
           {
                 for(x=1;x<=y;x++)
                      {
                           cout<<usr;                                                 
                       }
                  cout<<endl;
            }
         getch();
}

it looks like this:
*
**
***
****
*****


but i want it to look like this:
*
**
***
****
*****
****
***
**
*

any help would be appreciated

  • 2 Contributors
  • 3 Replies
  • 124 Views
  • 10 Hours Discussion Span
  • Latest Post Latest Post by Galf

Dani AI

Generated

The upward triangle in the original code is fine. The reason the second half printed as a single line of four stars is that the reverse pass you added only printed a star character repeatedly without producing one complete line per value — you need to print y stars then a newline for each decreased y. was right that you need a reversed outer pass; the missing piece in your attempt was an inner-per-line output (or an equivalent string-based approach).

A concise, modern C++ way that avoids an explicit inner loop is to use std::string to build each line and print it with a newline. For example:

#include <iostream>
#include <string>

int main() {
    const int n = 5;
    for (int i = 1; i <= n; ++i)
        std::cout << std::string(i, '*') << '\n';

    for (int i = n - 1; i >= 1; --i)
        std::cout << std::string(i, '*') << '\n';
}

Alternate approach: keep a single std::string, push_back('*') to grow and print each growing line, then pop_back() to shrink and print each shrinking line. That is memory-efficient and easy to reason about.

Quick troubleshooting notes: use int main() and return 0; (avoid nonstandard conio.h/getch), make sure loop counters are signed when you decrement them, and ensure you print a newline at the end of each line (cout << '\n' or cout << endl). These small fixes will produce the exact symmetric triangle you want.

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Member Avatar for Galf

Well, if you want to do it with loops, you just need to reverse your loop and you'll generate the other side of the pyramid. for (y=n-1;y>=1;y--) {...} should generate the other end.

Member Avatar for mani_singh

i tried that loop i put it right befor getch();
i put:

for (y=n-1;y>=1;y--)
	{
		cout<<usr;
	}

and what i got was:
*
**
***
****
*****
****

Member Avatar for Galf

Correct -- your first attempt was:

for(y=1;y<=n;y++)		// iterate over the number of lines you need to display
{
	for(x=1;x<=y;x++)	// outputs the desired number of '*'s
	{
		...
	}
}

I'm saying you need to reverse that process -- instead of counting 'up' lines, you need to count 'down' now, which is the outer loop I showed. You still need to provide another loop to output the right number of '*'s, though.

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