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Solved
Making a Right Angle with C Program...
Posted on 2007-12-06
Hi friends !
I want to get this output of a 90 degree tringle by using 'C' Language only. I am beginner in programming. Please see the desired output:
*
**
***
****
*****
******
*******
********
(The lines occupied by the triangle will depend on an input 'n' provided by the user.
And in the above picture, the size of the triangle depends on the height of the triange. I want user to input a number (n), which will determine, now many lines the trinagle will cover on the screen (Height).
I wrote this code...
#include<stdio.h>
#include<conio.h>
void main()
{
int i, j, n;
clrscr();
printf("Please enter the number:");
scanf("%d", &n);
i=1;
j=1;
for (i=1; i<=n; i++)
{
for (j=1; j<=i; j++)
{
printf("*\t\n");
}
}
getch();
}
When I compile and run it, it gives no error but it gives output like this...
for input n = 2, output is
*
*
*
for input n = 3, output is
*
*
*
*
*
likewise for input n = 5, 15 lines are occupied.
Please help me in getting my desired output.
Thanks
Hemant
I want to get this output of a 90 degree tringle by using 'C' Language only. I am beginner in programming. Please see the desired output:
*
**
***
****
*****
******
*******
********
(The lines occupied by the triangle will depend on an input 'n' provided by the user.
And in the above picture, the size of the triangle depends on the height of the triange. I want user to input a number (n), which will determine, now many lines the trinagle will cover on the screen (Height).
I wrote this code...
#include<stdio.h>
#include<conio.h>
void main()
{
int i, j, n;
clrscr();
printf("Please enter the number:");
scanf("%d", &n);
i=1;
j=1;
for (i=1; i<=n; i++)
{
for (j=1; j<=i; j++)
{
printf("*\t\n");
}
}
getch();
}
When I compile and run it, it gives no error but it gives output like this...
for input n = 2, output is
*
*
*
for input n = 3, output is
*
*
*
*
*
likewise for input n = 5, 15 lines are occupied.
Please help me in getting my desired output.
Thanks
Hemant
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10 Comments
Active 1 day ago
Hi JatinHemant,
In your code, the indexes i and j represent the height and width of the triangle.
For the width, you want to print the corresponding number of asterisks. Thats the simple loop:
for (j = 1; j < width; ++j)
printf ("*");
You want to print the new line character only AFTER printing the entire line. There's a couple of smaller issues, but solve that and the rest should fall into place. Try this for the loops:
for (i=1; i<=n; i++)
{
for (j=1; j<=i; j++)
{
printf("*");
}
printf ("\n");
}
That'll get you closer and let you work on the next little oddity.
Good Luck,
Kent
In your code, the indexes i and j represent the height and width of the triangle.
For the width, you want to print the corresponding number of asterisks. Thats the simple loop:
for (j = 1; j < width; ++j)
printf ("*");
You want to print the new line character only AFTER printing the entire line. There's a couple of smaller issues, but solve that and the rest should fall into place. Try this for the loops:
for (i=1; i<=n; i++)
{
for (j=1; j<=i; j++)
{
printf("*");
}
printf ("\n");
}
That'll get you closer and let you work on the next little oddity.
Good Luck,
Kent
Try :
for (j=1; j<=i; j++)
{
printf("*");
}
printf("\n");
instead of :
for (j=1; j<=i; j++)
{
printf("*\t\n");
}
for (j=1; j<=i; j++)
{
printf("*");
}
printf("\n");
instead of :
for (j=1; j<=i; j++)
{
printf("*\t\n");
}
Active 1 day ago
Hi Infinity08,
There's a twist. Usually it's me that has to learn to "type faster". :)
Kent
There's a twist. Usually it's me that has to learn to "type faster". :)
Kent
Refer to http://csusap.csu.edu.au/~sbuckley/maths/funpage/jokes1.htm#Longer for clarification on Pythagoras's theorem.
Active 4 days ago
After looking at your question history I concluded this was not a 'homework' type question, therefore I've provided a C++/STL solution example. If STL is not what you need you should be able to modify it a non-STL solution without too much effort: -
#include <iostream>
#include <string>
void main()
{
std::cout << "Please enter the number: ";
int n = 0;
std::cin >> n;
for (int i = 1 ; i <= n; ++i)
{
for (int j = 1 ; j <= i; ++j)
{
std::cout << "*";
}
std::cout << std::endl;
}
}
Please enter the number: 30
*
**
***
****
*****
******
*******
********
*********
**********
***********
************
*************
**************
***************
****************
*****************
******************
*******************
********************
*********************
**********************
***********************
************************
*************************
**************************
***************************
****************************
*****************************
******************************
My advice with respect to loops it to avoid [1,numberOfValues] and prefer [0,numberOfValues).
If you are not familiar with range notation, square brackets mean "inclusive" and round brackets mean "exclusive".
i.e. Prefer
for (int i = 0 ; i < n; ++i)
...and avoid...
for (int i = 1 ; i <= n; ++i)
It may seem like an unnecessary change, but you'll find that adopting the convention [0,numberOfValues), gets you ready for the joys of standard library algorithms of you go the way of C++.
Incidentally, question http:/Q_22981835.html needs to be closed on this. It is EE convention only to have one question open on a subject at a time.
If you are not familiar with range notation, square brackets mean "inclusive" and round brackets mean "exclusive".
i.e. Prefer
for (int i = 0 ; i < n; ++i)
...and avoid...
for (int i = 1 ; i <= n; ++i)
It may seem like an unnecessary change, but you'll find that adopting the convention [0,numberOfValues), gets you ready for the joys of standard library algorithms of you go the way of C++.
Incidentally, question http:/Q_22981835.html needs to be closed on this. It is EE convention only to have one question open on a subject at a time.
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