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In this tutorial, we will write a program to print a star pattern in C language. The star pattern for which we will write a C program is shown in the figure below
<img src="https://api.codewithharry.com/media/videoSeriesFiles/courseFiles/c-practice-programs-in-hindi-12/base64.png" />
Figure 1: Star Pattern
As shown in figure 1, consider the row as &ldquo;i&rdquo; and the columns as &ldquo;j&rdquo;. To deduce the formula about when to print the stars see that when the value of (i = 0) the star printed for the value of (j = 5); when the value of (i = 1) the stars printed for the values of (j = 4, 5, 6); similarly when the value of (i = 2) the stars printed for the values of (j = 3, 4, 5, 6, 7) and so on.
To print this star pattern we will use the formula (j&gt;=n-1-i &amp;&amp; j&lt;=n-1+i), if the condition (j&gt;=n-1-i &amp;&amp; j&lt;=n-1+i) will be true then we will print the star otherwise we will print the space; where &ldquo;i&rdquo; is the current index of row and &ldquo;j&rdquo; is the current index of column whereas &ldquo;n&rdquo; is the length of rows which will specify the range of the stars to print.
Note: The main thing to note here is that the star pattern shown in figure 1 is for n = 6
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C program to print the following pattern is shown below
<pre class="language-cpp"><code>int main(){

 int n = 6;

 for (int i = 0; i &lt; n; i++)

 {

 for (int j = 0; j &lt; 2*n; j++)

 {</code></pre>
Code Snippet 1: Main Program 1
As shown in the code snippet 1, we have to use the nested &ldquo;for&rdquo; loop to print such a pattern because the first loop will run for each row and the second loop will run for each column.
<ol>
<li>We have initialized an integer variable &ldquo;n&rdquo; and assigned value &ldquo;6&rdquo; to it</li>
<li>The first &ldquo;for&rdquo; loop will start from the value &ldquo;0&rdquo; and runs till (n); the value of &ldquo;n&rdquo; is &ldquo;6&rdquo; so the loop will run until the value of &ldquo;i&rdquo; becomes &ldquo;5&rdquo;</li>
<li>The second &ldquo;for&rdquo; loop will start from the value &ldquo;0&rdquo; and runs till (2*n); the value of &ldquo;n&rdquo; is &ldquo;6&rdquo; so the loop will run until the value of &ldquo;j&rdquo; becomes &ldquo;11&rdquo;</li>
</ol>
<pre class="language-cpp"><code>if(j&gt;=n-1-i &amp;&amp; j&lt;=n-1+i){

 printf("*");

 }

 else{

 printf(" ");

 }

 }

 printf("\n");

 }

 return 0;

}</code></pre>
Code Snippet 2: Main program 2
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As shown in the code snippet 2, we have used the formula (j&gt;=n-1-i &amp;&amp; j&lt;=n-1+i) in the &ldquo;if&rdquo; condition to determine whether to print star or print space, for example, if the value of &ldquo;i&rdquo; is &ldquo;0&rdquo; and the value of &ldquo;j&rdquo; is &ldquo;5&rdquo; and the value of &ldquo;n&rdquo; is &ldquo;6&rdquo; by solving the values &ldquo;6-1-0&rdquo; and &ldquo;6-1+0&rdquo; we get the value &ldquo;5&rdquo; which is equal to the value of &ldquo;j&rdquo; which is &ldquo;5&rdquo; so the star will be printed.
Similarly if the value of &ldquo;i&rdquo; is &ldquo;1&rdquo; and the value of &ldquo;j&rdquo; is &ldquo;3&rdquo; by solving the values &ldquo;6-1-1&rdquo; and &ldquo;6-1+1&rdquo; we get the value &ldquo;4&rdquo; and &ldquo;5&rdquo; which don&rsquo;t make the condition true so space will be printed.
The output of the following program is shown in figure 2
<img src="https://api.codewithharry.com/media/videoSeriesFiles/courseFiles/c-practice-programs-in-hindi-12/base64_C4O0MAg.png" />
Figure 2: Program Output