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Program 49:Symbol Pattern 2

Program 49:
 
#include<stdio.h>
main()
{
 int i,j,num;
 printf("Enter number of rows\n");
 scanf("%d",&num);
 for(i=1;i<=num;i++)
 {
  for(j=1;j<=i;j++)
  {
   printf("*");
     
  }
  
 printf("\n");
 }
}
Explanation:

  1. The program starts with initializing
    • num → To store number of rows to be printed
    • i,j Temporary variable.Where i is used for rows and j for columns
  2. printf("Enter number of rows\n");
     scanf("%d",&num);
    Taking number of rows from user
  3. Main Logic goes here:
    for(i=1;i<=num;i++)
     {
      for(j=1;j<=i;j++)
      {
       printf("*");
         
      }
      
     printf("\n");
     }
  4. As i=1.Lets take num=3 so we are going to print 3 rows
    • Iteration 1: As i=1 and i<=num → 1<=3 which is true so iteration continues
      • Sub Iterations:-
        • for(j=1;j<=i;j++)
            {
             printf("*");
               
            }
        • Sub Iteration 1:-j=1 and j<=i → 1<=1 which is true so sub iteration1 continues and prints '*' for 1 time
        • Sub Iteration 2:- j=2 and j<=i → 2<=1 which is false so 2nd loop terminates for 1st iteration of 1st loop
        • So till now '*' in the first row and first column is printed as shown in below screen shot
        • Now 'i' will be incremented by 1 and now i=2.
        •  printf("\n");------->Will print next line so now the cursor will be in 2nd row
    • Iteration 2: As i=2 and i<=num → 2<=3 which is true so iteration continues.Now we are in 2nd row
      • Sub Iterations:-
        • for(j=1;j<=i;j++)
            {
             printf("*");
               
            }
        • Sub Iteration 1:j=1 and j<=i → 1<=2 which is true so sub iteration1 continues and prints '*' for 1st time in the next row which is 2nd row and column 1
        • Sub Iteration 2:j=2 and j<=i → 2<=2 which is true so sub iteration2 continues and prints '*' for 2nd time in the 2nd row and column 2
        • Sub Iteration 3:j=3 and j<=i → 3<=2 which is flase so 2nd loop terminates for Iteration 2.
        • So till now 2 rows are printed as shown
        • Now 'i' will be incremented by 1 and now i=3.
        •  printf("\n");------->Will print next line so now the cursor will be in 3rd row
    • Iteration 3: As i=3 and i<=num → 3<=3 which is true so iteration continues.Now we are in 3rd row
      • Sub Iterations:-
        • for(j=1;j<=i;j++)
            {
             printf("*");
               
            }
        • Sub Iteration 1:j=1 and j<=i → 1<=2 which is true so sub iteration1 continues and prints '*' for 1st time in the next row which is 3rd row and column 1
        • Sub Iteration 2:j=2 and j<=i → 2<=2 which is true so sub iteration2 continues and prints '*' for 2nd time in the 3rd row and column 2
        • Sub Iteration 3:j=2 and j<=i → 2<=2 which is true so sub iteration3 continues and prints '*' for 3rd time in the 3rd row and column 3
        • Sub Iteration 4:j=3 and j<=i → 3<=2 which is flase 2nd loop terminates for Iteration 3.
        • So till now 3 rows are printed as shown
        • Now 'i' will be incremented by 1 and now i=4.
        •  printf("\n");------->Will print next line so now the cursor will be in 4th row
    • Iteration 4: As i=4 and i<=num → 4<=3 which is false so 1st loop will terminate and come out of loops.
    • So our final out put for num=3 is given below

Output:

Symbol Pattern 2


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