C Program to Print Diamond Pattern

In this article, you will learn and get code about printing of diamond pattern in following ways:

Diamond Pattern of Stars
Diamond Pattern of Numbers
Diamond Pattern of Alphabets

Print Diamond Pattern of Stars

Let's create a program that asks from user to enter the row-size of upper-half diamond to print the diamond pattern of stars. For example, if user enters 5 as row size, then a diamond of stars gets printed of size 5*2-1 or 9 rows.

// Diamond Pattern of Stars
// ----codescracker.com----

#include<stdio.h>
#include<conio.h>
int main()
{
    int i, j, row, space;
    printf("Enter Number of Rows: ");
    scanf("%d", &row);
    space = row-1;
    for(i=1; i<=row; i++)
    {
        for(j=1; j<=space; j++)
            printf(" ");
        space--;
        for(j=1; j<=(2*i-1); j++)
            printf("*");
        printf("\n");
    }
    space = 1;
    for(i=1; i<=(row-1); i++)
    {
        for(j=1; j<=space; j++)
            printf(" ");
        space++;
        for(j=1; j<=(2*(row-i)-1); j++)
            printf("*");
        printf("\n");
    }
    getch();
    return 0;
}

This program was build and run under Code::Blocks IDE. Here is its output:

Now enter the number of rows say 10 to print diamond pattern that expands upto row-1 line. That is, with row number 10, it will print diamond pattern with 9 lines as shown in the output given below:

Program Explained

If user enters 10 as size of diamond, then always remember these things to print upper half of diamond:

At first row, print 9 spaces, and one star
At second row, print 8 spaces, and three star
At third row, print 7 spaces, and five star
At fourth row, print 6 spaces, and seven star
At fifth row, print 5 spaces, and nine star
At sixth row, print 4 spaces, and eleven star
At seventh row, print 3 spaces, and thirteen star
At eighth row, print 2 spaces, and fifteen star
At ninth row, print 1 space, and seventeen star
At tenth row, print ninteen star

And for lower half of diamond:

At first row, print 1 space, and seventeen star
At second row, print 2 spaces, and fifteen star
At third row, print 3 spaces, and thirteen star
At fourth row, print 4 spaces, and eleven star
At fifth row, print 5 spaces, and nine star
At sixth row, print 6 spaces, and seven star
At seventh row, print 7 spaces, and five star
At eighth row, print 8 spaces, and three star
At ninth row, print 9 spaces, and one star

The dry run of above program goes like, supposing user input as 10:

10 gets initialized to the variable row
Using the statement
space = row-1;
9 gets initialized to space
Now program flow executes the for loop
At first run of for loop with i=1, the dry run goes like:
1. Inside the loop, 1 gets initialized to i and checks whether it is less than or equal to row or not
2. The condition evaluates to be true, because obviously the value of i (1) is less than the value of row (10)
3. Therefore program flow goes inside the loop's body and executes another for loop
4. This time, 1 gets initialized to j and checks whether it is less than or equal to the value of space (9) or not
5. The condition evaluates to be true
6. Therefore program flow goes inside the loop body and executes only one statement, that is,
  printf(" ");
  that prints a space
7. Now the variable j gets incremented and again program flow goes to the loop's condition part and checks whether the updated value or incremented value of j (2) is less than or equal to the value of space (9) or not
8. The condition again evaluates to be true. Therefore program flow again goes inside the loop, and prints another space
9. In this way printing of space continues until the value of j becomes greater than the value of space
10. Therefore, total of 9 space gets printed at first row
11. And then decrement the value of space to print one less space on next row (than previous one)
12. Now the program flow goes to the second for loop part (present inside the first outer for loop)
13. There, 1 gets initialized to j and checks whether it is less than or equal to the value of ((2*i)-1) or not
14. Because the value of i was 1 at first run, therefore compares j with (2*i)-1 or (2*1)-1 or 1
15. The condition evaluates to be true
16. Therefore program flow goes inside the loop and prints a star (*)
17. Then increments the value of j and compares again with ((2*i)-1)
18. This time, the value of j (2) is not less than or equal to (2*i)-1, that is 1
19. Therefore condition evaluates to be false, and the program flow exits the loop
20. And using newline character (\n), program flow now starts the next output (printing) thing from new line
At second run of for loop with i=2, dry run goes like:
- Using first for loop, prints 8 spaces. Process goes same from step no.1 to step no.10 as mentioned above with updated value of i
- Again decrements the value of space to print one less space next time. Now space=8
- Using second for loop, prints 3 stars. Process goes same from step no.12 to step no.19 as mentioned above with update value of i
- Then process step no.20
At third run of for loop with i=3, dry run goes like:
- Process the similar operation from step no.1 to step no.10 with new value of i (3). That is, prints 7 spaces
- Decrements the value of space. Now space=7
- Process the similar operation given from step no.12 to step no.19 with new value of i (3). That is, print 5 stars
- Process step no.20
Continue executing the loop, until the value of i becomes 11
Because 11 is not less than or equal to row (10), therefore condition evaluates to be false, and the program flow exits from this loop
Process in similar way to print lower half of diamond
With initial value of space as 1
Because, at first row of lower half, we only have to print single space

Print Diamond Pattern of Numbers

Here is another program of printing diamond pattern of numbers. In each row, the number starts with 1. The program is similar as of previous one. Except, in place of star, print the number using a variable say num (initialized with 1 at start of the program). Increment the value of num after each print.

Note - Always remember to initialize num with 1 after the statement printf("\n"); to start with 1 for each row

// Diamond Pattern of Numbers
// ----codescracker.com----

#include<stdio.h>
#include<conio.h>
int main()
{
    int i, j, row, space, num=1;
    printf("Enter Number of Rows: ");
    scanf("%d", &row);
    space = row-1;
    for(i=1; i<=row; i++)
    {
        for(j=1; j<=(space); j++)
            printf(" ");
        space--;
        for(j=1; j<=(2*i-1); j++)
        {
            printf("%d", num);
            num++;
        }
        printf("\n");
        num=1;
    }
    space = 1;
    for(i=1; i<=(row-1); i++)
    {
        for(j=1; j<=(space); j++)
            printf(" ");
        space++;
        for(j=1; j<=(2*(row-i)-1); j++)
        {
            printf("%d", num);
            num++;
        }
        printf("\n");
        num=1;
    }
    getch();
    return 0;
}

Here is the output produced by above program assuming that user has entered 5 as the size of diamond:

Print Diamond Pattern of Alphabets

This program is almost same as of previous one. In place of number, use a character using a variable say ch of char type. Initialize it with A. Rest of the things are similar with previous program.

// Diamond Pattern of Alphabets
// ----codescracker.com----

#include<stdio.h>
#include<conio.h>
int main()
{
    int i, j, row, space;
    char ch='A';
    printf("Enter Number of Rows: ");
    scanf("%d", &row);
    space = row-1;
    for(i=1; i<=row; i++)
    {
        for(j=1; j<=(space); j++)
            printf(" ");
        space--;
        for(j=1; j<=(2*i-1); j++)
        {
            printf("%c", ch);
            ch++;
        }
        ch='A';
        printf("\n");
    }
    space = 1;
    for(i=1; i<=(row-1); i++)
    {
        for(j=1; j<=(space); j++)
            printf(" ");
        space++;
        for(j=1; j<=(2*(row-i)-1); j++)
        {
            printf("%c", ch);
            ch++;
        }
        ch='A';
        printf("\n");
    }
    getch();
    return 0;
}

Here is its output assuming the user input 5:

Same Program in Other Languages

C Online Test

« Previous Program Next Program »