﻿ C Program: Check whether a number is a Strong Number or not - w3resource

# C Exercises: Check whether a number is a Strong Number or not

## C For Loop: Exercise-47 with Solution

Write a C program to check whether a number is a Strong Number or not.

Pictorial Presentation:

Sample Solution:

C Code:

``````#include <stdio.h>
void main()
{
int i, n, n1, s1=0,j;
long fact;

printf("\n\n  Check whether a number is Strong Number or not:\n ");
printf("----------------------------------------------------\n");

/* If sum of factorial of digits is equal to original number then it is Strong number */

printf("Input a number to check whether it is Strong number: ");
scanf("%d", &n);

n1 = n;

for(j=n;j>0;j=j/10)
{

fact = 1;
for(i=1; i<=j % 10; i++)
{
fact = fact * i;
}
s1 = s1 + fact;

}

if(s1==n1)
{
printf("\n%d is Strong number.\n\n", n1);
}
else
{
printf("\n%d is not Strong number.\n\n", n1);
}
}
```
```

Sample Output:

```  Check whether a number is Strong Number or not:
----------------------------------------------------
Input a number to check whether it is Strong number: 15

15 is not Strong number.
```

Flowchart:

C Programming Code Editor:

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## C Programming: Tips of the Day

You are seeing the ffffff because char is signed on your system. In C, vararg functions such as printf will promote all integers smaller than int to int. Since char is an integer (8-bit signed integer in your case), your chars are being promoted to int via sign-extension.

Since c0 and 80 have a leading 1-bit (and are negative as an 8-bit integer), they are being sign-extended while the others in your sample don't.

```char    int
c0 -> ffffffc0
80 -> ffffff80
61 -> 00000061
```
Here's a solution:
```char ch = 0xC0;
printf("%x", ch & 0xff);
```

This will mask out the upper bits and keep only the lower 8 bits that you want.

Ref : https://bit.ly/3vOLizM