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C Exercises: Convert a binary to a decimal using for loop and without using array

C For Loop: Exercise-42 with Solution

Write a program in C to convert a binary number into a decimal number without using array, function and while loop.

Pictorial Presentation:

Convert a binary to a decimal using for loop and without using array

Sample Solution:

C Code:

#include <stdio.h>

void main()
{       int n1, n,p=1;
	int dec=0,i=1,j,d;

     printf("\n\n  Convert Binary to Decimal:\n ");
     printf("-------------------------\n");

	printf("Input a binary number :");
	scanf("%d",&n);
	n1=n;
	for (j=n;j>0;j=j/10)
	{  
          d = j % 10;
            if(i==1)
                  p=p*1;
            else
                 p=p*2;

	   dec=dec+(d*p);
	   i++;
	}
        printf("\nThe Binary Number : %d\nThe equivalent Decimal  Number : %d \n\n",n1,dec);
}

Sample Output:

  Convert Binary to Decimal:                                                                                  
 -------------------------                                                                                    
Input a binary number :11001                                                                                  
                                                                                                              
The Binary Number : 11001                                                                                     
The equivalent Decimal  Number : 25 

Flowchart:

Flowchart : Convert a binary to decimal using for loop and without using array.

C Programming Code Editor:

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Next: Write a C program to find HCF (Highest Common Factor) of two numbers.

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C Programming: Tips of the Day

Printing hexadecimal characters in C:

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