Description
Get an octal number as the input from the use then convert it into equivalent binary number.
Input
10
Output
1000
C Program
#include<stdio.h>
#include<math.h>
void OBConvert(int oct)
{
int i = 0, dec = 0;
while (oct!=0)
{
int digit = oct % 10;
dec += digit * pow(8, i);
oct /= 10;
i++;
}
long long bin = 0;
int rem;
i = 1;
while(dec!=0)
{
rem = dec % 2;
dec /= 2;
bin += rem * i;
i *= 10;
}
printf("Binary Value: %lld",bin);
}
int main()
{
int oct;
printf("Enter an octal alue: ");
scanf("%d", &oct);
OBConvert(oct);
return 0;
}
C++ Program
#include <iostream>
#include<math.h>
using namespace std;
int OBConvert(int oct)
{
int i = 0, dec = 0;
while (oct!=0)
{
int digit = oct % 10;
dec += digit * pow(8, i);
oct /= 10;
i++;
}
long long bin = 0;
int rem;
i = 1;
while(dec!=0)
{
rem = dec % 2;
dec /= 2;
bin += rem * i;
i *= 10;
}
cout<<"Binary Value: "<<bin;
return 0;
}
int main()
{
int oct;
cout<<"Enter an octal value: ";
cin>>oct;
OBConvert(oct);
return 0;
}
Java Program
import java.util.Scanner;
public class Main
{
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
System.out.print("Enter an octal number: ");
int oct = sc.nextInt();
int dec = 0;
int n = 0;
while(oct > 0)
{
int temp = oct % 10;
dec += temp * Math.pow(8, n);
oct = oct/10;
n++;
}
int bin[] = new int[20];
int i = 0;
while(dec > 0)
{
int r = dec % 2;
bin[i++] = r;
dec = dec/2;
}
for(int j = i-1 ; j >= 0 ; j--)
System.out.print(bin[j]);
}
}
Python Program
import math
def OBConvert(octal):
dec = 0
i = 0
bin = 0
while(octal != 0):
dec += (octal%10) * math.pow(8,i)
i = i + 1
octal = int (octal / 10)
i = 1
while (dec != 0):
bin = bin + (dec % 2) * i;
dec = int(dec / 2);
i *= 10;
return bin
octal = int(input("Enter an octal number : "))
print(int(OBConvert(octal)))
