Description
Get a binary number as the input from the user and give the corresponding octal number as the output.
Input
1000
Output
10
C Program
#include<stdio.h>
#include<math.h>
void BOConvert(long long num)
{
int octal = 0, count = 1, i = 0, pos = 0;
int octalArr[32] = {0};
while(num != 0)
{
int digit = num % 10;
octal += digit * pow(2, i);
i++;
num /= 10;
octalArr[pos] = octal;
if(count % 3 == 0)
{
octal = 0;
i = 0;
pos++;
}
count++;
}
for (int j = pos; j >= 0; j--)
printf("%d",octalArr[j]);
}
int main()
{
long long binary;
printf("Enter binary number: ");
scanf("%lld", &binary);
BOConvert(binary);
return 0;
}
C++ Program
#include <iostream>
#include<math.h>
using namespace std;
int BOConvert(long bin)
{
int oct = 0, dec = 0, i = 0,rem;
while(bin != 0)
{
rem = bin % 10;
int res = rem * pow(2,i);
dec += res;
i++;
bin/=10;
}
i = 1;
while (dec != 0)
{
rem = dec % 8;
oct += rem * i;
dec /= 8;
i *= 10;
}
return oct;
}
int main()
{
long binary;
cout << "Enter a binary number: ";
cin >> binary;
int octal=BOConvert(binary);
cout << octal;
return 0;
}
Java
import java.util.Scanner;
public class Main
{
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
System.out.print("Enter a binary number : ");
int bin = sc.nextInt();
int dec = 0;
int n = 0;
while(bin > 0)
{
int temp = bin%10;
dec += temp*Math.pow(2, n);
bin = bin/10;
n++;
}
int oct[] = new int[20];
int i = 0;
while(dec > 0)
{
int r = dec % 8;
oct[i++] = r;
dec = dec / 8;
}
for(int j = i-1 ; j >= 0 ; j--)
System.out.print(oct[j]);
}
}
Python
bin = int(input("Enter the binary number: "))
octal = 0
octalnum = []
i = 0
mul = 1
c = 1
while bin!=0:
rem = bin % 10
octal = octal + (rem * mul)
if c%3==0:
octalnum.insert(i, octal)
mul = 1
octal = 0
c = 1
i = i+1
else:
mul = mul*2
c = c+1
bin = int(bin / 10)
if c!=1:
octalnum.insert(i, octal)
#print("\nEquivalent Octal Value = ", end="")
while i>=0:
print(str(octalnum[i]), end="")
i = i-1
print()
