# C++ Program to Convert Binary to Octal

In this article, you will learn and get code on binary to octal conversion in C++. The program is created with and without using function.

But before starting the program, if you are not aware about steps and formula used for the conversion, then you can refer to binary to octal. There you will get every required things.

## Binary to Octal in C++

To convert binary number to octal in C++ programming, you have to ask from user to enter any number in binary. And then convert it into its equivalent octal value as shown in the program given below.

The question is, write a program in C++ that receives any binary number and print its equivalent octal value. Here is its answer:

```#include<iostream>
using namespace std;
int main()
{
int binaryNum, octalDigit=0, octalNum;
int i=0, mul=1, chk=1, rem;
cout<<"Enter the Binary Number: ";
cin>>binaryNum;
while(binaryNum!=0)
{
rem = binaryNum%10;
octalDigit = octalDigit + (rem*mul);
if(chk%3==0)
{
octalNum[i] = octalDigit;
mul = 1;
octalDigit = 0;
chk = 1;
i++;
}
else
{
mul = mul*2;
chk++;
}
binaryNum = binaryNum/10;
}
if(chk!=1)
octalNum[i] = octalDigit;
cout<<"\nEquivalent Octal Value:  ";
for(i=i; i>=0; i--)
cout<<octalNum[i];
cout<<endl;
return 0;
}```

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

Now supply the binary number input say 1101110 and press `ENTER` key to print its equivalent value in octal, as shown in the snapshot given below:

Here is another sample run with user input as 11010010:

The dry run (of above program) with user input as 1101110 goes like:

• Initial values, i=0, mul=1, chk=1, octalDigit=0
• When user enters binary number say 1101110, then it gets initialized to binaryNum. Therefore binaryNum=1101110
• Now because the condition (of while loop) binaryNum!=0 or 1101110!=0 evaluates to be true, therefore program flow goes inside the loop
• Inside the loop, all the statements gets executed one by one, starting from the very first one
• That is, binaryNum%10 or 1101110%10 or 0 gets initialized to rem. So rem=0
• octalDigit + (rem*mul) or 0 + (0*1) or 0 gets initialized to octalDigit. So octalDigit=0
• Now the condition of if statement, that is chk%3==0 or 1%3==0 or 1==0 evaluates to be false, therefore program does not goes inside the if's body, rather it goes to its else's body
• Inside the else's body, mul*2 or 1*2 or 2 gets initialized to mul. Now mul=2
• And the value of chk gets incremented. Now chk=2
• Finally the last statement of while loop gets executed, that is, binaryNum/10 or 1101110/10 or 110111 gets initialized to binaryNum. Now binaryNum=110111
• Now the program flow goes back and evaluates the condition of while loop again with new value of binaryNum.
• Second time also the condition evaluates to be true, therefore program flow again goes inside the loop.
• This process continues, until the condition of while loop evaluates to be false. That is, the value of binaryNum becomes equal to 0
• In this way, here are the values after each evaluation of while loop:
1. rem=0, octalDigit=0, mul=2, chk=2, binaryNum=110111
2. rem=1, octalDigit=2, mul=4, chk=3, binaryNum=11011
3. rem=1, octalDigit=6, octalNum=6, mul=1, octalDigit=0, chk=1, i=1, binaryNum=1101. At this evaluation, the condition of if also evaluates to be true
4. rem=1, octalDigit=1, mul=2, chk=2, binaryNum=110
5. rem=0, octalDigit=1, mul=4, chk=3, binaryNum=11
6. rem=1, octalDigit=5, octalNum=5, mul=1, octalDigit=0, chk=1, i=2, binaryNum=1. Again the condition of if evaluates to be true
7. rem=1, octalDigit=1, mul=2, chk=2, binaryNum=0
• Because the value of binaryNum is 0 now, therefore the condition of while loop evaluates to be false, therefore the while loop evaluation gets ended
• Now the condition of if, that is chk!=1 gets evaluated. Because the condition chk!=1 or 2!=1 evaluates to be true, therefore octalDigit or 1 gets initialized to octalNum
• Now using the for loop the value of octalNum[] gets printed starting from its last to 0th index.
• That is the value of octalNum gets printed first. And then octalNum and octalNum gets printed. That equals or prints 156 as output

### Using user-defined Function

Let's create another program that does the same job using a user-defined function BinaryToOctal(). This function receives binary number as its argument and converts it into its equivalent octal value.

```#include<iostream>
using namespace std;
void BinaryToOctal(int);
int i=0, octalNum;
int main()
{
int binaryNum;
cout<<"Enter any Binary Number: ";
cin>>binaryNum;
BinaryToOctal(binaryNum);
cout<<"\nEquivalent Octal Value: ";
for(i=i; i>=0; i--)
cout<<octalNum[i];
cout<<endl;
return 0;
}
void BinaryToOctal(int binaryNum)
{
int octalDigit=0, mul=1, chk=1, rem;
while(binaryNum!=0)
{
rem = binaryNum%10;
octalDigit = octalDigit + (rem*mul);
if(chk%3==0)
{
octalNum[i] = octalDigit;
mul = 1;
octalDigit = 0;
chk = 1;
i++;
}
else
{
mul = mul*2;
chk++;
}
binaryNum = binaryNum/10;
}
if(chk!=1)
octalNum[i] = octalDigit;
}```

It will produce the same output as of previous program. You can also convert binary to octal in an indirect way. That is, first binary to decimal and then decimal to octal.

#### Same Program in Other Languages

C++ Online Test

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