# C++ Program to Convert Octal to Binary

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In this article, you will learn and get code on octal to binary conversion in C++. The first program is created that actually converts and then prints binary equivalent of given octal number. And the second program just prints the binary equivalent of each and every octal digit one by one.

Before going through these programs, if you're not aware about the simple steps and formula used for the conversion, then you can refer to octal to binary formula to get every required things.

There are two approaches that can be used to create a program for octal to binary conversion in C++:

1. Octal to Binary Direct Conversion
2. Octal to Binary Indirect Conversion

In indirect conversion, convert given octal number into its equivalent decimal value first. And then convert the decimal to its equivalent binary value. But here, the program is created on direct conversion only. Because you can approach indirect conversion with yourself. The program for indirect approach is already given in separate article.

## Octal to Binary in C++

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

The question is, write a program in C++ that converts octal to binary. The answer to this question is given below:

```#include<iostream>
#include<string.h>
using namespace std;
int main()
{
int octalNum, rev=0, rem, chk=0;
char binaryNum = "";
cout<<"Enter the Octal Number: ";
cin>>octalNum;
while(octalNum!=0)
{
rem = octalNum%10;
if(rem>7)
{
chk++;
break;
}
rev = (rev*10) + rem;
octalNum = octalNum/10;
}
if(chk==0)
{
octalNum = rev;
cout<<"\nEquivalent Binary Value: ";
while(octalNum!=0)
{
rem = octalNum%10;
switch(rem)
{
case 0: strcat(binaryNum, "000");
break;
case 1: strcat(binaryNum, "001");
break;
case 2: strcat(binaryNum, "010");
break;
case 3: strcat(binaryNum, "011");
break;
case 4: strcat(binaryNum, "100");
break;
case 5: strcat(binaryNum, "101");
break;
case 6: strcat(binaryNum, "110");
break;
case 7: strcat(binaryNum, "111");
break;
}
octalNum = octalNum/10;
}
cout<<binaryNum;
}
else
cout<<"\nInvalid Octal Digit!";
cout<<endl;
return 0;
}```

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

Now supply any octal number say 364 as input and then press `ENTER` key convert and print its binary equivalent as shown in the output given below:

The function strcat() is used to concatenate string. It takes two string as its argument. The value of second string gets appended at the end of first string.

The following block of code (included in above program):

```while(octalNum!=0)
{
rem = octalNum%10;
if(rem>7)
{
chk++;
break;
}
rev = (rev*10) + rem;
octalNum = octalNum/10;
}```

is used to check whether any digit of given octal number is greater than 7 or not. If any digit found as greater than 7, then that digit is not a valid octal digit. Because octal number can only be of 8 digits, that are from 0 to 7.

The while loop executes in following ways (supposing user input as 364):

• The condition octalNum!=0 or 364!=0 evaluates to be true, therefore program flow goes inside the loop
• There, octalNum%10 or 364%10 or 4 gets initialized to rem
• And the condition of if, i.e., rem>7 evaluates to be false, therefore program flow does not goes inside if's body, rather it goes to the statement that is available after the if's block of code
• That is, (rev*10)+rem or (0*10)+4 or 4 gets initialized to rev
• And the final statement of while loop, gets evaluated, that is octalNum/10 or 364/10 or 36 gets initialized to octalNum
• Program flow goes back and evaluates the condition of while loop again.
• The evaluation of while loop continues until the value of octalNum becomes equal to 0. That is the condition evaluates to be false
• Because, no any digit in 364 is greater than 7, therefore program flow never goes inside the if's body
• That means, the value of chk is 0 (initialized at start of the program)
• Therefore after the ending of while loop, we've checked whether the value of chk holds its initial value (0) or not
• If it holds its initial value, then process further to convert octal number to binary. Otherwise, print a message like An Invalid Octal Digit was Entered!
• Now let's perform the dry run of all block of codes inside the if(chk==0) block, given below

So the dry run of above program with user input 364 goes like (Inside the if block):

• binaryNum="", rev=463 (as reversed the number earlier using while loop
• The value of rev (463) gets initialized to octalNum. So octalNum=463
• Now the condition of while loop (inside if) gets evaluated
• That is, the condition, octalNum!=0 or 463!=0 evaluates to be true, therefore program flow goes inside the loop and octalNum%10 or 463%10 or 3 gets initialized to rem
• Now using switch, the value of rem gets compared with all of the 8 cases. That is, its value is equal to 0, 1, 2, ...., or 7
• Whatever its value is, concatenates its binary equivalent to binaryNum variable
• Because the value of rem is 3, therefore case 3:'s statement gets evaluated. That is,
`strcat(binaryNum, "011");`
gets evaluated. Therefore, binaryNum=011 (since initial value of binaryNum is empty). From next time, binary equivalent gets added after 011
• And octalNum/10 or 463/10 or 46 gets initialized to octalNum. So the new value of octalNum is 46
• Now program flow goes back and evaluates the condition of while loop again with new value of octalNum
• That is, the condition, octalNum!=0 or 46!=0 evaluates to be true and program flow again goes inside the loop
• The process continues, until the condition of while loop evaluates to be false
• On continuing the process of evaluating while loop's block of code, we'll get values as:
• rem=3, binaryNum=011, octalNum=46
• rem=6, binaryNum=011110, octalNum=4
• rem=4, binaryNum=011110100, octalNum=0
• Now print the value of binaryNum on output, that will be the binary equivalent (011110100) of given octal number (364)

### Print Binary Equivalent of Octal Number

Now let's create another program that only prints binary equivalent of given octal number by user at run-time without actually converting it.

```#include<iostream>
using namespace std;
int main()
{
int octalNum, rev=0, rem;
cout<<"Enter the Octal Number: ";
cin>>octalNum;
while(octalNum!=0)
{
rem = octalNum%10;
rev = (rev*10) + rem;
octalNum = octalNum/10;
}
octalNum = rev;
cout<<"\nEquivalent Binary Value: ";
while(octalNum!=0)
{
rem = octalNum%10;
switch(rem)
{
case 0: cout<<"000";
break;
case 1: cout<<"001";
break;
case 2: cout<<"010";
break;
case 3: cout<<"011";
break;
case 4: cout<<"100";
break;
case 5: cout<<"101";
break;
case 6: cout<<"110";
break;
case 7: cout<<"111";
break;
default: cout<<"--InvalidOctalDigit("<<rem<<")--";
break;
}
octalNum = octalNum/10;
}
cout<<endl;
return 0;
}```

Here is its sample run with user input 254 (octal number):

If user enters an octal number that contains one or more invalid octal digit say 2948. Then here is the output you'll see:

#### Same Program in Other Languages

C++ Online Test

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