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In this article, we've created some programs in Python, to convert any binary number entered by user at run-time to its equivalent octal value. Here are the list of programs:

- Binary to Octal with User-defined Code
- Binary to Octal using
**int()**and**oct()**Methods - Shortest Python Code for Binary to Octal Conversion

**Note - **Before starting these programs, if you're not aware about steps used for the conversion, then refer
to Binary to Octal Conversion Methods, Steps, Formula to get every required things.

To convert binary to octal number in Python, you have to ask from user to enter a number in binary number system to convert that number into octal number system as shown in the program given below.

The question is, **write a Python program to convert binary to octal using while loop**. Here is its answer:

print("Enter the Binary Number: ") binarynum = int(input()) octaldigit = 0 octalnum = [] i = 0 mul = 1 chk = 1 while binarynum!=0: rem = binarynum % 10 octaldigit = octaldigit + (rem * mul) if chk%3==0: octalnum.insert(i, octaldigit) mul = 1 octaldigit = 0 chk = 1 i = i+1 else: mul = mul*2 chk = chk+1 binarynum = int(binarynum / 10) if chk!=1: octalnum.insert(i, octaldigit) print("\nEquivalent Octal Value = ", end="") while i>=0: print(str(octalnum[i]), end="") i = i-1 print()

Here is its initial output:

Now supply any binary number as input say **11101** and press `ENTER`

key to convert it into
its equivalent octal value, then print the octal value on output as shown in the snapshot given below:

The **insert()** method is used to insert an element to the list. That is, the following statement:

`octalnum.insert(i, octaldigit)`

states that the value of **octaldigit** gets initialized to **octalnum[i]**.

The dry run of above program with user input **11101** goes like:

- Initial values,
**binarynum=11101**(entered by user),**octaldigit=0**,**i=0**,**mul=1**,**chk=1** - The condition (of
*while loop*)**binarynum!=0**or**11101!=0**evaluates to be true, therefore program flow goes inside the loop - Inside the loop, the first statement, that is

`rem = binarynum % 10`

gets executed **binarynum%10**or**11101%10**or**1**(last digit of 11101) gets initialized to**rem**. So**rem=1****octaldigit + (rem*mul)**or**0 + (1*1)**or**1**gets initialized to**octaldigit**. So**octaldigit=1**- Now the condition (of
*if*block)**chk%3==0**or**1%3==0**evaluates to be false, therefore program does not goes to its body, rather it goes to its**else**'s counterpart and**mul*2**or**1*2**or**2**gets initialized to**mul**. So**mul=2** **chk+1**or**1+1**or**2**gets initialized to**chk**. So**chk=2****int(binarynum/10)**or**int(11101/10)**or**1110**gets initialized to**binarynum**. So**binarynum=1110**- The condition of
**while loop**again gets evaluated with new value of**binarynum** - That is, the condition
**binarynum!=0**or**1110!=0**evaluates to be true, therefore program flow again goes inside the loop. This process continues until the condition evaluates to be false - In this way, the equivalent octal value of value stored in
**binarynum**gets stored in**octalnum**list after exiting from the loop

The following condition:

`if chk%3==0:`

is applied to check for three-three pair of binary digits. And the following block of code:

if chk!=1: octalnum.insert(i, octaldigit)

is used to insert the value of **octaldigit** to **octalnum[i]**, only if the value of **chk** is not
equal to **1**. The value of **chk** does not equal to **1** after exiting from the **while loop**
indicates that the program flow does not goes inside the **if**'s body before exiting from the loop.

Now print the value of **octalnum** list one by one starting from its last index. The **end=** is used to
skip printing of an automatic newline using **print()**

This is the modified version of previous program. This program uses **string** instead of **list**, which is
a better way in Python to convert any binary number to its equivalent octal value.

print("Enter the Binary Number: ", end="") bnum = int(input()) odig = 0 mul = chk = 1 onum = "" while bnum!=0: rem = bnum % 10 odig = odig + (rem * mul) if chk%3==0: onum = onum + str(odig) mul = chk = 1 odig = 0 else: mul = mul*2 chk = chk+1 bnum = int(bnum / 10) if chk!=1: onum = onum + str(odig) print("\nEquivalent Octal Value = ", onum[::-1])

Here is its sample run with user input, **11000111011** as binary number:

**Note - **The **str()** method is used to convert any type of value to a string type value.

This program uses **int()** and **oct()** methods to convert binary to octal. Using **input()** to receive
any input from user, treats the input as a string type value by default.

Therefore using **int()** with 2 as its second argument, converts the string to an integer type value with base two, that is, the
input gets converted into a binary number. And **oct()** converts the value of **onum** to its equivalent octal value.

print("Enter a Binary Number: ", end="") bnum = input() onum = int(bnum, 2) onum = oct(onum) print("\nEquivalent Octal Value = ", onum)

Here is its sample run with user input, **10111**:

**Note - **To print only **27**, that is, if you want to skip first two characters from octal number, then replace the following statement:

print("\nEquivalent Octal Value = ", onum)

with the statement given below:

print("\nEquivalent Octal Value = ", onum[2:])

Now the output with same user input looks like:

This is the shortest Python code to convert binary to octal. **[2:]** is used to print elements of **bnum** starting from its second index.

bnum = input() print(oct(int(bnum, 2))[2:])

Here is its sample run with user input, **1100110**:

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