Convert Binary to Octal in C++

Convert Binary to Octal in C++ (5 Programs).

IntermediateAdvanced Number ProgramsExample 9 of 13

convert-binary-to-octal.cpp

1#include <iostream>
2#include <cmath>
3using namespace std;
4
5int binaryToDecimal(long long binary) {
6    int decimal = 0, i = 0, remainder;
7    while (binary != 0) {
8        remainder = binary % 10;
9        binary /= 10;
10        decimal += remainder * pow(2, i);
11        ++i;
12    }
13    return decimal;
14}
15
16int decimalToOctal(int decimal) {
17    long long octal = 0;
18    int remainder, i = 1;
19    while (decimal != 0) {
20        remainder = decimal % 8;
21        decimal /= 8;
22        octal += remainder * i;
23        i *= 10;
24    }
25    return octal;
26}
27
28int main() {
29    long long binary;
30    cout << "Enter a binary number: ";
31    cin >> binary;
32    
33    int decimal = binaryToDecimal(binary);
34    long long octal = decimalToOctal(decimal);
35    
36    cout << "Binary: " << binary << " = Octal: " << octal << endl;
37    
38    return 0;
39}

Output

Enter a binary number: 1010
Binary: 1010 = Octal: 12

What's going on

The program converts a binary number to an octal number. For example:

Input binary: 1010

Output octal: 12

The program uses a two-step approach: first converts binary to decimal, then decimal to octal. This is straightforward and reuses the conversion algorithms we've already learned.

Example:

1010 (binary) → 10 (decimal) → 12 (octal)

1111 (binary) → 15 (decimal) → 17 (octal)

Header Files Used

1.#include <iostream>

Provides cout and cin for input/output operations.

2.#include <cmath>

Provides pow() function for calculating powers.

Used in the binary-to-decimal conversion.

Understanding the Two-Step Conversion

The program uses two helper functions:

Step 1: Binary to Decimal

Converts the binary number to decimal first.

Uses the algorithm: multiply each digit by 2^position and sum.

Step 2: Decimal to Octal

Converts the decimal result to octal.

Uses the algorithm: repeatedly divide by 8 and collect remainders.

Why Two Steps?

Simplest approach - reuses existing conversion algorithms.

Easy to understand and implement.

Can be optimized later with direct conversion.

Function 1: binaryToDecimal()

int binaryToDecimal(long long binary) {

int decimal = 0, i = 0, remainder;

while (binary != 0) {

remainder = binary % 10;

binary /= 10;

decimal += remainder * pow(2, i);

++i;

}

This function:

Takes a binary number as input.

Converts it to decimal using the standard algorithm.

Returns the decimal equivalent.

1return decimal;

How it works

(for binary = 1010):

Extracts digits from right to left.

Multiplies each digit by 2^position.

Sums all contributions: 0×1 + 1×2 + 0×4 + 1×8 = 10

Function 2: decimalToOctal()

int decimalToOctal(int decimal) {

long long octal = 0;

int remainder, i = 1;

while (decimal != 0) {

remainder = decimal % 8;

decimal /= 8;

octal += remainder * i;

i *= 10;

}

This function:

Takes a decimal number as input.

Converts it to octal using the standard algorithm.

Returns the octal equivalent.

1return octal;

How it works

(for decimal = 10):

Repeatedly divides by 8.

Collects remainders: 10÷8=1 rem 2, 1÷8=0 rem 1.

Builds octal: 2 + 1×10 = 12

Main Function - Combining Both Steps

long long binary;

cout << "Enter a binary number: ";

cin >> binary;

int decimal = binaryToDecimal(binary);

long long octal = decimalToOctal(decimal);

cout << "Binary: " << binary << " = Octal: " << octal << endl;

return 0;

}

1int main() {

Process Flow

1.User enters binary number (e.g., 1010).

2.Convert binary to decimal (1010 → 10).

3.Convert decimal to octal (10 → 12).

4.Display the result.

Direct Conversion Method (More Efficient)

Why Direct Conversion?

Two-step method works but is less efficient.

Direct conversion groups binary digits and converts directly.

More efficient: O(log n) vs O(log n) + O(log n).

How Direct Conversion Works

Group binary digits in sets of 3 (from right to left).

Each group of 3 binary digits = 1 octal digit.

Convert each group directly to octal.

Example

(binary 1010 to octal):

Group from right: 010 (pad with 0 if needed)

010 (binary) = 2 (octal)

Next group: 1 (pad to 001)

001 (binary) = 1 (octal)

Result: 12 (octal)

Why Groups of 3?

2³ = 8, so 3 binary digits can represent 0-7 (all octal digits).

This is the mathematical relationship between binary and octal.

Other Methods (Mentioned but not shown in code)

Method 2: Direct Conversion

Groups binary digits in sets of 3.

Converts each group directly to octal digit.

More efficient than two-step method.

Method 3: Using bitset

bitset<32> binaryBits(binary);

string binaryStr = binaryBits.to_string();

// Group and convert

Uses bitset for binary representation.

Groups and converts to octal.

1#include <bitset>

Method 4: Using String Manipulation

Converts binary to string.

Groups characters in sets of 3.

Converts each group to octal digit.

Method 5: Using Functions (Modular)

Encapsulates each conversion step in functions.

Makes code more organized and testable.

Easier to maintain and debug.

Displaying the Result

The program prints:

Output:

Binary: 1010 = Octal: 12

This clearly shows the conversion from binary to octal.

1cout << "Binary: " << binary << " = Octal: " << octal << endl;

Understanding Binary-Octal Relationship

Why 3 Binary Digits = 1 Octal Digit?

Binary is base-2, octal is base-8.

2³ = 8, so 3 binary digits can represent 0-7 (all octal digits).

This makes direct conversion possible.

Conversion Table

(3-bit binary to octal):

000 (binary) = 0 (octal)

001 (binary) = 1 (octal)

010 (binary) = 2 (octal)

011 (binary) = 3 (octal)

100 (binary) = 4 (octal)

101 (binary) = 5 (octal)

110 (binary) = 6 (octal)

111 (binary) = 7 (octal)

Direct Conversion Example

Binary: 1 0 1 0

Group: 1 | 010

001 | 010 (pad left group with 0)

Octal: 1 | 2

Result: 12 (octal)

When to Use Each Method

Two-Step Conversion

: Best for learning - clear, reuses known algorithms.

Direct Conversion

: Best for efficiency - faster, groups digits directly.

bitset Method

: Good for working with binary data - uses C++ bitset class.

String Manipulation

: Flexible - good for very large numbers.

Functions

: Best for code organization - modular and maintainable.

Best Practice

: Use two-step for learning, direct conversion for efficiency.

Important Considerations

Padding with Zeros

When grouping binary digits, leftmost group may need padding.

Pad with leading zeros to make groups of 3.

Example: 1010 → group as 001 and 010.

Large Binary Numbers

Very large binary numbers may cause overflow in intermediate steps.

Consider using string-based methods for very large numbers.

Direct conversion is more efficient for large numbers.

Validation

Ensure binary input contains only 0s and 1s.

This program assumes valid binary input.

return 0;

This ends the program successfully.

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