Python Decimal to Octal Conversion

What Is Decimal to Octal Conversion?

Decimal (base 10) is the everyday number system. Octal (base 8) uses only digits 0–7. Conversion means rewriting a nonnegative integer using powers of eight instead of powers of ten.

Rightmost octal digit = 8⁰ (value 1)
Next digit             = 8¹ (value 8)
Next digit             = 8² (value 64)
…

Example: 57₁₀
  57 ÷ 8 → rem 1, then 7
  reverse → 71₈
  Check: 7×8 + 1 = 57

How to Convert Decimal to Octal in Python

Four clear steps — then pick a remainder loop or the built-in formatter.

1. Accept a nonnegative integer n.
2. If n == 0, return "0" immediately.
3. Collect digits with n % 8 and n // 8, then reverse — or call oct(n)[2:].
4. Print or return the octal string.

Pseudocode

function DecimalToOctal(n):          // n >= 0
    if n == 0:
        return "0"

    digits = []
    while n > 0:
        digits.append(n mod 8)       // LSD first
        n = floor(n / 8)

    reverse(digits)
    return join(digits)

Cheat Sheet

Keep these facts handy while writing the program.

Live Preview

Enter a nonnegative integer to see remainders and the octal result.

Decimal    = 57
Remainders = 1, 7  (LSD → MSD)
Octal      = 71
Check      = 7×8 + 1 = 57

Worked Walkthrough — n = 57

Divide by 8. Stack remainders from right to left, then reverse for the final string.

57 ÷ 8 = 7 remainder 1   ← LSD (rightmost)
 7 ÷ 8 = 0 remainder 7   ← MSD (leftmost)

Remainders collected: 1, 7
Reverse for output:   7, 1
Octal:                71

Verify: 7×8¹ + 1×8⁰ = 56 + 1 = 57 ✓

Python Programs

Start with the remainder loop so you can explain every digit. Then use oct / format for a one-line production-style answer.

def decimal_to_octal(n: int) -> str:
    if n < 0:
        raise ValueError("Expected a nonnegative integer.")
    if n == 0:
        return "0"

    digits = []
    while n > 0:
        digits.append(str(n % 8))
        n //= 8

    return "".join(reversed(digits))


print(decimal_to_octal(57))  # 71
print(decimal_to_octal(0))   # 0
print(decimal_to_octal(64))  # 100

71
0
100

def decimal_to_octal(n: int) -> str:
    if n < 0:
        raise ValueError("Expected a nonnegative integer.")
    return oct(n)[2:]
    # or: return format(n, "o")


print(decimal_to_octal(57))  # 71
print(decimal_to_octal(0))   # 0
print(decimal_to_octal(8))   # 10

# Bonus: same idea for binary
print(bin(57)[2:])  # 111001
# 71₈ ↔ 111 001₂  (7=111, 1=001)

71
0
10
111001

Time and Space Complexity

Digit count is about log8 n for positive n.

Key Takeaways