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C Program to Convert Decimal to Octal

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Given a decimal number as input, we need to write a program to convert the given decimal number into an equivalent octal number. i.e convert the number with base value 10 to base value 8. The base value of a number system determines the number of digits used to represent a numeric value. For example, the binary number system uses two digits 0 and 1, the octal number system uses 8 digits from 0-7 and the decimal number system uses 10 digits 0-9 to represent any numeric value.

Examples:

Input : 16
Output: 20

Input : 10
Output: 12

Input : 33
Output: 41

Algorithm:  

  1. Store the remainder when the number is divided by 8 in an array.
  2. Divide the number by 8 now
  3. Repeat the above two steps until the number is not equal to 0.
  4. Print the array in reverse order now.

For Example: 

If the given decimal number is 16. 

Step 1: Remainder when 16 is divided by 8 is 0. Therefore, arr[0] = 0. 
Step 2: Divide 16 by 8. New number is 16/8 = 2. 
Step 3: Remainder, when 2 is divided by 8, is 2. Therefore, arr[1] = 2. 
Step 4: Divide 2 by 8. New number is 2/8 = 0. 
Step 5: Since number becomes = 0. 

Stop repeating steps and print the array in reverse order. Therefore, the equivalent octal number is 20.

The below diagram shows an example of converting the decimal number 33 to an equivalent octal number.  

decToOctal

Below is the implementation of the above idea.  

C




// C program to convert decimal
// number to octal number
#include <stdio.h>
  
// function to convert decimal to octal
void decToOctal(int n)
{
  
    // array to store octal number
    int octalNum[100];
  
    // counter for octal number array
    int i = 0;
    while (n != 0) {
  
        // storing remainder in octal array
        octalNum[i] = n % 8;
        n = n / 8;
        i++;
    }
  
    // printing octal number array in reverse order
    for (int j = i - 1; j >= 0; j--)
        printf("%d", octalNum[j]);
}
  
// Driver Code
int main()
{
    int n = 33;
  
    // Function Call
    decToOctal(n);
  
    return 0;
}


Output

41

Time Complexity: O(log N) 

Space Complexity: O(N) since creating array to store octal numbers

Another Approach: (O(1) space Complexity)

This problem can also be solved without using an array  using the following algorithm:

  • Initialize octal num to 0 and countVal to 1 and the decimal number as n
  • Find the remainder when decimal number divided by 8
  • Update octal number by octalNum + (remainder * countval)
  • Increase countval by countval*10
  • Divide decimal number by 8
  • Repeat from the second step until the decimal number is zero

Below is the implementation of the above idea:

C




// C program to convert decimal
// number to octal number
#include <stdio.h>
  
// function to calculate the octal value of the given
// decimal number
void decimaltoOctal(int deciNum)
{
  
    int octalNum = 0, countval = 1;
    int dNo = deciNum;
  
    while (deciNum != 0) {
        // decimals remainder is calculated
        int remainder = deciNum % 8;
  
        // storing the octalvalue
        octalNum += remainder * countval;
  
        // storing exponential value
        countval = countval * 10;
        deciNum /= 8;
    }
    printf("%d", octalNum);
}
  
// Driver Code
int main()
{
    int n = 33;
  
    // Function Call
    decimaltoOctal(n);
    return 0;
}


Output

41

Time Complexity: O(log N)

Auxiliary Space: O(1)



Last Updated : 26 Oct, 2022
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