# C Program to Print all digits of a given number

Given a number **N**, the task is to write a C program to print all digits of the number **N** in their original order.

**Examples:**

Input:N = 12Output:1, 2

Input:N = 1032Output:1, 0, 3, 2

**Method 1:** The simplest way to do is to extract the digits one by one and print it.

- Extract the last digit of the number
**N**by**N%10**, and store that digit in an array(say**arr[]**). - Update the value of
**N**by**N/10**and repeat the above step till**N**is not equals to**0**. - When all the digits have been extracted and stored, traverse the array from the end and print the digits stored in it.

Below is the implementation of the above approach:

`// C program of the above approach` ` ` `#include <stdio.h>` `#define MAX 100` ` ` `// Function to print the digit of` `// number N` `void` `printDigit(` `int` `N)` `{` ` ` `// To store the digit` ` ` `// of the number N` ` ` `int` `arr[MAX];` ` ` `int` `i = 0;` ` ` `int` `j, r;` ` ` ` ` `// Till N becomes 0` ` ` `while` `(N != 0) {` ` ` ` ` `// Extract the last digit of N` ` ` `r = N % 10;` ` ` ` ` `// Put the digit in arr[]` ` ` `arr[i] = r;` ` ` `i++;` ` ` ` ` `// Update N to N/10 to extract` ` ` `// next last digit` ` ` `N = N / 10;` ` ` `}` ` ` ` ` `// Print the digit of N by traversing` ` ` `// arr[] reverse` ` ` `for` `(j = i - 1; j > -1; j--) {` ` ` `printf` `(` `"%d "` `, arr[j]);` ` ` `}` `}` ` ` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 3452897;` ` ` ` ` `printDigit(N);` ` ` `return` `0;` `}` |

**Output:**

3 4 5 2 8 9 7

**Time Complexity:** O(log_{10} N)**Auxiliary Space:** O(log_{10} N)

**Method 2: Using Recursion**

- Recursively iterate till N become 0:
**Base Case:**If the value of**N**is**0**, exit from the function.if(N==0) return ;

**Recursive Call:**If the base case is not met, recursively call for next iteration by updating**N**to**N/10**and print the value of last digit extracted(say**r**) from the number**N**, after recursive call.recursive_function(N/10); print(r);

Below is the implementation of the above approach:

`// C program of the above approach` ` ` `#include <stdio.h>` ` ` `// Function to print the digit of` `// number N` `void` `printDigit(` `int` `N)` `{` ` ` `int` `r;` ` ` ` ` `// Base Case` ` ` `if` `(N == 0) {` ` ` `return` `;` ` ` `}` ` ` ` ` `// Extract the last digit` ` ` `r = N % 10;` ` ` ` ` `// Recursive call to next` ` ` `// iteration` ` ` `printDigit(N / 10);` ` ` ` ` `// Print r` ` ` `printf` `(` `"%d "` `, r);` `}` ` ` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 3452897;` ` ` ` ` `printDigit(N);` ` ` `return` `0;` `}` |

**Output:**

3 4 5 2 8 9 7

**Time Complexity:** O(log_{10} N)**Auxiliary Space:** O(1)

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