Here is a C program to reverse digits of a number using recursion. Reversing the digits of a number means reversing the sequence of digits in a number. After reversing, least significant digit will become most significant digit of number and most significant digit becomes least significant digit and so on.
Let the total number of digits in a number is N. So a digit at ith position from left will become ith digit from right or N-i+1th digit from left.
For Example
Befor reverse : 25346After reverse : 64352
- N%10 returns the last digit(least significant) of N. For Example : 354%10 = 4.
- N/10 return the number after removing least significant digit of N(rightmost digit of N). For Example: 2345/10 = 234.
- log10(N) + 1 returns the number of digits in N. log10(N) is logarithm of N with base 10. For Example: log10(2311) + 1 = 4.
- Let getReversedNumber(N) is a function, which returns reverse of N. Then, we can use recursion to reverse the digits if a number using below mention recursive equation.
- getReversedNumber(N) = (N%10)X pow(10, (log10(N/10) + 1)) + getReversedNumber(N/10)
For Example
C program to reverse the digits of a number recursively
This program first take an integer as input form user, then reverse it's digits using modulus(%), division(/), multiplication(*) and logarithmic function inside a recursive function. After reversing it prints the reversed number.
#include <stdio.h>
#include <math.h>
int getReversedNumber(int number);
int main(){
int number, reverse = 0;
printf("Enter a number\n");
scanf("%d", &number);
reverse = getReversedNumber(number);
printf("Reversed number : %d\n", reverse);
return 0;
}
int getReversedNumber(int number){
int lastDigit, numberOfDigits, sign = 1;
if(number < 0){
number = number * -1;
sign = -1;
}
if(number < 10)
return number*sign;
lastDigit = number % 10;
number = number / 10;
numberOfDigits = log10(number) + 1;
return (lastDigit*pow(10,numberOfDigits) +
getReversedNumber(number))*sign;
}
Output
Enter a number 6542 Reversed number : 2456
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