Checking if a number is a palindrome - c++

Question

I've tried to check whether a number is a palindrome with the following code:
unsigned short digitsof (unsigned int x)
{
unsigned short n = 0;
while (x)
{
x /= 10;
n++;
}
return n;
}
bool ispalindrome (unsigned int x)
{
unsigned short digits = digitsof (x);
for (unsigned short i = 1; i <= digits / 2; i++)
{
if (x % (unsigned int)pow (10, i) != x % (unsigned int)pow (10, digits - 1 + i))
{
return false;
}
}
return true;
}
However, the following code isn't able to check for palindromes - false is always returned even if the number is a palindrome.
Can anyone point out the error?
(Please note: I'm not interested to make it into a string and reverse it to see where the problem is: rather, I'm interested to know where the error is in the above code.)

Answer 1

I personally would just build a string from the number, and then treat it as a normal palindrome check (check that each character in the first half matches the ones at length()-index).

Answer 2

x % (unsigned int)pow (10, i) is not the ith digit.

Answer 3

The problem is this:
x % (unsigned int)pow (10, i)
Lets try:
x =504405
i =3
SO I want 4.
x % 10^3 => 504405 %1000 => 405 NOT 4
How about
x / (unsigned int)pow (10, i -1) % 10

Answer 4

Just for more info! The following two functions are working for me:
double digitsof (double x)
{
double n = 0;
while (x > 1)
{
x /= 10;
n++;
}
return n;
}
bool ispalindrome (double x)
{
double digits = digitsof (x);
double temp = x;
for(double i = 1; i <= digits/2; i++)
{
float y = (int)temp % 10;
cout<<y<<endl;
temp = temp/10;
float z = (int)x / (int)pow(10 , digits - i);
cout<<(int)z<<endl;
x = (int)x % (int)pow(10 , digits - i);
if(y != z)
return false;
}
return true;
}

Answer 5

Code to check if given number is palindrome or not in JAVA
import java.util.*;
public class HelloWorld{
private static int countDigits(int num) {
int count = 0;
while(num>0) {
count++;
num /= 10;
}
return count;
}
public static boolean isPalin(int num) {
int digs = HelloWorld.countDigits(num);
int divderToFindMSD = 1;
int divderToFindLSD = 1;
for (int i = 0; i< digs -1; i++)
divderToFindMSD *= 10;
int mid = digs/2;
while(mid-- != 0)
{
int msd = (num/divderToFindMSD)%10;
int lsd = (num/divderToFindLSD)%10;
if(msd!=lsd)
return false;
divderToFindMSD /= 10;
divderToFindLSD *= 10;
}
return true;
}
public static void main(String []args) {
boolean isPalin = HelloWorld.isPalin(1221);
System.out.println("Results: " + isPalin);
}
}

Answer 6

I have done this with my own solution which is restricted with these conditions
Do not convert int to string.
Do not use any helper function.
var inputNumber = 10801
var firstDigit = 0
var lastDigit = 0
var quotient = inputNumber
while inputNumber > 0 {
lastDigit = inputNumber % 10
var tempNum = inputNumber
var count = 0
while tempNum > 0 {
tempNum = tempNum / 10
count = count + 1
}
var n = 1
for _ in 1 ..< count {
n = n * 10
}
firstDigit = quotient / n
if firstDigit != lastDigit {
print("Not a palindrome :( ")
break
}
quotient = quotient % n
inputNumber = inputNumber / 10
}
if firstDigit == lastDigit {
print("It's a palindrome :D :D ")
}