I simply write a program to reverse a number but the output not meet my expectation

Question

#include<iostream>
using namespace std;
class sample{
    int x,y;
    public:
    void rev();
};
void sample::rev(){
    cout<<"Enter a no:";
    cin>>x;
    int r,n;
    while(x!=0){
    r=x%10;
    n=n*10+r;
    x=x/10;
    }
    cout<<n;
}
int main(){
    sample A;
    A.rev();
    return 0;
}

If i give input a number like : 10,it need to give me the rev no: 01,but it give only 1...how can i solve it?

integers can not retain the number of leading 0s. It's not stored in any way in the int. An integer is a fixed number of bits always. For leading 0s you have to account for that yourself and handle it in the output: [https://stackoverflow.com/questions/1714515/how-can-i-pad-an-int-with-leading-zeros-when-using-cout-operator](https://stackoverflow.com/questions/1714515/how-can-i-pad-an-int-with-leading-zeros-when-using-cout-operator) — drescherjm, Nov 11 '22 at 14:53
@drescherjm While correct, it's still possible to do this with integers. — sweenish, Nov 11 '22 at 14:58
`01`, `1`, or even `000001` are all equal. Numbers don't have leading zeros. — Lukas-T, Nov 11 '22 at 14:58
You do not initialise n before you use it -> hence undefined behavoir — Mike Vine, Nov 11 '22 at 14:58
rev now has two responsibilities, handle user input and reverse the number. Better make that two functions. — Pepijn Kramer, Nov 11 '22 at 15:04

sweenish · Answer 1 · 2022-11-11T15:20:32.503

In your code, this line is problematic: n=n*10+r;

Since n was not given an initial value, the result of the right side expression is likely not what you think it is.

And as stated, integer types can't have leading zeroes. So trying to build a reversed int is not going to work.

There are a ton of ways to go about doing this. Unfortunately, you chose a method that can't work.

What comes to mind first is recursion.

#include <iostream>

void print_reversed(int n) {
  int next = n / 10;
  std::cout << n % 10;
  if (next != 0) {
    print_reversed(next);
  } else {
    std::cout << '\n';
  }
}

int main() {
  int number;
  std::cout << "Input: ";
  std::cin >> number;

  print_reversed(number);
}

Note that the recursive solution only prints one digit at a time. Leading zeroes are not an issue.

Or, take the input as a string.

#include <algorithm>
#include <iostream>
#include <string>

int main() {
  std::string number;
  std::cout << "Input: ";
  std::getline(std::cin, number);

  /*
   * Ideally you would validate the string before continuing.
   */

  std::ranges::reverse(number);
  std::cout << number << '\n';
}

std::ranges::reverse() requires C++20. There is a version you can call that takes a pair of iterators in earlier standards.

Or, break the number into its digits and store them in a container for printing.

#include <iostream>
#include <vector>

int main() {
  int number;
  std::cout << "Input: ";
  std::cin >> number;

  std::vector<int> digits;
  while (number != 0) {
    digits.push_back(number % 10);
    number /= 10;
  }

  for (auto i : digits) {
    std::cout << i;
  }
  std::cout << '\n';
}

Note that the digits are automatically stored in reverse order; there is no need to do anything after breaking it apart.

My preference is the recursive solution, with "input as a string" as a close second. I wouldn't really use this third method, but it still works.

For all three solutions, if I enter 10, the print-out is 01.

I simply write a program to reverse a number but the output not meet my expectation

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