reverse the position of integer digits? - c++

Question

i have to reverse the position of integer like this
input = 12345
output = 54321
i made this but it gives wrong output e.g 5432
#include <iostream>
using namespace std;
int main(){
int num,i=10;
cin>>num;
do{
cout<< (num%i)/ (i/10);
i *=10;
}while(num/i!=0);
return 0;
}

Answer 1

Here is a solution
int num = 12345;
int new_num = 0;
while(num > 0)
{
new_num = new_num*10 + (num % 10);
num = num/10;
}
cout << new_num << endl;

Answer 2

Your loop terminates too early. Change
}while(num/i!=0);
to
}while((num*10)/i!=0);
to get one more iteration, and your code will work.

Answer 3

If you try it once as an example, you'll see your error.
Input: 12
first loop:
out: 12%10 = 2 / 1 = 2
i = 100
test: 12/100 = 0 (as an integer)
aborts one too early.
One solution could be testing
(num % i) != num
Just as one of many solutions.

Answer 4

Well, remember that integer division always rounds down (or is it toward zero?) in C. So what would num / i be if num < 10 and i = 10?

Answer 5

replace your while statement
with
while (i<10*num)

Answer 6

If I were doing it, I'd (probably) start by creating the new value as an int, and then print out that value. I think this should simplify the code a bit. As pseudocode, it'd look something like:
output = 0;
while (input !=0)
output *= 10
output += input % 10
input /= 10
}
print output
The other obvious possibility would be to convert to a string first, then print the string out in reverse:
std::stringstream buffer;
buffer << input;
cout << std::string(buffer.str().rbegin(), buffer.str().rend());

Answer 7

int _tmain(int argc, _TCHAR* argv[])
{
int x = 1234;
int out = 0;
while (x != 0)
{
int Res = x % (10 );
x /= 10;
out *= 10;
out += Res;
}
cout << out;
}

Answer 8

This is a coding assignment for my college course. This assignment comes just after a discussion on Operator Overloading in C++. Although it doesn't make it clear if Overloading should be used for the assignment or not.

Answer 9

The following code works for a two-digit number only.
#include<iostream>
using namespace std;
int main() {
int n;
cin >> n;
cout << (n%10) << (n/10);
return 0;
}

Answer 10

int a,b,c,d=0;
cout<<"plz enter the number"<<endl;
cin>>a;
b=a;
do
{
c=a%10;
d=(d*10)+c;
a=a/10;
}
while(a!=0);
cout<<"The reverse of the number"<<d<<endl;
if(b==d)
{
cout<<"The entered number is palindom"<<endl;
}
else
{
cout<<"The entered number is not palindom"<<endl;
}
}

Answer 11

template <typename T>
T reverse(T n, size_t nBits = sizeof(T) * 8)
{
T reverse = 0;
auto mask = 1;
for (auto i = 0; i < nBits; ++i)
{
if (n & mask)
{
reverse |= (1 << (nBits - i - 1));
}
mask <<= 1;
}
return reverse;
}
This will reverse bits in any signed or unsigned integer (short, byte, int, long ...). You can provide additional parameter nBits to frame the bits while reversing.
i. e.
7 in 8 bit = 00000111 -> 11100000
7 in 4 bit = 0111 -> 1110

Answer 12

public class TestDS {
public static void main(String[] args) {
System.out.println(recursiveReverse(234));
System.out.println(recursiveReverse(234 ,0));
}
public static int reverse(int number){
int reversedNumber = 0;
int temp = 0;
while(number > 0){
//use modulus operator to strip off the last digit
temp = number%10;
//create the reversed number
reversedNumber = reversedNumber * 10 + temp;
number = number/10;
}
return reversedNumber;
}
private static int reversenumber =0;
public static int recursiveReverse(int number){
if(number <= 0){
return reversenumber;
}
reversenumber = reversenumber*10+(number%10);
number =number/10;
return recursiveReverse(number);
}
public static int recursiveReverse(int number , int reversenumber){
if(number <= 0){
return reversenumber;
}
reversenumber = reversenumber*10+(number%10);
number =number/10;
return recursiveReverse(number,reversenumber);
}
}

Answer 13

I have done this simply but this is applicable upto 5 digit numbers but hope it helps
#include<iostream>
using namespace std;
void main()
{
int a,b,c,d,e,f,g,h,i,j;
cin>>a;
b=a%10;
c=a/10;
d=c%10;
e=a/100;
f=e%10;
g=a/1000;
h=g%10;
i=a/10000;
j=i%10;
cout<<b<<d<<f<<h<<j;
}`