reverse the position of integer digits?

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         


        

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

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 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

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 5

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 6

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;
}`