Recursive Function to Reverse an Integer

[badmofo]

I have an assignment to write a recursive function that takes one integer variable and output the integer in reverse order to the screen. I've tried converting it to a char array(c++ "string" class not allowed) but could not get a counter to work. I can reverse an int array with and "upper" and "lower" variable no problem, but I'm stuck without them.

???? reverse(int num) //This is what the assignment calls for.

I'm not looking for anyone to write it for me but I would appreciate some tips.
Thanks in advance.

[freudian_slip]

Why would you need to use a char array? using int array should be fine.

reverse (num)
   array[count] = num
   count ++
   get a new num
   reverse (num)
   print out array[count]
   count--

It might be logically flawed somewhere..just make sure you have a base case (where you would stop calling the reverse function).

[dougy83]

If you're just displaying the reversed thing to the screen, there is no need to store any numbers in any arrays.

Each call to the function will print out a single digit (the digit of number in the units place). If the number is more than 9, the function will call itself with number/10 as the argument. This will recursively print out the number in reverse. When all digits have been printed, the function will stop, and return, due to the above stated call condition.

[ddanbe]

raw scetch of how I should start to do it:
revprint(int n)
{
if n==0 return
cout<<n%10
revprint(n/10)
}

[mrboolf]

I'd do it that way too, just I'd add a little if-else-abs() trick to handle n<0 (otherwise it would print -567 reversed as -7-6-5).

something like

if(n<0) {
     cout << "-" << abs(n%10);
     revprint(n/-10);
}
else {
     cout << n%10;
     revprint(n/10);
}

[badmofo]

Thanks everyone! I got it working using "n % 10" I was making it too complicated as usual

[ddanbe]

So you can mark this as solved.

[anuj khasgi]

if we give a input which has a digit 0 in between it then loop will end there and output will be incorrect
for eg 1034
the loop will end at the second digit 0 and otput will be incorrect

what to do in this case??

[Ali_40]

dear i think this would do the trick. its a function, all u need to do is call it in your main program(e.g. flip_integer(12345) will show 54321)

void flip_integer(int number)
{
        int var=0;
        int var2=0;
        var=number;

        var2=var%10;
        var=(var-var2)/10;
        cout<<var2;

        if(var!=0)
        {
        flip_integer(var);
        }
}

hope this helps!