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Recursive Function to Reverse an Integer
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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.
???? 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.
Why would you need to use a char array? using int array should be fine.
It might be logically flawed somewhere..just make sure you have a base case (where you would stop calling the reverse function).
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).
Last edited by freudian_slip; Nov 4th, 2008 at 12:54 am.
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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.
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.
raw scetch of how I should start to do it:
revprint(int n)
{
if n==0 return
cout<<n%10
revprint(n/10)
}
revprint(int n)
{
if n==0 return
cout<<n%10
revprint(n/10)
}
Today is a gift, that's why it is called "The Present".
Make love, no war. Cave ab homine unius libri.
Danny
Make love, no war. Cave ab homine unius libri.
Danny
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
something like
C++ Syntax (Toggle Plain Text)
Last edited by mrboolf; Nov 4th, 2008 at 10:32 am.
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