Help with recursive function

Hi all,

I am trying to solve the following task:
-Write a recursive function that prints an integer with decimal separators. For example, 12345678 should be printed as 12,345,678.

The problem is that I don't know how to modify the integer in such way. I was thinking to convert it into a string, do an algorithm and then turn the string back to integer.

Please give me some pointers or advises on how to do such modification.

Thanks in advance.

In fact you can't do that at all with the integer variable alone. It stores the integral number as such, but no number formatting information whatsoever. (In fact, even though it's commonly output in decimal for easy reception by humans, the integer varable isn't even decimal.)

As I understand your assignment, it's asking you to convert the integer into a string in an appropriate way (as you already suggesed) and then print that. You're not even required to preserve the string holding the conversion result; you can simply discard it after printing. And, in particular, you're not required to put the formatted integer value back into the original integer variable, which would be impossible anyway, as pointed out above.

Quote:

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Originally Posted by Symus

-Write a recursive function that prints an integer with decimal separators. For example, 12345678 should be printed as 12,345,678.

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An alternative to building a string is to split the integer numerically and print the parts separated by commas to standard output.

The standard way to strip off individual digits from a number is to use the % (modulo) operator followed by integer division. For a decimal number (base 10) it would look like this,

12345678 % 10 = 8
12345678 / 10 = 1234567

The number has effectively been split like this,

12345678 = 1234567*10 + 8

In this particular case you could instead split into 3-digit chunks,

12345678 % 1000 = 678
12345678 / 1000 = 12345

In the recursive function you decide what number is to be printed and if it's to be preceded by a comma. Then the function calls itself with the reduced number until 0 is reached. Finally when the recursion winds back the actual printing takes place according to the information gathered in each incarnation of the recursive function. In this process the number will be split from right to left and then printed with commas from left to right.

Phew, it's harder to describe than to code. :)

Thanks for pointing that out, nuzzle. :) If I would have to code that and I had the choice, I'd have taken the iterative approach (using the modulo operator, as you described), which would build the string to print back-to-front and only print it when done. However, it was rather late here when I wrote my post, and that probably was the reason why I overlooked that string reversal can elegantly be done implicitly by doing the printing in the recursion unwinding phase.

The way you describe it no string varable would be required at all, and, strictly speaking, to encapsulate the objective "print that integer" in a black-box funtion taking just the int parameter, for my original approach I'd actually have needed two functions: a recursive one that does the int-to-string conversion and one that takes a single int parameter, calls the recursive conversion function, and eventually prints the conversion result.

Hello guys,

Thanks for your help and suggestions. I managed to solve this task without converting to string. Here is my code if someone is interested:

Code:

---

#include <iostream>
#include <cstdlib>

using namespace std;

void separate(int number)
{
    // If the number is below 1000 - print it because it holds exactly or below 3 digits
    if (number < 1000)
    {
        cout << number;
    }
    else
    {
        int temp = number % 1000; // Variable which holds the remainder of the calculation (the last 3 digits of the integer)
        number = number / 1000; // Remove the last 3 digits and pass the remaining of the integer again to the function
        separate(number);
        cout << "," << temp;
    }
}

int main()
{
    int input;
    cout << "Enter your number :" << endl;
    cin >> input;
    separate(input);
    cout << endl;

    system("PAUSE");
    return 0;
}

---

I really like your approach for its pragmatism of just separating the thousands groups yourself, leaving conversion of the individual digits inside each group to the C++ standard library. :cool: However, your implementation isn't really perfect yet, which you can easily see when you're entering a simple number such as ten millions. ;)

Quote:

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Originally Posted by Symus

Hello guys,

Thanks for your help and suggestions. I managed to solve this task without converting to string. Here is my code if someone is interested:

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What data did you test this with?

Try 1001. You will see that this doesn't work.

Regards,

Paul McKenzie

OK, why is this not working? :(

And how can I fix it?

Code:

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cout << "," << temp;

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cout as default does not output leading 0's to pad fill a number to a set width! You have to tell cout what width to use and what fill character to use if you want leading '0's output as you do here. Hint. look up manipulators.

I finally managed to resolve the situation. And here it is the final code:

Code:

---

#include <iostream>
#include <cstdlib>
#include <iomanip>

using namespace std;

void separate(int number)
{
    // If the number is below 1000 - print it because it holds exactly or below 3 digits
    if (number < 1000)
    {
        cout << number;
    }
    else
    {
        int temp = number % 1000; // Variable which holds the remainder of the calculation (the last 3 digits of the integer)
        number = number / 1000; // Remove the last 3 digits and pass the remaining of the integer again to the function
        separate(number);
        cout << "," << setfill('0') << setw(3) << temp;
    }
}

int main()
{
    int input;
    cout << "Enter your number :" << endl;
    cin >> input;
    separate(input);
    cout << endl;

    system("PAUSE");
    return 0;
}

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