getting the length of an dynamic array back into main...

[Bdanielz]

Working on a exercise out and I would like a little help with one part of my code.. Not really looking for people to solve the exercise for me.

Using a function to turn an int - into a array for futher processing.

What I think needs to happen is (guessing and checking here) is call the length
of the dynamic array as a pointer, as the function I am using is already returning an array? And I thought arrays could not return via functions, so I am feeling a little confused here. I guess if some one could fill in my comments in my code, with valid code It would help me understand what is happening here.

// Function Source http://stackoverflow.com/questions/515612/convert-an-integer-number-into-an-array for function 

#include <iostream>
#include <math.h>
char * convertNumberIntoArray(unsigned int number);

using namespace std;

int main(int argc, const char * argv[])
{

    int digit;
    cout << "Please enter in a digit: ";
    cin >> digit;
    cout << endl;

    convertNumberIntoArray(digit);

    // I would like to get the arr back into main
    // and do somthing like this...

    //    for (int i = 0; i < arr.length; i++) {
    //        # Do stuff..
    //    }



    return 0;
}

char * convertNumberIntoArray(unsigned int number) {
    int length = (int)floor(log10((float)number)) + 1;
    char * arr = new char[length];
    int i = 0;
    do {
    	arr[i] = number % 10;
    	number /= 10;
    	i++;
    } while (number != 0);
    return arr;
}

[Josue Molina]

Indeed, a function cannot return an array, however, it can return a pointer to one.

convertNumberIntoArray returns a pointer to a character, namely, the first element of the array it allocates.

In main, you call convertNumberIntoArray, but don't store its return value. Do something like this:

char* arr = convertNumberIntoArray(digit);

Sadly, a pointer to an array and an array are two different animals: you cannot determine the length of a dynamically allocated array; you're going to have to calculate it manually or deduce it logically.

[Bdanielz]

thank you.. This gives me some traction. Will report back with my solution.

[Bdanielz]

Another problem (:)) that is jumping out at me now. Is I also have to turn the new charr array

char* arr = convertNumberIntoArray(digit);

back into ints in order to perform math/arithmetic on the array members.

[Bdanielz]

This is going to get messy.

input
12345

code

#include <math.h>
char * convertNumberIntoArray(unsigned int number);

using namespace std;

int main(int argc, const char * argv[])
{

    int digit;
    cout << "Please enter in a digit: ";
    cin >> digit;
    cout << endl;

    char* arr = convertNumberIntoArray(digit);

    cout << (int)arr[0] << " " << (int)arr[1]<< " " << (int)arr[2] << " "<< (int)arr[3] << " "<< (int)arr[4] << " "<< (int)arr[5] << " " << endl;


    return 0;
}

char * convertNumberIntoArray(unsigned int number) {
    int length = (int)floor(log10((float)number)) + 1;
    char * arr = new char[length];
    int i = 0;
    do {
    	arr[i] = number % 10;
    	number /= 10;
    	i++;
    } while (number != 0);
    return arr;
}

output

5 4 3 2 1 127 

Thank you everyone for your help!

[Josue Molina]

If you have no need for a character array, then don't use one; change your function:

int* convertNumberIntoArray(int number)
{
    int length = (int)floor(log10((float)number)) + 1;
    int* arr = new int[length];
    int i = 0;
    
    for (int i = 0; number != 0; ++i)
    {
        arr[i] = number % 10;
        number /= 10;
    }

    return arr;
}

This one returns an array of integers instead.

[Bdanielz]

Wow. Thank you for pointing that out. That is the trouble with using other peoples code, that I don't fully understand. (have not arrived at dynamic arrays yet - In my studying - working through conditional programs currently. So I am getting ahead of my self a bit. )

Thank you again!