Storing data into array

Question

I am a self learner, and I was going through a free class online. I am trying to put the values of the everyOther into an array, so I can access it to later. I have looked around the internet, but am not able to find anything resourceful. Can you show me how to store the output values of everyOther in to an array. Thanks in advance.

#include <cs50.h>
#include <stdio.h>
#include <math.h>

long countDigit(long long n);

int main(void)
{
    long n;

   //This is asking for the input
    do
    {
        n = get_long("Number: ");
    }
    while(!(countDigit(n)>13));

    //Checksum math
    long everyOther = 0;

    while(n > 0)
    {
        long lastNumber = n/10;
        everyOther = lastNumber % 10;
        n = n / 100;
        printf("%li\n", everyOther);
    }

}

//This function helps us with the counting of the number
long countDigit(long long n) {
  return floor(log10(n) + 1);
}

Answer 1

C++ array, once created with certain length, cannot be expanded or shrank. This is an advantage when it comes to maximizing the space efficiency, but more than often poses a restraint on your code.

In your case, since you are taking a user input with an unpredicted length (>13), you risk creating an array without enough space.

You have two options:

1. Make a rule on the user's input and create array accordingly, for example, if you make sure that the there are maximum of 20 digits, then the array length could not exceed 20 with your

   long lastNumber = n/10; everyOther = lastNumber % 10;code.

What you can do to put everyother to array is to first create an array with length e.g, 20, by long foo [20] then transfer the everyother into the array each time with foo[i] = everyother remember to keep a counter on i to iterate to the next cell.

2. Instead of array, use an vector, which is basically dynamic array that the system take care of the space restraint for you. This is not the most efficient but is fool-proof. To use this, #include <vector> first, then simply declare an vector with vector<long> foo; then add everyother with foo.push_back(everyother) on each iteration of the while loop.

To reuse both of these method, simply call foo[index_of_position] to extract the everyother information stored.

Remember that every datatype you try to store in these structures has to match the type they are created with (most of the time). e.g, a vector<int> would not accept a string object.

Answer 2

If I were you I would #include <vector> at the top, then declare std::vector<long> vectorArray;. You can then put the values into the vector, which is a dynamically sized array. It should be said though that this is a C++ feature, and will not work in C, but since this post is tagged with both I am assuming it will be fine.

// #include <cs50.h>
#include <iostream>
#include <stdio.h>
#include <math.h>
#include <vector>

long countDigit(long long n);

int main(void)
{
    long n = 0; // initialise n

    //This is asking for the input
    do
    {
        std::cin >> n; // using this as input
    }
    while(!(countDigit(n)>13));

    //Checksum math
    long everyOther = 0;

    std::vector<long> vectorArray;
    
    while(n > 0)
    {
        long lastNumber = n/10;
        everyOther = lastNumber % 10;
        vectorArray.push_back( everyOther ); // this appends everyOther to the end of the array
        n = n / 100;
        printf("%li\n", everyOther);
    }

    for ( auto x : vectorArray )
        std::cout << x << ' ';      // sample printout of vector's contents
    
}

//This function helps us with the counting of the number
long countDigit(long long n) {
  return floor(log10(n) + 1);
}

Answer 3

Using std::vector:

#include <cs50.h>
#include <math.h>
#include <stdio.h>
#include <vector>

long countDigit(long long n);

int main(void) {
  long n;

  // This is asking for the input
  do {
    n = get_long("Number: ");
  } while (!(countDigit(n) > 13));

  // Checksum math
  long everyOther = 0;

  std::vector<long> destination;

  while (n > 0) {
    long lastNumber = n / 10;
    everyOther = lastNumber % 10;
    n = n / 100;
    printf("%li\n", everyOther);

    destination.push_back(everyOther);
  }
}

// This function helps us with the counting of the number
long countDigit(long long n) { return floor(log10(n) + 1); }

As you are iterating over the digits of a random number, it is hard to use std::array efficiently, as it has a size set at compile-time. You could however make an array of a gigantic size, and use it as a buffer.

#include <array>
#include <cs50.h>
#include <math.h>
#include <stdio.h>

#define BUFFERSIZE 65536

long countDigit(long long n);

int main(void) {
  long n;

  // This is asking for the input
  do {
    n = get_long("Number: ");
  } while (!(countDigit(n) > 13));

  // Checksum math
  long everyOther = 0;

  std::array<long, BUFFERSIZE> destination;

  size_t i = 0;
  while (n > 0) {
    long lastNumber = n / 10;
    everyOther = lastNumber % 10;
    n = n / 100;
    printf("%li\n", everyOther);

    destination[i++] = everyOther;
  }
}

// This function helps us with the counting of the number
long countDigit(long long n) { return floor(log10(n) + 1); }