Q: How to use a bidimensional array to calculate and store frequencies of values from other array? I'm beggining to study C programming,so I don't know how to do basic things (may be what I do has no-sense).
What I have to do is to get from the user an undefined number of integers from 1-9 and load it to an array, when the user entered 0 or 10 the loop finishes and the array was completed, so I have: 
 int qualifications[SIZEQ] and a function to load it, it works.
Now I have to calculate the frequency of each number introduced on this array. I suppose that there are a lot of ways to do it simple (I don't know any), I'm trying to do it with another array, bidimensional, with 2 rows and 10 cols, using 1 row to contain the values from 0 to 9 to compare each number to the numbers of array qualifications[]. And the other row to store the number of times that each number is present on qualifications. I don't know how to do it, I try:
// 1st row of the freq array contains the 1-9 possible qualifications
// 2nd row to count the times that every value appears in qualifications array
// example: 1st row {0,1,2,3...} if 1 repeated 3 times:
//          2nd row {0,2,0,0...}  

int freq[2][10]={{0,1,2,3,4,5,6,7,8,9},{0,0,0,0,0,0,0,0,0,0}};

    int  j=0;   //var to load 0-9 positions of freq[1][j]

    for (int i = 0; i < SIZEQ; i ++)
    {
        if(qualifications[i]==freq[0][j]){
//what i want to do is to compare from 0 each cell of
//qual with freq with 1-9 numbers

        freq[1][j] = freq[1][j] + freq[1][j];
        //if detected a coincidence, increment the field of the row to
        //count repetitions
        //so, if qualifications[i] it's now '2' when compared to 
        //freq[0][2] (what it's 2) freq[1][2] it's now 1.  
        }
        j=i; 

        if(j>=9){ //to avoid that j be bigger than freq column size
        j=0;
        }

    }
    j=1; //skip 0
        while(j<=9){
            //j print 1-9 numbers and freq[1][j] print the number of
            //times it is repeated 
                cout << "Qualification " << j << " repeated " << freq[1][j] << " times." << endl;
            j++;
        }
}


A: In your original code you are not checking each 'qualification' against each possible value in 'freq'.  
I think something like this is what you're looking for.  
#include <iostream>
using namespace std;

int main()
{
  //make an example 'qualifications'
  int SIZEQ = 5;
  const int SIZEQ = 5;
  int qualifications[SIZEQ] = {1,1,2,3,4};




  int freq[2][10]={{0,1,2,3,4,5,6,7,8,9},{0,0,0,0,0,0,0,0,0,0}};

  for (int i = 0; i < SIZEQ; i ++) // loop through each qualification
  {
    for (int j = 0; j < 10; j++)   // loop through each frequency to see if it matches
    {
      if(qualifications[i]==freq[0][j])
      {
        freq[1][j] += 1;
      }
    }

  }
  int k;
  k=1; //skip 0
  while(k<=9){
    //j print 1-9 numbers and freq[1][j] print the number of
    //times it is repeated
    cout << "Qualification " << k << " repeated " << freq[1][k] << " times." << endl;
    k++;
  }



  return 0;
}

the output should be: 
Qualification 1 repeated 2 times.
Qualification 2 repeated 1 times.
Qualification 3 repeated 1 times.
Qualification 4 repeated 1 times.
Qualification 5 repeated 0 times.
Qualification 6 repeated 0 times.
Qualification 7 repeated 0 times.
Qualification 8 repeated 0 times.
Qualification 9 repeated 0 times.

