Convert one function procesr(n,C) in matrice[i][j] ant print it in the screen

[kristov]

Hi,I made a program which gives all possible combinations in the screen of DEVC++.

So I want to print the same result in the screen, but in this case like one matrice[i][j], because then my program have continue with matrice, not with function. Many Thanks!

The function : procesar give me the next result in the screen:

123
 213
 231
 321
 312
 132

All these dates ( rows&cols) I want to present with one matrice[6][3]
cout<<" The converted matrice is :"<<matriz[i][j];

Because the previos results it appear like one function:procesar(n,C)
The code of the combination function is :

#include <iostream.h>
#include <conio.h> 
using namespace std;
 
int procesar(int n, int C[])
 {
     int i;
     int inc = 0;
     int it = n * (n - 1);
 
     while (inc < it) 
     {
         for (i = n - 1; i > 0; i--) 
         {
             int aux = C[i-1];
             C[i-1] = C[i];
             C[i] = aux;
             for (int j = 0; j < n; j++)
 
                 cout <<" "<< C[j];
             cout << endl;
             inc++;
           }}}
int main()
 {
     int n;
 
     cout << "How many numbers of digits you want to use for a combination: ";
     cin >> n;
 
     int C[n];
 
     for (int i = 0; i < n; i++) 
     {
         cout << "Input digit: " << i + 1 << ": ";
         cin >> C[i];
     }
 
     procesar(n, C);
 
 
 getch();
 
 
     return 0;
 }

[weaknessforcats]

You need a function like this one you wrote

int procesar(int n, int C[])
{ etc...
 

but takes a matrice argument.There are a number of ways to do this (read http://bytes.com/topic/c/insights/77...rrays-revealed ).

The easiest way is to have an argument for the first dimension, the second dimension and the address of element
[0][0]. Since the matrice contains int, the argument would be an int*:

int procesar(int i, int j, int* c)
{ etc...
 

In this function you have an outer loop than goes from 0 to i and inside this loop is a secnd loop that goes from 0 to j.

Unside this second loop you can print c[i][j] but you have to calculate it since c is only an int* making c a one dimensional array.

The calculation is that i (a row) has j elements. So i*j gives the row where j is the col. c[i*j + j] is the element. This si what you print inside the secnd loop.

When you exit the inner loop you can print the \n in the outer loop before going back to the top of the outer loop.

[kristov]

Thank´s a lot weaknessforcats ,I will try to make solution with your advise.I will reply you soon,let´s see what I can make about this problem,thank´s onceagain.....

[kristov]

Thank´s a lot weaknessforcats ,I will try to make solution with your advise.I will reply you soon,let´s see what I can make about this problem,thank´s onceagain.....
By the whay, I am making another works,that´s why I reply to you not right now,ok.

[kristov]

Excuseme weaknessforcats , can you show me the code for my case in dettails,because I didn´t understood it.

[weaknessforcats]

Did you read the article?

The solution you want is there.