Error: "can not convert 'int *' to 'int **"

Miguel, it turns out that &numaboa[2] is the address of the index cell 2 . That is, &numaboa[0] gives the same as numaboa (address of the first cell), &numaboa[1] gives in the same as numaboa+1 (address of the second cell) and &numaboa[2] gives in the same as numaboa+2 , points to a cell that does not even exist.)

In addition, the int v[] parameter indicates that v is a pointer because the array (or vector) symbol is treated as a constant pointer that is worth the address of the first cell, that is, the int *vetor[2] indicates that vetor is a pointer type for another pointer, this is for integer.

So, know that if you intend to use cell values then you can pass numaboa as an argument to a parameter of type int* (that is, function can be void somadiferenca( int *vetor ) or void somadiferenca( int vetor[] ) or void somadiferenca( int vetor[2] ) ) and then access the two cells using vetor[0] and vetor[1] .

A valid way to implement everything is the following, which somadiferenca accepts pointer, somadiferenca(numaboa) actually does pointer passing and inside the function makes cell access via pointer parameter.

#include <iostream>
using namespace std;

int somadiferenca( int *vetor ){        
    int x = vetor[0] + vetor[1] ;
    int y = vetor[0] - vetor[1] ;
    if( y<0 ){       
        cout << x << endl << (-y) ;
    }
    else {
        cout << x << endl << y;
    }    
}

int main(){
   int numaboa[2] ;
   cin >> numaboa[0] ;
   cin >> numaboa[1] ;
   somadiferenca( numaboa ) ;
   return 0 ;
}

Remembering some basic things:

1) int numaboa[2] is the definition of two temporary local cells in the function that are consecutively allocated in memory (execution stack);

2) numaboa is the constant (unchanging) pointer that always points to these two cells ( numaboa is the address of the first, via index addresses other) seeing them as array;

3) numaboa[i] is an array cell and numaboa+i is the same as &(numaboa[i]) , ie the address of the cell numaboa[i] ;

4) Parameters int vector[] and int *vector are the same, both pointers, after all symbols of arrays like vetor allocated in the stack are treated as pointers.

Okay? Any questions?

What about:

#include <iostream>

using namespace std;

void somadiferenca( int vetor[2] )
{
    int x = vetor[0] + vetor[1];
    int y = vetor[0] - vetor[1];

    if( y < 0 )
        y = -y;

    cout << "Soma: " << x << endl;
    cout << "Diferenca: " << y << endl;
}

int main()
{
    int numaboa[2];

    cout << "a: ";
    cin >> numaboa[0];

    cout << "b: ";
    cin >> numaboa[1];

    somadiferenca( numaboa );

    return 0;
}

Or:

#include <iostream>

using namespace std;

void somadiferenca( int * vetor )
{
    int x = *vetor + *(vetor+1);
    int y = *vetor - *(vetor+1);

    if( y < 0 )
        y = -y;

    cout << "Soma: " << x << endl;
    cout << "Diferenca: " << y << endl;
}

int main()
{
    int numaboa[2];

    cout << "a: ";
    cin >> numaboa[0];

    cout << "b: ";
    cin >> numaboa[1];

    somadiferenca( &numaboa[0] );

    return 0;
}