Error: "cannot Convert 'int*' to 'int**"

Question

I am trying to send the address of an array to a created function (which has as parameter a pointer that is also an array). The error is as follows:

main.cpp: In function 'int main()':
main.cpp:30:28: error: cannot convert 'int*' to 'int**' for argument '1' to 'int somadiferenca(int**)'
   somadiferenca(&numaboa[2]);    
exit status 1

I am compiling with repl.it, but the error is basically the same when compiling elsewhere.

The code is as follows::

#include <iostream>
using namespace std;

int somadiferenca(int *vetor[2])
{        
    int x;
    int y;

    x = (*vetor[0]) + (*vetor[1]);
    y = (*vetor[0]) - (*vetor[1]);

    if(y<0)
    {                
        y = -y;
        cout << x << endl << y;
    }
    else
    {
        cout << x << endl << y;
    }    
}
int main()
{
   int numaboa[2];

   cin >> numaboa[0];
   cin >> numaboa[1];

   somadiferenca(&numaboa[2]);

  return 0;

}

Answer 1

Miguel, it turns out that &numaboa[2] is the address of the index cell 2. You mean, &numaboa[0] is the same as numaboa (address of the first cell), &numaboa[1] is the same as numaboa+1 (address of the second cell) and &numaboa[2] is the same as numaboa+2 (I mean, it points to a cell that doesn’t even exist).

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

So, know that if the intention is to use the values of the cells then you can pass numaboa as argument in a type parameter 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 as follows, that somadiferenca accepts pointer, the somadiferenca(numaboa) in fact does pointer pass and within the function does cell access via pointer type 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 allocated consecutively in memory (execution stack);

2) numaboa is the constant (unalterable) pointer that always points to these two cells (numaboa is worth the address of the first, via index is addressed to others) seeing them as array;

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

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

Okay? Any doubt?

Answer 2

How 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;
}