returning dynamically created array from function

Question

I'm trying to return the solutions of the pq-formula as a dynamically created array. What's the correct way to do that? This is my function:

double *pq (double a, double b)
{
 double x1=(-1)*(a/2)-sqrt((a/2)*(a/2)-b);
 double x2=(-1)*(a/2)+sqrt((a/2)*(a/2)-b);
 double *arr[]=(double *)malloc(2*sizeof(double));
 arr[2]={{x1}, {x2}};

 return arr;

}

Also, why do I get an 'expected an expression' error on arr[2]={{x1}, {x2}}; ?

My main function:

    int main ()
{
    double *arr[2]={0}, a=0.00, b=0.00;

    scanf("%lf %lf", a,b);

    if ((a*a)-(b*a)>=0)
    {
        for (int i=0; i<2; i++)
        {
            arr[i] = pq(a,b);
        }   
    }

    else
    {
        printf("Es gibt keine reellen L\224sungen.");
    }
 
    for (int i=0; i<2;i++)
    {
        printf("%lf", arr[i]);
    }


    return 0;
}

Your arr variable is the wrong type, and even if corrected you'll breach your array (indexing is zero based, so [1] is the max index for an array of length 2). — WhozCraig
– WhozCraig, Commented Jul 20, 2021 at 11:39
Consider returning a struct with an array of two doubles and an int counting the number of real solutions and avoid dynamical allocation at all. — Bob__
– Bob__, Commented Jul 20, 2021 at 11:44

Vlad from Moscow · Accepted Answer · 2021-07-20 11:48:11Z

1

Instead of these lines

 double *arr[]=(double *)malloc(2*sizeof(double));
 arr[2]={{x1}, {x2}};

 return arr;

you need to write within the function pq

double *arr = malloc( 2 * sizeof( double ) );

if ( arr != NULL )
{
    arr[0] = x1;
    arr[1] = x2;
}

return arr;

And in main

double *arr = NULL;
double a = 0.0, b = 0.0;

scanf("%lf %lf", &a, &b );
                 ^^^^^^
if ((a*a)-(b*a)>=0)
{
    arr = pq( a, b );
}
else
{
    printf("Es gibt keine reellen L\224sungen.");
}

if ( arr != NULL )
{
    for (int i=0; i<2;i++)
    {
        printf( "%f", arr[i] );
                 ^^^
    }
}

free( arr );

answered Jul 20, 2021 at 11:48

Vlad from Moscow

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Comments

Adrian Mole · Accepted Answer · 2021-07-20 11:46:45Z

You can't initialize such a dynamically allocated array en bloc. Instead, assign values to each element, in turn. With a little reordering of your function, you can even remove the need for your intermediate (x1 and x2) variables:

double *pq (double a, double b)
{
    double *arr = malloc(2*sizeof(double)); // No need to cast!
    arr[0] = (-1)*(a/2)-sqrt((a/2)*(a/2)-b);
    arr[1] = (-1)*(a/2)+sqrt((a/2)*(a/2)-b);
    return arr;
}

On the casting of the return value of the malloc function, see: Do I cast the result of malloc?

Also, you have to change the way your main function works; don't declare a local array and try to assign data after the call; just use the 'array' returned from your function, as the elements' values will already be there:

int main ()
{
    double a=0.00, b=0.00;

    scanf("%lf %lf", &a, &b); // Note the address (&) operators!

    if ((a*a)-(b*a)>=0)
    {
        double *arr = pq(a, b);
        for (int i=0; i<2; i++)
        {
            printf("%lf", arr[i]);
        }
        free(arr); // Don't forget to free the memory!
    }
    else
    {
        printf("Es gibt keine reellen L\224sungen.");
    }
    return 0;
}

Cory Kramer · Accepted Answer · 2021-07-20 11:39:36Z

0

The problem is with this line

arr[2]={{x1}, {x2}};

arr[2] = is assigning to the 3rd element of the array, which is out of bounds. And you cannot use that brace syntax to assign to an array slice like that. Instead

arr[0] = x1;
arr[1] = x2;

answered Jul 20, 2021 at 11:39

Cory Kramer

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1 Comment

WhozCraig Over a year ago

double *arr[] isn't doing them any favors either.

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