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

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 );

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

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;

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returning dynamically created array from function

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