Passing Arrays as Parameters and Returning Them in C++

// createArray_1 returns the array as a return value
double* createArray_1( ) {
    return new double [ 10 ];
}

// createArray_2 returns the array from the parameter list
// (using a reference parameter)
void createArray_2( double*& arr ) {
    arr = new double [ 10 ];
}

// createArray_3 returns the array from the parameter list
// (without using a reference parameter but simulating
// pass-by-reference using a pointer)
void createArray_3( double** arr ) {
    *arr = new double [ 10 ];
}

// What is wrong with the following two functions?
// void incorrectCreateArray_1( double* arr ) {
// arr = new double [ 10 ];
//}
// double* incorrectCreateArray_2( ) {
// double arr[ 10 ];
// return arr;
// }

And we have the main function:

int main() {
   double* D;
   D = createArray_1();
   delete [] D;
   createArray_2( D );
   delete [] D;
   createArray_3( &D );
   delete [] D;
   return 0;
}

Can you help me understand why create_array2 and create_array3 are correct whereas incorrectCreateArray_1 and incorrectCreateArray_2 are wrong?

To me, incorrectCreateArray_1 should be fine because we are passing a pointer, and then assign a double array of size 10 to it, which seems correct.

On the other hand, incorrectArray_2 returns a double pointer, which should be fine because arr points to a double array, which also seems correct.

>Solution :

void incorrectCreateArray_1( double* arr ) 
{
    arr = new double [ 10 ];
}

is incorrect because you receive a pointer (uninitialized) and you make it point somewhere else. But only the arr pointer, local to this function, gets changed. The calling code still keeps its (uninitialized) pointer.

double* incorrectCreateArray_2( ) 
{
    double arr[ 10 ];
    return arr;
}

is incorrect because you return a pointer to a local object (arr) which will not be valid to access after the function returns.