Thread: please help : arrays/parameters

This would work if 'wg_hr_py' in the calling code was a pointer to a pointer to a double but I doubt it is. Remove the '*' before it in the call to 'calc_pay' and see if it works then.

> This would work if 'wg_hr_py' in the calling code was a pointer to a pointer to a double
No.

wg_hr_py in the caller should be declared as
  double wg_hr_py[MAX_ROW][MAX_COL];
and the call is simply
  calc_pay(emp_counter, wg_hr_py);

It would work!

First, the '*' dereferences the pointer to the pointer and just the pointer is passed to the function. However, I don't think it is a pointer to a pointer so the '*' should be removed.

Jax: in August I posted a question about arrays also, so I just thought you might want to have a look at them. The answers are from "Prelude" and "ali c".

> It would work!
Ok, given
double wg_hr_py[MAX_ROW][MAX_COL];

Show me how you would prototype and write a function using whatever combination and number of *'s you want, and how you would then call that function.

I meant if wg_hr_py was a pointer to a pointer to a double:

double **wg_hr_py;

Suppose that some space is allocated at the pointer position.

If the calc_pay function was called like this:

calc_pay(emp_counter, *wg_hr_py);

It would work. The pointer to the double is passed.

> Suppose that some space is allocated at the pointer position.
How?
I can allocate memory to this pointer in 2 ways.

One way is consistent with the way the pointer is declared, but incompatible with the way the function is declared

The other way is inconsistent with the way the pointer is declared, but compatible with the way you call the function.

double ** is not the same as double[][], and they don't magically become the same thing when you call a function.

Both are pointers!

> Both are pointers!
Yes they are, but what types are they?

double **ptr;
When passed to a function has a type of double**

double arr[x][y];
When passed to a function has a type double (*)[y]