incompatible pointer type?

Hey,
I need to pass a **pointer of an array as a function argument, but for some reason I keep getting the error "warning: assignment from an incompatible pointer type". As you can tell I'm pretty new to using C, so I'll post my code and hope someone can point out the blatant error :).

double x[2][2],y[2][2];
double **u;
double **v;

x[0][0]=x[1][0] = 4;
x[0][1]=x[1][1] = -3;

y[0][0]=y[1][0] = 1;
y[0][1]=y[1][1] = -1;

u = &x; //this is where the compiler error occurs.
v = &y;

functioncall(u,v); //these arguments need to be of the type const double**

Cheers for any help,
Kieran

You should use

Code:

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u = x;

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Because u and x are both pointers to pointers. x[2][2] is like **x.

By using * you dereference a variable (you get the value the pointer is pointing to).
By using & you get the address of a variable (address is the location in memory).

So using

Code:

---

u = &x;

---

you make u point to x as if it is declared as ***x. Pointers can be hard in the beginning.

Quote:

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Originally Posted by Ideswa

You should use

Code:

---

u = x;

---

Because u and x are both pointers to pointers. x[2][2] is like **x.

By using * you dereference a variable (you get the value the pointer is pointing to).
By using & you get the address of a variable (address is the location in memory).

So using

Code:

---

u = &x;

---

you make u point to x as if it is declared as ***x. Pointers can be hard in the beginning.

---

I've already tried that, still get the same error :S.

The name of an array is not actually a pointer. It's close for most purposes, but it's not the same thing. It's actually a pointer to an array of type T. So you're incorrect when you try to pass it as a pointer to a pointer. You probably want something like this.

Quzah.