Thread: Advanced pointer use

Originally Posted by mvanrijnbach

I just don't understand why a is a pointer to an array of 5 float elements. I would think it is an array to n*5 elements. Does anybody see what I'm overlooking here?

It just happens to be the syntax of how to declare a pointer to an array. Think about the book's point that *a is the array (although it does not exist until later), thus a is a pointer to an array.

By the way, you can simplify this:

Code:

a = (float(*)[5])malloc(n*sizeof(float[5]));

to:

Code:

a = malloc(n * sizeof(*a));

Originally Posted by boxden

You know one dimension in your initialization (5). For every element of your original a you are allocating another n size array of which you return the address.

Not quite. It is more like space is allocated for n arrays of 5 floats, and a points to the first of these n arrays.

Originally Posted by boxden

So you are actually working with pointers to pointers.

No, pointer to an array, not pointer to a pointer.

Originally Posted by boxden

The name of an array is a pointer?

Nope. In many contexts, an array is converted to a pointer to its first element. That would be so for a[i], but then it would not be correct to try and pass a where a float** is required.

Yes, I am not sure if this is what laserlight was referring to, but the subject of pointer declarations and dereferencing is often a criticized design choice in C.

A lot of beginners are confused by a similar notation that means two different things: when you define a pointer:

Code:

int *p;

;

The SYMBOL * just identifies the fact that p is a pointer to type int.

Code:

*p = 5;

Later on in the code if you do this, you are saying "write to the address pointed by p the value 5".

What helped me greatly when learning C was my teacher always asking me what a pointer is when I got confused. The answer was always : "A POINTER IS AN ADDRESS".

In your case:

Code:

float (*a)[5];

*a identifies that a is a pointer (because it is a definition).
(i.e. an address to a type float [5], so to an array of 5 floats, which are obviously stored consecutively in memory because that's what an array does).

Perhaps this will help. The data is stored exactly as if one had done this:

Code:

typedef struct foo {
    float bar[5];
};

struct foo *a;

The only difference is that instead of having to write

Code:

a[i].bar[j]

you can omit 'bar' and just write

Code:

a[i][j]

As you can see, 'a' can point to some number of items that each contain 5 floats. You can also see from such a definition that this is not compatible with float** because there is no double-pointer involved.
Just to reiterate, the second dimension is not a pointer, it is the array itself, with a defined size.