1

I have followed the following webpage:

http://www.1024cores.net/home/lock-free-algorithms/reader-writer-problem/improved-lock-free-seqlock

the source is like following:

struct data_t 
{ 
    int seq; // sequence number 
    int data [1024]; // user data 
}; 

struct seqlock_t 
{ 
    data_t* current; 
    data_t pool [16]; // 16 user objects are held 
}; 

seqlock_t sl;

The structure is quite simple, what confuses me is the following:

data_t* d0 = sl.current; // load-consume
int idx = (sl.current - sl.pool + 1) % 16; 
data_t* d = &sl.pool[idx];

The sl.current is a pointer, sl.pool is? What current - pool can achieve? in c language view, how should I explain this statement? 

int idx = (sl.current - sl.pool + 1) % 16;

Edit :

Thanks for all information , it help a lot !!! in my own coding style would use int idx = (sl.current - &[sl.pool[0]) + 1) % 16; now I know &(sl.pool[0]) is the same with sl.pool !!! I should figure it out , the following example , what I read before , showes pointer/array ....

void display(int *p)
{
    int idx ;
    printf("(%d)(%d)\n",*p,p[1]) ;
}
int main()
{
    int i,Arr[10] ;
    for(i=0;i<10;i++)
        Arr[i] = i + 100;
    display(Arr) ;
}

Yes , I can use display(Arr) and display(&Arr[0]) , they are the same ~!!

Improve this question
2
  • pool is an array of data_t. So, pool can be used for pointer operations. Commented Oct 22, 2013 at 9:08
  • 1
    It occurs to me that a strong article on pointer math is C must be somewhere on SO, but for the life of me I can't find one I'd recommend. If anyone reading this has a good candidate by all means toss a link to it; as I think the OP can use it right about now. Edit: ok. this is a pretty damn good one. Commented Oct 22, 2013 at 9:19

3 Answers 3

Reset to default
6

sl.current is a pointer. sl.pool is an array, and by writing like that it is equal to &sl.pool[0]. So it points to the first element of the array. We can do some pointer operations such as subtraction, in that case to obtain some index.

I find this link http://www.eskimo.com/~scs/cclass/notes/sx10b.html quite simple to understand some pointer arithmetic.

Improve this answer
Sign up to request clarification or add additional context in comments.

5 Comments

sl.pool is not a pointer. It is an array. They're not the same. A pointer is a variable that holds an address; an array is a variable that effectively is an address. Were sl.pool a pointer, sl.pool++ would be valid; it isn't.
I do not understand the first sentence. It reads as if sl.pool and sl.pool[0] were the same - which isn't true.
@glglgl I concur. They're not the same. &sl.pool[0], sl.pool + 0 and sl.pool are all equivalent; sl.pool[0] is equivalent to exactly none of those.
@Abhineet I know how pointer arithmetics works. But the statement in that sentence is just not true.
@rcs Perfect. Thus, +1 from me.
2 
data_t* d0 = sl.current; // load-consume

gets the pointer into a local variable or even a register in order to access easier.

int idx = (sl.current - sl.pool + 1) % 16;

It is assumed that sl.current points to one of the elements of the array sl.pool. So sl.current - sl.pool gets the respective distances in terms of data_t *, so it gets the current index. + 1 advances to the next index and with % 16 you achieve that a value of 16 - which would be illegal - points to 0 instead.

data_t* d = &sl.pool[idx];

This points to the new one; I suppose later comes a sl.current = d or something like that.

Improve this answer

1 Comment

The part of this concept that I'm quite-sure it a struggle for newbs to pointer math is the typed-pointer differencing. It is the contrapositive to scalar additions to pointers. You did pretty good in your "in terms of data_t clause, but even with that I think the OP may be scratching their head. Imnsho, there really is no easy way to describe it that leads to revelation quickly. Samples (lots of them) usually help. (+1, btw).
1

From what I can deduce by the given information::

int idx = (sl.current - sl.pool + 1) % 16; 
data_t* d = &sl.pool[idx];

s1.pool is an array. So, negating s1.pool will give some value, say int x. Now ((x+1) % 16) will give a valid array bound index for s1.pool which is stored in idx. And thus, they found a valid array index for second line processing.

Improve this answer

Comments

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.