傳遞的結構與gcc內置插件用於原子訪問

Question

我有兩種結構，就像下面：

template <class T> 
struct pointer_t 
{ 
    T *ptr; 
    uintptr_t tag; 
}; 


template <class T> 
struct Node 
{ 
    T data; 
    pointer_t<Node<T> > next; 
}; 

，現在我想例如pointer_t<Node<T> > newNext傳遞給__sync_bool_compare_and_swap()功能。根據該函數的原型我傳：

__sync_bool_compare_and_swap((unsigned long*) &newTail.ptr->next, oldNext, newNext)

的問題是，如果我不投newNext到unsigned long我會得到：

error: ‘struct pointer_t<Node<int> >’ used where a ‘long unsigned int’ was expected 
     if (__sync_bool_compare_and_swap((unsigned long*) &newTail.ptr->next, newNext, newNext)) 

，如果我將它轉換爲unsigned長則：

if (__sync_bool_compare_and_swap((unsigned long*) &newTail.ptr->next, (unsigned long) oldNext, (unsigned long) newNext)) 

我會得到：

error: ‘struct pointer_t<Node<int> >’ used where a ‘long unsigned int’ was expected. 

有人可以解釋我，你可以使用__sync_bool_compare_and_swap這兩個結構？

感謝

Answer 1

#include <iostream> 
#include <bitset> 
using namespace std; 

template <class T> 
struct pointer_t 
{ 
    T *ptr; 
    uintptr_t tag; 
}; 


template <class T> 
struct Node 
{ 
    T data; 
    pointer_t<Node<T> >* next; 
}; 

int main() 
{ 
    Node<int> *newTail = new Node<int>(); 
    pointer_t<Node<int> > *oldNext = newTail->next; 


    Node<int> *newNext = new Node<int>(); 

    pointer_t<Node<int> >* newNextPtr = new pointer_t<Node<int> >(); 
    newNextPtr->ptr=newNext; 

    if (__sync_bool_compare_and_swap(&newTail->next, oldNext, newNextPtr)) { 
     std::cout<<"Gotcha!\n"; 
     } 
} 

這並不能真正解決您的指針標記問題。如果你想實現指針標記，那麼只需從指針中竊取一些未使用的位，並設置這些位來標記指針，然後取消它們以重置指針。不要忘記在解引用之前取消標記位。

#define get_markedness(p) (((ptr_int) (p)) & 3) 
#define get_unmarked_reference(p) ((void *) (((ptr_int) (p)) & (~3))) 
#define get_marked_reference(p,m) ((void *) (((ptr_int) (p)) | m)) 

Answer 2

你可以看一下下面的鏈接，如果你發現指針標示有趣

stealing bits from a pointer

http://concurrencyfreaks.blogspot.se/2014/03/harriss-linked-list.html

https://timharris.uk/papers/2001-disc.pdf