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a-) The compare and swap() instruction can be used to design lock-free data structures such as stacks, queues, and lists. The program example shown in Figure 6.18  presents a possible solution to a lock-free stack using CAS instructions, where the stack is represented as a linked list of Node elements with top representing the top of the stack. Is this implementation free from race conditions?   b-) Describe how the signal() operation associated with monitors differs from the corresponding operation defined for semaphores.

a-) The compare and swap() instruction can be used to design lock-free data structures such as stacks, queues, and lists. The program example shown in Figure 6.18  presents a possible solution to a lock-free stack using CAS instructions, where the stack is represented as a linked list of Node elements with top representing the top of the stack. Is this implementation free from race conditions?   b-) Describe how the signal() operation associated with monitors differs from the corresponding operation defined for semaphores.

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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a-) The compare and swap() instruction can be used to design lock-free data structures such as stacks, queues, and lists. The program example shown in Figure 6.18  presents a possible solution to a lock-free stack using CAS instructions, where the stack is represented as a linked list of Node elements with top representing the top of the stack. Is this implementation free from race conditions?

 

b-) Describe how the signal() operation associated with monitors differs from the corresponding operation defined for semaphores.

 

 

 

 

typedef struct node { value_t data; struct node *next; } Node; Node *top; // top of stack void push(value_t item) { Node *old node; Node *new_node; new node = malloc(sizeof (Node)); new node->data = item; do { old node = top; new node->next = old_node; while (compare and swap (top,old node,new node) != old node); } value_t pop() { Node *old node; Node *new_node; do { old node = top; if (old node == NULL) return NULL; new node = old node->next; while (compare_and_swap(top,old_node, new_node) != old_node); return old node->data; } Figure 6.18 Lock-free stack for Exercise 6.15.
Transcribed Image Text:typedef struct node { value_t data; struct node *next; } Node; Node *top; // top of stack void push(value_t item) { Node *old node; Node *new_node; new node = malloc(sizeof (Node)); new node->data = item; do { old node = top; new node->next = old_node; while (compare and swap (top,old node,new node) != old node); } value_t pop() { Node *old node; Node *new_node; do { old node = top; if (old node == NULL) return NULL; new node = old node->next; while (compare_and_swap(top,old_node, new_node) != old_node); return old node->data; } Figure 6.18 Lock-free stack for Exercise 6.15.
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