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.

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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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: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.
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