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 the top representing the top of the stack. Is this implementation free from race conditions? typedef struct node {value_t data;struct node *next;} Node;Node *top; // top of stackvoid push(valu0_t itom) {Node +old_node;Node +new_node;new node = malloc(sizeof (Node));nev node->data = item;do {old nodo = top;now nodo->noxt = old_nodo;}while (compare and.swap(top, old.node,new.node) != oldnode);}value_t pop() {Node +old node;Node *nev 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

Solution :: Let's see first what is compare and swap() Instructions:: Answer:: compare-and-swap…

Answered: The compare and swap() instruction can…

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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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 the top representing the top of the stack. Is this implementation free from race conditions?

typedef struct node {
value_t data;
struct node *next;
} Node;
Node *top; // top of stack
void push(valu0_t itom) {
Node +old_node;
Node +new_node;
new node = malloc(sizeof (Node));
nev node->data = item;
do {
old nodo = top;
now nodo->noxt = old_nodo;
}
while (compare and.swap(top, old.node,new.node) != oldnode);
}
value_t pop() {
Node +old node;
Node *nev 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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