如果两个线程同时调用 fork() 会发生什么
What happens if two threads call fork() simultaneously
我有一个多线程进程。我已经使用 __register_atfork(blocksigprof,restoresigprof,NULL,NULL); 注册了准备函数和父处理程序;
功能。
现在让我们假设两个线程同时调用 fork。我在 blocksigprof 中有一个计数器递增,在 restoresigprof 中有一个计数器递减。
考虑到上述情况,blocksigprof 和 restoresigprof 是否总是成对调用?
__register_atfork.
中是否存在固有的锁定机制
#define NUM_THREADS 8
static int go=0;
static int exec = 1;
static int ev_od = 0;
static void *
test_thread (void *arg) {
int j;
pid_t c, d;
while(!go) // Wait, so that all threads are here.
continue;
// All will fork, hopefully at same time because of go signal wait.
while(exec) {
c = fork();
if (c < 0) {
printf("SANJAY: fork() failed.\n");
exit(1);
} else if (c == 0) { // Child
exit(0);
}
else { // parent
d = waitpid(c, NULL, 0);
}
}
return NULL;
}
extern int __register_atfork(void (*)(void),void (*)(void),void (*)(void),void *);
static sigset_t s_new;
static sigset_t s_old;
static int count = 0;
static void blocksigprof(void){
count++;
#ifdef SYS_gettid
pid_t tid = syscall(SYS_gettid);
if (tid % 2) {
printf("sleep.\n");
usleep(1);
}
#else
#error "SYS_gettid unavailable on this system"
#endif
printf("Pre-fork. Count should be one. %d\n", count);
}
static void restoresigprof(void){
printf("Post-fork. Count should be one. %d\n", count);
count--;
}
int
main () {
pthread_t t[NUM_THREADS];
void *ptr;
long size = 500 * 1024 * 1024;
int i, m;
volatile int result = 0;
int g_iters = 100;
(void) __register_atfork(blocksigprof,restoresigprof,NULL,NULL);
// Increase size, so fork takes time.
printf("SANJAY: Increasing process size.\n");
ptr = malloc(size);
memset(ptr, 0, size);
ptr = malloc(size);
memset(ptr, 0, size);
ptr = malloc(size);
memset(ptr, 0, size);
ptr = malloc(size);
memset(ptr, 0, size);
ptr = malloc(size);
memset(ptr, 0, size);
ptr = malloc(size);
memset(ptr, 0, size);
ptr = malloc(size);
memset(ptr, 0, size);
ptr = malloc(size);
memset(ptr, 0, size);
ptr = malloc(size);
memset(ptr, 0, size);
// Create threads.
for (i = 0; i < NUM_THREADS; ++i) {
pthread_create(&t[i], NULL, test_thread, NULL);
}
printf("SANJAY: Killing time.\n");
// Kill time, so that all threads are at same place post it, waiting for go. 100M cycles.
for (m = 0; m < 1000000; ++m)
for (i = 0; i < g_iters; ++i )
result ^= i;
// Give all threads go at same time.
printf("SANJAY: Let threads execute.\n");
go = 1;
usleep(10000000); // Wait for 10 sec.
exec = 0;
// Wait for all threads to finish.
for (i = 0; i < NUM_THREADS; ++i) {
pthread_join(t[i], NULL);
}
printf("SANJAY: Done.\n");
return 0;
}
pthread_atfork
规范不要求其实现序列化对 prepare
和 parent
处理程序的调用,因此一个安全的假设是没有同步。
glibc
implementation does lock an internal mutex 防止多个线程并行进入处理程序。但是,这是一个实现细节。代码中的注释说这样的实现不是 POSIX-compliant 因为 POSIX 要求 pthread_atfork
是 async-signal-safe,并且在那里使用互斥量使得它不是 async-signal-safe .
为了使您的代码健壮,我建议使用原子或互斥来保护您的共享状态免受竞争条件的影响。
我有一个多线程进程。我已经使用 __register_atfork(blocksigprof,restoresigprof,NULL,NULL); 注册了准备函数和父处理程序; 功能。 现在让我们假设两个线程同时调用 fork。我在 blocksigprof 中有一个计数器递增,在 restoresigprof 中有一个计数器递减。
考虑到上述情况,blocksigprof 和 restoresigprof 是否总是成对调用? __register_atfork.
中是否存在固有的锁定机制#define NUM_THREADS 8
static int go=0;
static int exec = 1;
static int ev_od = 0;
static void *
test_thread (void *arg) {
int j;
pid_t c, d;
while(!go) // Wait, so that all threads are here.
continue;
// All will fork, hopefully at same time because of go signal wait.
while(exec) {
c = fork();
if (c < 0) {
printf("SANJAY: fork() failed.\n");
exit(1);
} else if (c == 0) { // Child
exit(0);
}
else { // parent
d = waitpid(c, NULL, 0);
}
}
return NULL;
}
extern int __register_atfork(void (*)(void),void (*)(void),void (*)(void),void *);
static sigset_t s_new;
static sigset_t s_old;
static int count = 0;
static void blocksigprof(void){
count++;
#ifdef SYS_gettid
pid_t tid = syscall(SYS_gettid);
if (tid % 2) {
printf("sleep.\n");
usleep(1);
}
#else
#error "SYS_gettid unavailable on this system"
#endif
printf("Pre-fork. Count should be one. %d\n", count);
}
static void restoresigprof(void){
printf("Post-fork. Count should be one. %d\n", count);
count--;
}
int
main () {
pthread_t t[NUM_THREADS];
void *ptr;
long size = 500 * 1024 * 1024;
int i, m;
volatile int result = 0;
int g_iters = 100;
(void) __register_atfork(blocksigprof,restoresigprof,NULL,NULL);
// Increase size, so fork takes time.
printf("SANJAY: Increasing process size.\n");
ptr = malloc(size);
memset(ptr, 0, size);
ptr = malloc(size);
memset(ptr, 0, size);
ptr = malloc(size);
memset(ptr, 0, size);
ptr = malloc(size);
memset(ptr, 0, size);
ptr = malloc(size);
memset(ptr, 0, size);
ptr = malloc(size);
memset(ptr, 0, size);
ptr = malloc(size);
memset(ptr, 0, size);
ptr = malloc(size);
memset(ptr, 0, size);
ptr = malloc(size);
memset(ptr, 0, size);
// Create threads.
for (i = 0; i < NUM_THREADS; ++i) {
pthread_create(&t[i], NULL, test_thread, NULL);
}
printf("SANJAY: Killing time.\n");
// Kill time, so that all threads are at same place post it, waiting for go. 100M cycles.
for (m = 0; m < 1000000; ++m)
for (i = 0; i < g_iters; ++i )
result ^= i;
// Give all threads go at same time.
printf("SANJAY: Let threads execute.\n");
go = 1;
usleep(10000000); // Wait for 10 sec.
exec = 0;
// Wait for all threads to finish.
for (i = 0; i < NUM_THREADS; ++i) {
pthread_join(t[i], NULL);
}
printf("SANJAY: Done.\n");
return 0;
}
pthread_atfork
规范不要求其实现序列化对 prepare
和 parent
处理程序的调用,因此一个安全的假设是没有同步。
glibc
implementation does lock an internal mutex 防止多个线程并行进入处理程序。但是,这是一个实现细节。代码中的注释说这样的实现不是 POSIX-compliant 因为 POSIX 要求 pthread_atfork
是 async-signal-safe,并且在那里使用互斥量使得它不是 async-signal-safe .
为了使您的代码健壮,我建议使用原子或互斥来保护您的共享状态免受竞争条件的影响。