c-如何找到生产者将字符传递到缓冲区的轮次

Question

我有一个生产者-消费者程序，它逐个字符读取文件，并将内容放入缓冲区。

我需要帮助输出生产者函数为将字符传递到缓冲区而进行的循环数。循环意味着一次或多次连续写入缓冲区而不会因等待而中断(由于队列已满(。

#include <pthread.h>
#include <semaphore.h>
#include <stdlib.h>
#include <stdio.h>
/*
This program provides a possible solution for producer-consumer problem using mutex and semaphore.
I have used 5 producers and 5 consumers to demonstrate the solution. You can always play with these values.
*/
#define MaxItems 5 // Maximum items a producer can produce or a consumer can consume
#define BufferSize 5 // Size of the buffer
sem_t empty;
sem_t full;
int in = 0;
int out = 0;
int buffer[BufferSize];
pthread_mutex_t mutex;
void *producer(void *pno)
{   
int item;
for(int i = 0; i < MaxItems; i++) {
item = rand(); // Produce an random item
sem_wait(&empty);
pthread_mutex_lock(&mutex);
buffer[in] = item;
printf("Producer %d: Insert Item %d at %dn", *((int *)pno),buffer[in],in);
in = (in+1)%BufferSize;
pthread_mutex_unlock(&mutex);
sem_post(&full);
}
}
void *consumer(void *cno)
{   
for(int i = 0; i < MaxItems; i++) {
sem_wait(&full);
pthread_mutex_lock(&mutex);
int item = buffer[out];
printf("Consumer %d: Remove Item %d from %dn",*((int *)cno),item, out);
out = (out+1)%BufferSize;
pthread_mutex_unlock(&mutex);
sem_post(&empty);
}
}
int main()
{   
pthread_t pro[5],con[5];
pthread_mutex_init(&mutex, NULL);
sem_init(&empty,0,BufferSize);
sem_init(&full,0,0);
FILE *fp = fopen("file.txt", "r");
if (fp != NULL) {
if (fseek(fp, 0L, SEEK_END) == 0) {
/* Get the size of the file. */
p1.BUFFER_SIZE = ftell(fp);
if (p1.BUFFER_SIZE == -1) { /* Error */ }
/* Allocate our buffer to that size. */
p1.item = malloc(sizeof(char) * (p1.BUFFER_SIZE + 1));
/* Go back to the start of the file. */
if (fseek(fp, 0L, SEEK_SET) != 0) { /* Error */ }
/* Read the entire file into memory. */
size_t newLen = fread(p1.item, sizeof(char), p1.BUFFER_SIZE, fp);
if ( ferror( fp ) != 0 ) {
fputs("Error reading file", stderr);
} else {
p1.item[newLen++] = ''; /* Just to be safe. */
}
}
int a[5] = {1,2,3,4,5}; //Just used for numbering the producer and consumer
for(int i = 0; i < 5; i++) {
pthread_create(&pro[i], NULL, (void *)producer, (void *)&a[i]);
}
for(int i = 0; i < 5; i++) {
pthread_create(&con[i], NULL, (void *)consumer, (void *)&a[i]);
}
for(int i = 0; i < 5; i++) {
pthread_join(pro[i], NULL);
}
for(int i = 0; i < 5; i++) {
pthread_join(con[i], NULL);
}
pthread_mutex_destroy(&mutex);
sem_destroy(&empty);
sem_destroy(&full);
return 0;

}

Answer 1

通常，由于信号量的值最初为零，因此无法确定调用sem_wait()返回所需的时间是否延长。sem_wait()功能不传递该信息。

然而，您可以做的一件事是从sem_trywait()开始，如果它不能立即减少目标信号量，它将失败而不是阻塞。在这种情况下，您可以增加一个计数器，然后继续执行常规的sem_wait()。示例：

int full_count = 0;
void *producer(void *pno) {   
int my_num = *(int *)pno;
for(int i = 0; i < MaxItems; i++) {
int item = rand(); // Produce an random item
int result = sem_trywait(&empty);
if (result == -1) {
if (errno == EAGAIN) {
full_count += 1;
result = sem_wait(&empty);
// handle any error ...
} else {
// handle other error ...
}
}
pthread_mutex_lock(&mutex);
buffer[in] = item;
printf("Producer %d: Insert Item %d at %dn", my_num, buffer[in], in);
in = (in + 1) % BufferSize;
pthread_mutex_unlock(&mutex);
sem_post(&full);
}
}

请注意，不能保证在调用sem_wait()的情况下，它实际上会阻塞，因为信号量可以在trywait和wait之间递增。但它确实告诉你，如果执行了wait而不是trywait，那么wait就会被阻止。在这种情况下，生产者确实由于缓冲区已满而被延迟，即使它没有在sem_wait()中阻塞任何延迟。