操作系统经典同步互斥问题——哲学家就餐仅仅允许4个人同时就餐#include <iostream>#include <mutex>#include <cstdio>#include <thread>#include <semaphore.h>using namespace std;#define THINK(i) PRintf("ph[%d] is thinking...\n", i)#define EAT(i) printf("ph[%d] eats.\n", i)void P(mutex &mt){ mt.lock();}void V(mutex &mt){ mt.unlock();}void P(sem_t* sem){ if(sem_wait(sem)) perror("P error!");}void V(sem_t* sem){ if(sem_post(sem)) perror("V error!");}// 加入unistd.h出现问题，似乎与thread的兼容性比较差，于是重写void delay(){ int sum=0; for(int i = 0; i < 10000000; i++) sum += i;}mutex fork[5];sem_t room;void init(){ sem_init(&room, 0, 4);}void philosopher (int i){ for(int j = 0; j < 5; j++) { THINK(i); P(&room); P(fork[i]); P(fork[(i+1)%5]); EAT(i); V(&room); V(fork[i]); V(fork[(i+1)%5]); }}int main(){ init(); thread t[] = { thread(philosopher, 0), thread(philosopher, 1), thread(philosopher, 2), thread(philosopher, 3), thread(philosopher, 4), }; for(int k = 0; k < 5; k++) t[k].join(); return 0;}