假设我初始化了两个全局信号量数组，
semaphore empty[someNum];
和
semaphore full[someNum];
和someNum初始化为
const int someNum = 3;（全球）
我有一个名为init()的方法，里面是for-loop来帮助索引这些数组。
for (index=0; index<someNum; index++) num[index]=0;
我的目标是在数组中使用wait和signal等命令，例如，如果num已满，那么我不希望我的生产者将值放入它
在init()内部我知道我需要初始化循环体中的所有信号量。到目前为止我已经尝试了
empty[index] = create_semaphore(1)
full[index] = create_semaphore(0)
并且在方法内部我想要执行wait和signal我尝试过的操作
wait(empty) ;
但我收回了错误。我知道这是因为我使用的是数组的名称而不是特定的信号量。
所以我的问题是，如何在for-loop中正确索引我的数组以指定一个信号量？
感谢您提供的所有输入！如果您有任何问题，请随时要求澄清！请耐心等待，信号量对我来说是一个新概念。
注意
命令create_semaphore存在于与此关联的另一个程序中。
修改
foodPass.cpp
#include <iostream>
#include <sched.h>
#include <time.h>
#include <pthread.h>
#include <string>
#include "sem.h"
using namespace std ;
/* ######################################## */
/* Misc Data Types */
/* ######################################## */
/* A data type - a struct with an int field
to represent a child ID at the program level. */
struct threadIdType
{
int id ;
};
/* ######################################## */
/* Global Variables */
/* ######################################## */
//const int numTrivets = 6 ;
const int numTrivets = 3 ;
const int numDiners = numTrivets - 1 ;
const int maxDishNames = 13 ;
//const int numDishNames = 13 ;
const int numDishNames = 5 ;
int trivet[numTrivets] ;
string dishName[maxDishNames];
/* Here declare the semaphores and other variables you will
need to synchronize threads. */
sim_semaphore empty[numTrivets] ;
sim_semaphore full[numTrivets] ;
/* child_t are global variables to represent the
dynamically-created threads. */
pthread_t child_t[numTrivets] ;
/* ######################################## */
/* "Special" Global Variables */
/* ######################################## */
/* Code in sem.cpp "expects" the two variables below to be here.
This particular program does not use "checking." */
/* "Checking" is just a flag that you set to 1 if you want lots of
debugging messages and set to 0 otherwise. The semaphore code in
sem.cpp imports "checking". Therefore the semaphore operations
will write lots of messages if you set checking=1. */
int checking ;
/* In some programs, we use the "stdoutLock" variable declared below to
get intelligible printouts from multiple concurrent threads that write
to the standard output. (There has to be something to prevent the
output of the threads from interleaving unintelligibly on the standard
output, and we can't use semaphores if the semaphore code is writing
messages too.)
To print a message to standard output, a thread first locks standard
output, then writes, then unlocks standard output. See files sem.cpp
or conc.cpp for examples of code that write messages in this manner.
WARNING: DON'T change how the locking of standard output is done
until you've thought a WHOLE lot about the consequences. In
particular, using semaphores to do the job of stdoutLock can cause
"infinite recursion" under certain circumstances. The reason is that
the semaphore code itself imports "stdoutLock" and writes messages
when the "checking" variable is set to 1. */
pthread_mutex_t stdoutLock ;
/* ################################################## */
/* init */
/* ################################################## */
void init()
{
int index ;
srandom(time((time_t *) 0)); /* INITIALIZE RANDOM NUMBER GENERATOR */
checking = 0 ;
/* Initialize the "special lock" that is used only to get
exclusive access to the screen. */
if ( 0!=pthread_mutex_init(&stdoutLock, NULL) )
{ cout << "MUTEX INITIALIZATION FAILURE!" << endl;
exit(-1) ;}
/* Initialize the trivets to indicate that each contains "no
dish." */
for (index=0; index<numTrivets; index++) trivet[index]=0;
/* Here initialize the semaphores and other variables you use
for synchronization. */
for (index=0; index<numTrivets; index++) full[index] = create_sim_sem(0) ;
for (index=0; index<numTrivets; index++) empty[index] = create_sim_sem(1) ;
/* Give some mnemonic names to the dishes. The first name is
used for an empty trivet. The last name denotes the check
(bill) for the meal. This is coded so no changes are needed
here as long as the value of "numDishNames" is between 2 and
13. */
dishName[0]="no dish";
dishName[1]="vegetable soup" ;
dishName[2]="bread and butter" ;
dishName[3]="beets and chickpeas" ;
dishName[4]="hardboiled eggs" ;
dishName[5]="calf tongue" ;
dishName[6]="baked potato" ;
dishName[7]="string beans" ;
dishName[8]="rack of lamb" ;
dishName[9]="salad" ;
dishName[10]="coffee" ;
dishName[11]="flan" ;
dishName[numDishNames-1]="check" ;
}
/* ################################################## */
/* DelayAsMuchAs */
/* ################################################## */
void delayAsMuchAs (int limit)
{
int time, step;
time=(int)random()%limit;
for (step=0;step<time;step++) sched_yield() ;
}
/* ################################################## */
/* Server */
/* ################################################## */
/*
The mother thread spawns a child thread that executes this
function. This function carries out the job of the server
at the restaurant.
*/
void * Server(void * ignore)
{
int i, j, delayLimit=100 ;
for (i=1; i<numDishNames; i++)
{
/* I delay a random time before I "feel like" placing
another dish on the table.*/
delayAsMuchAs(delayLimit);
/* When the trivet is available, I place the dish on the
trivet to my right. */
/* Here do a synchronization task. One thing you need to
do is be sure that you are not going to place a dish on
a trivet that alreay has a dish on it. *DO NOT* just
busy-wait until you see that the trivet is empty. */
wait_sem(empty[i]) ;
trivet[0]=i; // put dish #i onto trivet #0
pthread_mutex_lock(&stdoutLock) ;
cout << "Server places " << dishName[trivet[0]]
<< " on trivet #0." << endl ;
pthread_mutex_unlock(&stdoutLock);
/* Here you may want to a synchronization task --
something that "opens the door" for diner #0 to get
access to the new dish. */
signal_sem(full[i]) ;
}
pthread_exit ((void *)0) ;
}
/* ################################################## */
/* Diner */
/* ################################################## */
/*
The mother thread spawns child threads that execute this
function. This function carries out the job of one of the
diners at the restaurant.
*/
void * Diner(void * postnPtr)
{
/* Type cast the parameter to recover "position" -- which
tells me the position at which I am seated at the
table. */
int position = ((threadIdType *)(postnPtr))->id ;
int i, j, delayLimit=100 ;
for (i=1; i<numDishNames; i++)
{
/* I delay a random time before I "feel like" picking up the next
dish.*/
delayAsMuchAs(delayLimit);
/* When available, I pick up the next new dish on my left. */
/* Here do a synchronization task. One thing you need to
do is be sure that there is a new dish on the trivet to
your left now, and that the person on your left has
"let go" of it. */
wait_sem(full[i]);
/* I declare what I am doing */
pthread_mutex_lock(&stdoutLock) ;
cout << "Diner number "<< position ;
if (i<numDishNames-1) cout << " enjoys ";
else if (position<numDiners-1) cout << " examines " ;
else cout << " examines and pays " ;
cout << dishName[trivet[position]] << endl ;
pthread_mutex_unlock(&stdoutLock);
/* I delay a random time to simulate the time it takes for me to
serve myself some of what is on the dish -- or look at the
check. */
delayAsMuchAs(delayLimit);
/* When available, I place the dish on the trivet to my right. */
/* Here do a synchronization task. One thing you need to
do is be sure that the trivet on your right does not
have a dish on it now.*/
wait_sem (empty[i]);
pthread_mutex_lock(&stdoutLock) ;
cout << "Diner number "<< position << " moves "
<< dishName[trivet[position]] << " from trivet #"
<< position << " to trivet #" << position+1 << endl;
pthread_mutex_unlock(&stdoutLock);
/* transfer the dish on my left to trivet on my right */
trivet[position+1]=trivet[position] ;
/* mark trivet on my left as empty */
trivet[position]=0;
/* Here do a synchronization task. You have transferred a
dish from your left to your right. The person on your
left will need to find out that the trivet on your left
is now empty. The person on your right will need to
find out that the trivet on your right now has a new
dish on it. */
signal_sem(empty[i]);
signal_sem(full[i]);
}
delete((threadIdType *)(postnPtr)) ;
pthread_exit ((void *) 0) ;
}
/* ################################################## */
/* Busser */
/* ################################################## */
/*
The mother thread spawns children and then executes this
function. This is convenient because this function should
be the last to exit. This function carries out the job of
the busser at the restaurant.
*/
void * Busser (void * ignore)
{
int i, j, delayLimit=100 ;
for (i=1; i<numDishNames; i++)
{
/* I delay a random time before I "feel like" bussing another
dish.*/
delayAsMuchAs(delayLimit);
/* When another dish is on the trivet to my right I remove it. */
/* Here do a synchronization task. One thing you need to
do is be sure that there is a new dish on the trivet to
your left now, and that the person on your left has
"let go" of it. */
wait_sem (full[i]) ;
pthread_mutex_lock(&stdoutLock) ;
cout << "Busser removes "
<< dishName[trivet[numTrivets-1]] << " from trivet #"
<< numTrivets-1<< "." << endl ;
pthread_mutex_unlock(&stdoutLock);
trivet[numTrivets-1]=0; // remove the dish.
/* Here do a synchronization task. The person on your left
will need to find out that the trivet on your left is
now empty. */
signal_sem (empty[i]);
}
return ignore ;
}
/* ################################################## */
/* Main */
/* ################################################## */
int main()
{
init();
cout << endl << endl;
cout << "Welcome to the restaurant!" << endl ;
cout << numDiners << " will be dining." << endl ;
cout << "The meal will consist of " << numDishNames-2
<< " dishes." << endl;
cout << "Bon appetite!" << endl ;
cout << endl << endl;
int i;
/* This is a pointer to a struct that contains an int
field - it is a convenient data type to use as the
parameter to the child function. */
threadIdType * idPtr ;
for (i=0; i<numDiners; i++)
{
/* This records the current index as this child's ID */
idPtr = new threadIdType ;
idPtr->id = i ;
if (0!=pthread_create(&child_t[i], NULL, Diner, (void *) idPtr))
{cout << "THREAD CREATION FAILURE!" << endl; exit(-1) ;}
if (0!=pthread_detach(child_t[i]))
{cout << "THREAD DETACHMENT FAILURE!" << endl ; exit(-1) ;}
}
if (0!=pthread_create(&child_t[numDiners], NULL, Server, (void *) 0))
{cout << "THREAD CREATION FAILURE!" << endl; exit(-1) ;}
if (0!=pthread_detach(child_t[numDiners]))
{cout << "THREAD DETACHMENT FAILURE!" << endl ; exit(-1) ;}
Busser((void *) 0) ;
cout << endl << endl;
cout << "Thank you for coming!" << endl ;
cout << endl << endl;
return 0 ;
}
答案 0 :(得分：1)
好的，我发现了我的问题。
虽然我在init()，
for (index=0; index<numTrivets; index++) full[index] = create_sim_sem(0) ;
for (index=0; index<numTrivets; index++) empty[index] = create_sim_sem(1) ;
我的问题实际上来自wait和signal命令。
在server()中是纯生产者，只负责第一个数组索引0所以命令只是
wait_sem(empty[0])和signal_sem(full[0])
在diner()中，它有点复杂。它充当生产者和消费者之间的混合体，这意味着我们不能只使用0和1，因此我们看看trivet[]如何索引单个信号量并借用它，然后我们得到
wait_sem(full[position])
此外，diner()还有两个位置需要担心。因此，我们必须将下一个排名设为独占位置，以便我们初始化trivet[position+1]，并在我们处理右侧的waits和signals时再次借用它。
最后，我们busser()纯粹作为消费者行事。当他的位置满满时我们只会发信号给他消耗。由于三脚架的数量可以互换，（numTrivets）我们不能简单地放入1，因为下次可能会有超过2个点。因此，我们只需要输入像
wait_sem(full[numTrivets-1])然后signal_sem(empty[numTrivets-1])
然后我得到了我想要的输出。
感谢所有给我输入的人我希望我的答案证明对所有像我一样困惑的人都有用