【发布时间】：2015-11-02 06:24:34
【问题描述】：
假设我初始化了两个全局信号量数组，
semaphore empty[someNum];
和
semaphore full[someNum];
而someNum被初始化为
const int someNum = 3;（全球）
我将有一个名为 init() 的方法，里面是一个 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 之类的命令，例如，如果num 已满，那么我不希望我的生产者将值放入其中。
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过
empty[index] = create_semaphore(1)
full[index] = create_semaphore(0)
在方法中，我想执行 wait 和 signal 的操作，例如
wait(empty) ;
但我得到了错误。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，如何正确索引到我的 for-loop 中的数组以指定一个信号量？
感谢您的所有意见！如果您有任何问题，请随时要求任何澄清！请耐心等待，信号量对我来说是一个新概念。
注意
命令create_semaphore 存在于与此相关的另一个程序中。
编辑
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 ;
}
【问题讨论】：
-
你用的是什么库？如果您想等待特定的信号量，那么您必须为其提供索引。我不确定那是你想要的。 POSIX 信号量基于计数器，因此（如果我理解正确的话）可以使用初始化为正确值的单个信号量来执行此操作。
-
好的，我已将我的库添加到我的编辑中
-
嗯，我想你知道直接问题的答案，即你不能将数组作为 sem 传递。如果你确定你有正确的实现，那么它就像迭代它们并单独等待它们一样简单。但这些都是阻塞电话，所以我很犹豫在没有看到你具体做什么的情况下提出更多建议。
-
如果你要索引一个 sem 数组，那么它就是
wait(full[i]):i= 你想要的索引。我认为您认为它是错误的感觉是对的，但很难说清楚。它可能是正确的，但这取决于它的使用方式。假设 sem 告诉您“这是一次可以访问资源或关键部分的数量”。如果您只想要一个，则将其初始化为 1。但如果您想要多个，则将其初始化为n。当 sem decs 为 -
full[index]和empty[index]是各自数组之外的“元素”。正如你的教练所说，使用循环。 （已经向您展示了如何操作 - 查看trivet初始化循环。）
标签： c++ arrays indexing semaphore