【发布时间】：2023-03-06 08:07:02
【问题描述】：
我正在为分配构建速率单调调度程序，我需要让所有线程在同一个处理器上运行。我不确定我做错了什么，但我不断收到同样的错误（见下图）。我一直在查看一堆 Linux 文档，试图弄清楚该怎么做，但我就是做不到。我真的很感激一些帮助。提前致谢。
我的 Scheduler.cpp 文件：
#define _GNU_SOURCE
#include <iostream>
#include <pthread.h>
#include <unistd.h>
#include <mutex>
#include <condition_variable>
#include <semaphore.h>
#include <sched.h>
#include <stdio.h>
#include <assert.h>
#include <stdlib.h>
using namespace std;
sem_t semaphore;
sem_t mutex1;
sem_t mutex2;
sem_t mutex3;
sem_t mutex4;
// initialze variables
int i = 0;
int overrun1 = 0;
int overrun2 = 0;
int overrun3 = 0;
int overrun4 = 0;
int doWork();
void* p1(void *param);
void* p2(void *param);
void* p3(void *param);
void* p4(void *param);
int main(int argc, char const *argv[])
{
cpu_set_t cpus;
CPU_ZERO(&cpus);
CPU_SET(1, &cpus);
sem_init(&mutex1, 0, 0);
sem_init(&mutex2, 0, 0);
sem_init(&mutex3, 0, 0);
sem_init(&mutex4, 0, 0);
// initialze all threads
pthread_t thread1;
pthread_t thread2;
pthread_t thread3;
pthread_t thread4;
// actually create all threads
pthread_create(&thread1, NULL, p1, NULL);
pthread_create(&thread2, NULL, p2, NULL);
pthread_create(&thread3, NULL, p3, NULL);
pthread_create(&thread4, NULL, p4, NULL);
while (i < 160)
{
if (i == 0) // initial case. at time 0 schedule all threads
{
sem_post(&mutex1);
sem_post(&mutex2);
sem_post(&mutex3);
sem_post(&mutex4);
}
else if (i % 16 == 0) // last case which happens every 16 units which schedules all threads again
{
sem_post(&mutex1);
sem_post(&mutex2);
sem_post(&mutex3);
sem_post(&mutex4);
}
else if (i % 4 == 0) // case that happens every 4 units of time
{
sem_post(&mutex1);
sem_post(&mutex2);
sem_post(&mutex3);
}
else if (i % 2 == 0) // case that happens every other unit of time
{
sem_post(&mutex1);
sem_post(&mutex2);
}
else if (i % 1 == 0) // case that happens every unit of time
{
sem_post(&mutex1);
}
i++; // increment i to go through the loop again
}
// join all threads back to the main one
pthread_join(thread1, NULL);
pthread_join(thread2, NULL);
pthread_join(thread3, NULL);
pthread_join(thread4, NULL);
return 0;
}
// doWork function
int doWork()
{
int lousyArray[10][10];
int product = 1;
for (int i = 0; i < 10; i++)
{
for (int j = 0; j < 10; j++)
{
lousyArray[i][j] = 1;
}
}
for (int k = 0; k < 1; k++)
{
for (int j = 0; j < 10; j++)
{
for (int i = 0; i < 10; i++)
{
product *= lousyArray[i][j];
}
}
}
return 1;
}
void* p1(void *param)
{
bool thread1FinishFlag = false;
while(1)
{
sem_wait(&mutex1);
thread1FinishFlag = false;
for (int i = 0; i < 1; i++)
{
doWork();
}
//increment counter here
thread1FinishFlag = true;
}
}
void* p2(void *param)
{
bool thread2FinishFlag = false;
while(1)
{
sem_wait(&mutex2);
thread2FinishFlag = false;
for (int i = 0; i < 2; i++)
{
doWork();
}
//increment counter here
thread2FinishFlag = true;
}
}
void* p3(void *param)
{
bool thread3FinishFlag = false;
while(1)
{
sem_wait(&mutex3);
thread3FinishFlag = false;
for (int i = 0; i < 4; i++)
{
doWork();
}
//increment counter here
thread3FinishFlag = true;
}
}
void* p4(void *param)
{
bool thread4FinishFlag = false;
while(1)
{
sem_wait(&mutex4);
thread4FinishFlag = false;
for (int i = 0; i < 16; i++)
{
doWork();
}
//increment counter here
thread4FinishFlag = true;
}
}
void nsleep()
{
struct timespec delay;
delay.tv_sec = 0;
delay.tv_sec = 100000000L;
nanosleep(&delay, NULL);
}
【问题讨论】：
-
请edit您的问题提供minimal reproducible example。
-
@Barmar 不确定你是不是在开玩笑，但无论如何：发布包含一些错误的整个源文件几乎不是最小。
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@BaummitAugen 我相信你明白，发帖人并不总是知道哪些部分是重要的，所以他们很难只缩小到那个部分。更严重的问题是，当他们发布的代码不完整或无法验证时，他并没有犯这个错误。
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@Barmar 在这种情况下，我们谈论的是编译错误。编译器指向有问题的确切行。做出比“仅仅转储所有代码”更简单的努力可以而且必须得到提问者的期望，即使它“很难”。
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@BaummitAugen 来吧，你知道事情没那么简单。错误消息中未提及的行可能存在各种依赖关系。