C++ pthread_连接错误，从线程生成线程

我目前正在尝试从其他线程生成线程。 当我的线程试图加入时，我会遇到一个错误

这是我的主要职能：

int main( int argc, char *argv[] ) {
    std::cout<< "start" << std::endl;
    init();
    std::cout<<"finished init" << std::endl;
    t1=clock();
    pthread_t threads[THREAD_COUNT];

    for (int i = 0; i < THREAD_COUNT; i++) {
        pthread_create(&threads[i], NULL, &threadMain, (void*)((long)i));
    }

    for (int i = 0; i < THREAD_COUNT; i++) {
        printf("joining %d \n" , i);
        pthread_join(threads[i], NULL);
    }
    timeEnd();


    return(0);
}

每当我在ubuntu机器上运行它时，就会出现一个分段错误。 使用gdb，在pthread_连接中放置一个断点并单步执行最终会给我一个segfault

在mac上运行它，我得到以下输出：

/a.out

开始

完成初始化

加入0

a、 out（3473,0x10c61e000）malloc:*对象0x7f97fa5016f0的错误：未分配要释放的指针 *在malloc\u error\u break中设置断点以进行调试

中止陷阱：6

编辑

一些定义：

#define INNER_THREADS 2
#define THREAD_COUNT 10

struct desc{
  long outterThread;
  long innerThread;
  volatile int* tix;
  volatile int* c;
  volatile int* r;
};

struct desc* vec[THREAD_COUNT*2];

---

long thread=（long）*thread不应仅用于“（long）thread。我指的是threadMain函数的第一行

请通过添加缺少的详细信息来完成您的程序：threadBody（）、init（）、clock（）和timeEnd（），如果它们与所面临的问题无关，请删除它们

我可以通过删除/假设缺少一些详细信息来准备代码：

它很好用

$ ./a.out | sort
joining outer thread 0
joining outer thread 1
joining outer thread 2
joining outer thread 3
joining outer thread 4
joining outer thread 5
joining outer thread 6
joining outer thread 7
joining outer thread 8
joining outer thread 9
Outer thread 0 Inner thread 0
Outer thread 0 Inner thread 1
Outer thread 1 Inner thread 0
Outer thread 1 Inner thread 1
Outer thread 2 Inner thread 0
Outer thread 2 Inner thread 1
Outer thread 3 Inner thread 0
Outer thread 3 Inner thread 1
Outer thread 4 Inner thread 0
Outer thread 4 Inner thread 1
Outer thread 5 Inner thread 0
Outer thread 5 Inner thread 1
Outer thread 6 Inner thread 0
Outer thread 6 Inner thread 1
Outer thread 7 Inner thread 0
Outer thread 7 Inner thread 1
Outer thread 8 Inner thread 0
Outer thread 8 Inner thread 1
Outer thread 9 Inner thread 0
Outer thread 9 Inner thread 1
$ 

valgrind也很高兴，除了一些内存泄漏

值得一提的是，“2”（在vec[i+thread*2]和vec[thread\u COUNT*2]中）的用法应该替换为内部线程

---

主要的问题是threadBody（）函数中有什么？

free（）在哪里？这看起来很糟糕：

vec[i+thread*2]->tix=tix；

：您正在制作动态分配内存地址的多个副本；我打赌您是

free（）

tix多次。@YSC我没有释放它们。我需要线程拥有自己的tix、c和r内存，这样它们的内部线程就可以使用它。这看起来很可疑：

memset（（void*）tix，0，sizeof（tix））；

，但未更新的行是

t.c:146

？@YSC line

t.c:146

是

pthread\u连接（线程[I]，NULL）在threadBody中，我正在更新c、r和tix中的值，所有内线程都应该可以访问这些值。
#define INNER_THREADS 2
#define THREAD_COUNT 10

struct desc{
  long outterThread;
  long innerThread;
  volatile int* tix;
  volatile int* c;
  volatile int* r;
};

struct desc* vec[THREAD_COUNT*2];

$ ./a.out | sort
joining outer thread 0
joining outer thread 1
joining outer thread 2
joining outer thread 3
joining outer thread 4
joining outer thread 5
joining outer thread 6
joining outer thread 7
joining outer thread 8
joining outer thread 9
Outer thread 0 Inner thread 0
Outer thread 0 Inner thread 1
Outer thread 1 Inner thread 0
Outer thread 1 Inner thread 1
Outer thread 2 Inner thread 0
Outer thread 2 Inner thread 1
Outer thread 3 Inner thread 0
Outer thread 3 Inner thread 1
Outer thread 4 Inner thread 0
Outer thread 4 Inner thread 1
Outer thread 5 Inner thread 0
Outer thread 5 Inner thread 1
Outer thread 6 Inner thread 0
Outer thread 6 Inner thread 1
Outer thread 7 Inner thread 0
Outer thread 7 Inner thread 1
Outer thread 8 Inner thread 0
Outer thread 8 Inner thread 1
Outer thread 9 Inner thread 0
Outer thread 9 Inner thread 1
$