如何将int传递为“ void *”线程启动功能？ [英] How to pass an int as "void *" to thread start function?

问题描述

我本来是斐波那契变量数组的全局变量，但发现这是不允许的。我需要进行基本的多线程处理并处理竞争条件，但是我无法克服在pthread create中将int作为void参数的问题。我尝试使用没有运气的常量指针。出于某些奇怪的原因，void *通过了第一个布尔测试，但在以下情况下没有通过：

I originally had a global variable for my fibonacci variable array, but found out that is not allowed. I need to do elementary multithreading and handle race conditions, but I can't get past feeding an int as a void argument in pthread create. I've tried using a constant pointer with no luck. For some strange reason the void* gets past the first boolean test but not the else if:

  $ gcc -o fibonacci fibonacci.c
    fibonacci.c:22:16: warning: comparison between pointer and integer ('void *' and 'int')
      else if (arg == 1)
               ~~~ ^  ~
    1 warning generated.

我的代码一团糟，我真的很困惑，因为我已经重写了很多次了。如果将线程运行函数中的所有args转换为int，则会得到分段错误11，这很有意义。所有尝试通过地址传递i索引并对其取消引用的尝试均以失败告终，因为它是空值，不能用作整数。您还能提出其他建议吗？

My code is a mess and I am getting really confused because I have rewritten it so many times. If I cast all the args in my thread run function as ints I get a segmentation fault 11, which makes sense. All attempts at passing the i index by address and dereferencing it have failed, as it is a void and can't be used as an int. Can you suggest something else?

#include<stdio.h> //for printf
#include<stdlib.h>  //for malloc
#include<pthread.h> //for threading

#define SIZE 25 //number of fibonaccis to be computed
int *fibResults;  //array to store fibonacci results

void *run(void *arg)  //executes and exits each thread
{
  if (arg == 0)
  {
    fibResults[(int)arg] = 0;
    printf("The fibonacci of %d= %d\n", (int)arg, fibResults[(int)arg]);    
    pthread_exit(0); 
 }

  else if (arg == 1)
  {
    fibResults[(int)arg] = 1;
    printf("The fibonacci of %d= %d\n", (int)arg, fibResults[(int)arg]);   
    pthread_exit(0);  
  }
  else
  {
    fibResults[(int)arg] = fibResults[(int)arg -1] + fibResults[(int)arg -2];
    printf("The fibonacci of %d= %d\n", (int)arg, fibResults[(int)arg]);
    pthread_exit(0);
  }
}

//main function that drives the program.
int main()
{
  pthread_attr_t a;
  fibResults = (int*)malloc (SIZE * sizeof(int));
  pthread_attr_init(&a);  

  for (int i = 0; i < SIZE; i++)
  {
    pthread_t thread;
    pthread_create(&thread, &a, run,(void*) &i);
    printf("Thread[%d] created\t", i); 
    fflush(stdout);
    pthread_join(thread, NULL);
    printf("Thread[%d] joined & exited\t", i); 
  }
  return 0;
}

推荐答案

在 run（）函数，您应该执行以下操作：

In the run() function you should do:

void *run(void *ptrarg)  //executes and exits each thread
{
  int arg = *((int *)ptrarg);
  if (arg == 0)
  ....
  ....

和其余的 run（），您无需投放 arg 。将（int）arg 替换为 arg 。

and in rest of the run(), you don't need to cast the arg. Replace (int)arg with arg.

编辑：

创建线程时将参数传递给 fun（）的方式可能会导致竞赛条件，因为所有线程都将使用相同的指针。请检查@Jonathan的答案以避免出现此问题。

The way you are passing the argument to fun() while creating threads may cause race condition because all threads will be using same pointer. Check the @Jonathan's answer to avoid this problem.

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