如何投放的整数空指针？ [英] How to cast an integer to void pointer?

问题描述

虽然使用线程在C工作，我面对的警告

While working with Threads in C, I'm facing the warning

警告：强制转换为指针，从大小不同的整数

"warning: cast to pointer from integer of different size"

在code是如下：

#include<stdio.h>
#include<sys/types.h>
#include<stdlib.h>
#include<pthread.h>
void *print(void *id)
{
 int a=10;
 printf("My thread id is %ld\n",pthread_self());
 printf("Thread %d is executing\n",id);
 return (void *) 42;
}

int main()
{
 pthread_t th[5];
 int t;
 int i;
 int status;
 void *ret;
 for(i=0;i<5;i++)
 {
   status=pthread_create(&th[i],NULL,print,(void *)i); //Getting warning at this line
   if(status)
   {
    printf("Error creating threads\n");
    exit(0);
   }
   pthread_join(th[i],&ret);
   printf("--->%d\n",(int *)ret);
 }
 pthread_exit(NULL);
}

任何人可以解释如何整数传递给接收（无效*）作为参数的函数吗？

Can anybody explain how to pass an integer to a function which receives (void * ) as a parameter?

推荐答案

这是通过整数新pthreads的，如果这是你所需要的罚款方式。你只需要燮preSS的警告，这将做到这一点：

This is a fine way to pass integers to new pthreads, if that is what you need. You just need to suppress the warning, and this will do it:

#include <stdint.h>

void *threadfunc(void *param)
{
    int id = (intptr_t) param;
    ...
}

int i, r;
r = pthread_create(&thread, NULL, threadfunc, (void *) (intptr_t) i);

讨论

这可能会冒犯你的感情，但它的很短的，没有竞争条件（如，如果你使用你不得不＆放大器;我 ）。以书面形式额外的code几十行没有任何意义只是为了得到一堆编号线程。

Discussion

This may offend your sensibilities, but it's very short and has no race conditions (as you'd have if you used &i). No sense in writing a few dozen lines of extra code just to get a bunch of numbered threads.

下面是一个坏版本数据争：

Here is a bad version with a data race:

#include <pthread.h>
#include <stdio.h>

#define N 10

void *thread_func(void *arg)
{
    int *ptr = arg;
    // Has *ptr changed by the time we get here?  Maybe!
    printf("Arg = %d\n", *ptr);
    return NULL;
}

int main()
{
    int i;
    pthread_t threads[N];
    for (i = 0; i < N; i++) {
        // NO NO NO NO this is bad!
        pthread_create(&threads[i], NULL, thread_func, &i);
    }
    for (i = 0; i < N; i++) {
        pthread_join(threads[i], NULL);
    }
    return 0;
}

现在，什么，当我与线程Sanitizer运行它发生？

Now, what happens when I run it with the thread sanitizer?

（此外，检查出它是如何打印出5两次......）

(Also, check out how it prints "5" twice...)

==================
WARNING: ThreadSanitizer: data race (pid=20494)
  Read of size 4 at 0x7ffc95a834ec by thread T1:
    #0 thread_func /home/depp/test.c:9 (a.out+0x000000000a8c)
    #1 <null> <null> (libtsan.so.0+0x000000023519)

  Previous write of size 4 at 0x7ffc95a834ec by main thread:
    #0 main /home/depp/test.c:17 (a.out+0x000000000b3a)

  Location is stack of main thread.

  Thread T1 (tid=20496, running) created by main thread at:
    #0 pthread_create <null> (libtsan.so.0+0x0000000273d4)
    #1 main /home/depp/test.c:18 (a.out+0x000000000b1c)

SUMMARY: ThreadSanitizer: data race /home/depp/test.c:9 thread_func
==================
Arg = 1
Arg = 2
Arg = 3
Arg = 4
Arg = 5
Arg = 6
Arg = 7
Arg = 8
Arg = 9
Arg = 5
ThreadSanitizer: reported 1 warnings

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