Question

在C语言中使用Threads时，我正面临警告

＆＃34;警告：从不同大小的整数＆＃34;

转换为指针

代码如下

#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);
}

有人可以解释如何将一个整数传递给接收（void *）作为参数的函数吗？

Answer 1

这是将整数传递给新pthread的好方法，如果这是你需要的。您只需要取消警告，这样就可以了：

#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，但它非常短并且没有竞争条件（如果你使用&i那么就是如此）。为了获得一堆编号的线程，编写几十行额外代码是没有意义的。

数据竞赛

这是一个带有数据竞争的糟糕版本：

#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，当我使用线程清理程序运行它时会发生什么？

（另外，看看它如何打印“5”两次......）

==================
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

Answer 2

你可以这样做：

#include <stdio.h>
#include <sys/types.h>
#include <stdlib.h>
#include <pthread.h>
struct th {
    pthread_t thread;
    int id;
    int ret;
};

void *print(void *id) {
    int a=10;
    struct th *self = (struct th *) id;
    printf("My thread id is %ld\n",pthread_self());
    printf("Thread %d is executing\n",self->id);
    self->ret = random();
    return;
}

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

    }
    pthread_exit(NULL);
}

将输出：

My thread id is 4496162816
My thread id is 4497870848
My thread id is 4498944000
My thread id is 4498407424
Thread 0 is executing
Thread 1 is executing
My thread id is 4499480576
Thread 3 is executing
Thread 2 is executing
0--->1804289383
Thread 4 is executing
1--->846930886
2--->1714636915
3--->1681692777
4--->1957747793

传递一个唯一的指针到每个线程不会竞争，你可以在结构中获取/保存任何类型的信息

Answer 3

你可以将int值作为void指针传递，如(void *)&n，其中n是整数，并在函数中接受void指针作为参数，如void foo(void *n);，最后在函数convert void指向int之类， int num = *(int *)n;。这样你就不会得到任何警告。

Answer 4

变化：

status=pthread_create(&th[i],NULL,print,(void *)i);

为：

status=pthread_create(&th[i],NULL,print,(reinterpret_cast<void*>(i));

reinterpret_cast使int成为指针的大小，警告将停止。基本上它是（void *）i。

的更好版本

如何将整数转换为void指针？

4 个答案:

讨论

数据竞赛