如何将整型转换为void指针？
在C中使用线程时，我正面临警告
“警告：从不同大小的整数转换为指针”
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 %ldn",pthread_self());
printf("Thread %d is executingn",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 threadsn");
exit(0);
}
pthread_join(th[i],&ret);
printf("--->%dn",(int *)ret);
}
pthread_exit(NULL);
}
任何人都可以解释如何将一个整数传递给一个接收（void *）作为参数的函数吗？
如果这是您需要的，这是将整数传递给新pthread的一种很好的方法。 你只需要抑制警告，这将做到这一点：
#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
这可能会冒犯你的敏感度，但它非常短暂，没有竞争条件（如果你使用&i就会有这种情况）。 写几十行额外的代码只是为了得到一串编号的线程没有意义。
数据竞赛
这是一个数据争夺的糟糕版本：
#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 = %dn", *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，当我使用线程清理器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
你可以做这样的事情：
#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 %ldn",pthread_self());
printf("Thread %d is executingn",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 threadsn");
exit(0);
}
}
for (i=0;i<5;i++) {
pthread_join(th[i].thread,&ret);
printf("%d--->%dn",th[i].id,th[i].ret);
}
pthread_exit(NULL);
}
会输出：
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
将一个唯一的指针传递给每个线程不会竞争，并且您可以在th结构中获取/保存任何类型的信息
你可以将int的值作为void指针传递，如(void *)&n ，其中n是整数，并且在函数accept void指针中作为参数，如void foo(void *n); 最后在函数内部将void指针转换为int类似， int num = *(int *)n; 。 这样你就不会有任何警告。