我理解pthread_join manual我的程序应该输出我提供给print_xs
的相同值,但获得的输出是:
Message from printxs 11131 a
Message from printxs 11234 b
32766 -16
32766 0
该计划:
#include<stdio.h>
#include<pthread.h>
typedef struct paramsThread{
char c;
int count;
} threadPara;
void* print_xs(void* unused){
threadPara *tp = (threadPara *)unused;
int i=tp->count;
printf("Message from printxs %d %c\n", tp->count, tp->c);
return (void*)tp;
}
int main(){
pthread_t thread1, thread2;
threadPara t1,t2,t3,t4;
t1.c = 'a';
t2.c = 'b';
t1.count = 11131;
t2.count = 11234;
t3.count=0;
pthread_create(&thread1, NULL, &print_xs, &t1);
pthread_create(&thread2, NULL, &print_xs, &t2);
pthread_join(thread1,(void*)&t3);
printf("%d %d\n", t3.count, t3.c);
pthread_join(thread2,(void*)&t4);
printf("%d %d\n", t4.count, t4.c);
return 0;
}
有人可以解释为什么会这样吗?
答案 0 :(得分:5)
如果你read a pthread_join
reference or manual page,你会看到第二个参数是一个指向的指针。这是一种在C中模拟按引用传递的方法。
线程函数返回的指针被复制到另一个指针。
解决方案是使用t3
和t4
的指针代替:
threadPara t1, t2;
threadPara *t3, *t4;
// ...
pthread_join(thread1,(void**)&t3); // Pass pointer to the pointer, emulating pass by reference
printf("%d %d\n", t3->count, t3->c);
pthread_join(thread2,(void**)&t4);
printf("%d %d\n", t4->count, t4->c);
如果您打印指针或使用调试器,那么您会看到t3
指向t1
(即t3 == &t1
),t4
指向t2
和{{1}}。
答案 1 :(得分:0)
线程函数返回一个地址void*
。
pthread_join()
收到到的地址,其中2 nd 参数(void**
)指向。< / p>
所以要在代码中使用它而不是
pthread_join(thread1, (void*)&t3);
DO
{
void * pv;
pthread_join(thread1, &pv);
if (NULL != pv)
{
t3 = *((threadPara*) pv);
}
}
作为旁注:在C中, no 需要强制转换为void
- 来自/到其他指针的指针,所涉及的指针会被隐式转换。
所以这段代码
void* print_xs(void* unused){
threadPara *tp = (threadPara *)unused;
int i=tp->count;
printf("Message from printxs %d %c\n", tp->count, tp->c);
return (void*)tp;
}
很可能是这样写的:
void* print_xs(void* unused){
threadPara *tp = unused;
int i=tp->count;
printf("Message from printxs %d %c\n", tp->count, tp->c);
return tp;
}