pthreads不能从main（）函数访问变量吗？

Question

int main()
{
   int i; 
   pthread_t t; 
}

Can t not see i? t is created inside main, right? That means it must be using the same shared memory main() is using? How do I make it see i without making ia global variable?

Answer 1

What? t is a thread, it doesn't really "see" anything. Strictly, it's a variable that represents a thread -- you haven't actually created a thread -- but assuming you do create one, it runs in the same process as main() , so it shares memory space in that sense, but it doesn't share the scope of main. The functions which run in that thread can see whatever variables are in scope for those functions.

You could pass a pointer to i as the user data pointer to pthread_create . Or if you need to access more than just i , you could pass a pointer to some structure which contains (among other things) a pointer to i , and so on.

Example code:

#include <pthread.h>
#include <iostream>
#include <cstring>

void *thread_entry_point(void *data) {
    int *idata = static_cast<int*>(data);
    std::cout << "thread: i = " << *idata << "\n";
    *idata = 23;
    return const_cast<char*>("might as well return something");
}

int main() {
    int i = 12;
    pthread_t thr;

    int err = pthread_create(&thr, 0, thread_entry_point, &i);
    if (err == 0) {
        void *result;
        pthread_join(thr, &result);

        std::cout << "main: result = " << static_cast<const char*>(result) << "\n";
        std::cout << "main: i = " << i << "\n";
    } else {
        std::cout << "error creating thread: " << err << " " << std::strerror(err) << "\n";
    }
}

Answer 2

pthreads aren't special. For instance, the following code has the same "problem":

void foo()
{
  i = 5;
}
int main()
{
   int i; 
   foo();
}

Surely foo is called by main , so they're even on the same thread. Yet foo doesn't see the int in main . The solution is simple: if foo needs an int , main should pass that:

void foo(int& i)
{
  i = 5;
}
int main()
{
   int i; 
   foo(i);
}

With threads, the situation is the same: pass what you need to share.