Pthreads, confusion with pthread_join(pthread_t, void**)
-
07-07-2021 - |
题
I can't understand why does pthread_join
takes as 2nd argument void**
for the return value, whereas pthread_exit
, which is given the return value, has the return value argument as void*
.
解决方案
pthread_join waits for the thread to end, and the resulting value from pthread_exit is stored into *value_ptr. If you want to ignore the result, you can pass NULL for the value_ptr. This is the common C practice of simulating pass by reference by passing a pointer to a variable. See Passing by reference in C
The pthread_join returns 0 as the function return value on success; then you know that the thread has been joined, and you can access the value from *value_ptr.
void *value = NULL;
if (pthread_join(thread, &value) == 0) {
// thread has ended, and the exit value is available in
// the value variable
}
其他提示
Essentially, pthread_join()
wants to return two peices of information:
- a success/failure indication
- whatever the thread returned (which has a type
void*
)
In C, the typical way a function 'returns' two separate values is to have the function return one of those values normally and to 'return' the ofther value in a location provided by the caller where the caller passes in a pointer to that location. So pthread_join()
:
- returns success/failure as the function's value
- returns the thread's
void*
result in a caller-provided location that the caller passes avoid**
to.
Note that in pthread_join()
's case, the caller-provided location is optional. NULL can be passed in id the caller isn't interested in that result. Making the caller-provided location optional is a common, by by no means universal, idiom.
pthread_exit()
doesn't need to use a void**
as it's argument because that argument isn't a result of the function. So it can simply take the value directly.