Question

In hello world example of pthread it is stated:

#include <pthread.h>
#include <stdio.h>

void * print_hello(void *arg)
{
  printf("Hello world!\n");
  return NULL;
}

int main(int argc, char **argv)
{
  pthread_t thr;
  if(pthread_create(&thr, NULL, &print_hello, NULL))
  {
    printf("Could not create thread\n");
    return -1;
  }

  if(pthread_join(thr, NULL))
  {
    printf("Could not join thread\n");
    return -1;
  }
  return 0;
}

As you can see print_hello in pthread_create() has no argument, however in the definition, it looks like void * print_hello(void *arg)

What does that mean?

Now assume I have this implementation

void * print_hello(int a, void *);
int main(int argc, char **argv)
{
  pthread_t thr;
  int a = 10;
  if(pthread_create(&thr, NULL, &print_hello(a), NULL))
  {
    printf("Could not create thread\n");
    return -1;
  }
  ....
  return 0;
}
void * print_hello(int a, void *arg)
{
  printf("Hello world and %d!\n", a);
  return NULL;
}

Now I get this error:

too few arguments to function print_hello

So how can I fix that?

Answer 1

pthread passes one argument of type void * to thread function, So you can pass a pointer to any type of data that you want as fourth argument of pthread_create function, look at the example below that fixes your code.

void * print_hello(void *);
int main(int argc, char **argv)
{
    pthread_t thr;
    int a = 10;
    if(pthread_create(&thr, NULL, &print_hello, (void *)&a))
    {
        printf("Could not create thread\n");
        return -1;
    }
    ....
    return 0;
}

void * print_hello(void *arg)
{
    int a = (int)*arg;
    printf("Hello world and %d!\n", a);
    return NULL;
}