How can the function Thello get its arguments?

Question

#include "thread.h"

int x = 0;

void Thello(int id) {
  usleep(id * 100000);
  printf("Hello from thread #%c\n", "123456789ABCDEF"[x++]);
}

int main() {
  for (int i = 0; i < 10; i++) {
    create(Thello);
  }
}

when executing it , I get this:

Hello from thread #1
Hello from thread #2
Hello from thread #3
Hello from thread #4
Hello from thread #5
Hello from thread #6
Hello from thread #7
Hello from thread #8
Hello from thread #9
Hello from thread #A

"thread.h" is a lib

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <stdatomic.h>
#include <assert.h>
#include <unistd.h>
#include <pthread.h>

#define NTHREAD 64
enum { T_FREE = 0, T_LIVE, T_DEAD, };
struct thread {
  int id, status;
  pthread_t thread;
  void (*entry)(int);
};

struct thread tpool[NTHREAD], *tptr = tpool;

void *wrapper(void *arg) {
  struct thread *thread = (struct thread *)arg;
  thread->entry(thread->id);
  return NULL;
}

void create(void *fn) {
  assert(tptr - tpool < NTHREAD);
  *tptr = (struct thread) {
    .id = tptr - tpool + 1,
    .status = T_LIVE,
    .entry = fn,
  };
  pthread_create(&(tptr->thread), NULL, wrapper, tptr);
  ++tptr;
}

void join() {
  for (int i = 0; i < NTHREAD; i++) {
    struct thread *t = &tpool[i];
    if (t->status == T_LIVE) {
      pthread_join(t->thread, NULL);
      t->status = T_DEAD;
    }
  }
}

__attribute__((destructor)) void cleanup() {
  join();
}

while I do create(Thello) , How can Thello() get its arguments?