c - 了解 C 中回调的情况

Question

我想了解这段代码的实际含义，尤其是。函数放在花括号中的最后一部分。 broadcast_open 函数是否以某种方式调用函数 broadcast_recv？如果是，如何？

static void broadcast_recv(struct broadcast_conn *c, const rimeaddr_t *from)
{
  printf("broadcast message received from %d.%d: '%s'\n",
         from->u8[0], from->u8[1], (char *)packetbuf_dataptr());
}

static const struct broadcast_callbacks broadcast_call = {broadcast_recv};
static struct broadcast_conn broadcast;

PROCESS_THREAD(example_broadcast_process, ev, data)
{

  broadcast_open(&broadcast, 129, &broadcast_call);

  ...

}

void broadcast_open(struct broadcast_conn *c, uint16_t channel, const struct broadcast_callbacks *u)   
{ 
      abc_open(&c->c, channel, &broadcast); 
      c->u = u; 
      channel_set_attributes(channel, attributes); 
} 

Answer 1

似乎 broadcast_callbacks 是一个定义如下的结构:

struct broadcast_callbacks
{
    void (*callback)(struct broadcast_conn *, const rimeaddr_t *from);
};

然后是行

static const struct broadcast_callbacks broadcast_call = {broadcast_recv};

创建一个新的结构对象，其成员指向 broadcast_recv 函数。该成员现在可用于调用 broadcast_recv(这可能是 broadcast_open 的一部分)。