Question

我有以下代码：

/* callback taking a pointer to a "generic" msg as argument */
typedef void (*Callback_t)(void *msg);

/* initialize callback pointer */
Callback_t const Callback[] =
{
  ProcessMsgAction0,  /**< 0: MSGID_ACTION_0 */
  ProcessMsgAction1,  /**< 1: MSGID_ACTION_1 */
  /* ... */
};

/* Call the callback according to msgid */
msg = GetMessage();
Callback[msg->id](msg);

/* implement callback for each message id */
void ProcessMsgAction0(action0_t *msg) { /* ... */ }
void ProcessMsgAction1(action1_t *msg) { /* ... */ }

我虽然void *被认为是通用指针。因此，无论我给出的任何地址都不会产生警告，但上述情况action0_t*与void*不同，因此产生以下警告：initialization from incompatible pointer type

如果我想抑制警告，我唯一的解决办法是将ProcessMsgAction参数更改为void *？

使用的编译选项：-ansi -pedantic -Wall -Wextra

Answer 1

I though void* was considered as generic pointer

这是事实，但这个概念并未扩展到函数类型中的匹配参数。

Callback_t的简单类型转换有效，see this functional example on ideone.

带有void指针参数的回调产生警告

1 个答案: