c - 如何将参数传递给函数指针

Question

我正在使用 c 中的函数指针，因为我需要自定义 API 库的回调机制。用一个简单的例子总结一下：

*userfunction*(SY_msg msg)
{
  /* do something */
};

SY_msg的大小为 1024 字节。因此堆栈中有 1024 个字节。

指向userfuncion()的指针作为 calback_wrapper[] 中的第一个元素存在。

here is an example of use:
// (...) some code
    SY_msg* msg;
    msg = SYmalloc (sizeof(SY_msg)); /* it takes 1024 bytes in heap */
// (...) some code
    calback_wrapper[0] (*msg); /*  1204 are passed by value  */
    /* during userfunction() execution , 1024 unused bytes are present in the heap */
    free (msg); /* now finally heap is free */
// (...) some code

但我想拥有以下内容：

// (...) some code
    SY_msg* msg;
    msg = SYmalloc (sizeof(SY_msg)); /* it takes 1024 bytes in heap */
// (...) some code
    memcpy(someplace,msg,sizeof(SY_msg); /*  where "someplace" is a point in the stack referred by the argument of userfunction()  */
    free (msg); /*  heap is free */
    calback_wrapper[0] (*someplace); /* is starts userfunction() execution */
// (...) some code

有可能找到“某个地方”的地址吗？我的编译器是 gcc。

Answer 1

什么让你做不到

// (...) some code
SY_msg msg, * pmsg;
pmsg = SYmalloc (sizeof(SY_msg)); /* it takes 1024 bytes in heap */
// (...) some code using pmsg instead of msg
memcpy(&msg, pmsg, sizeof(SY_msg)); /*  where "someplace" is a point in the stack referred by the argument of userfunction()  */
free (pmsg); /*  heap is free */
calback_wrapper[0] (msg); /* is starts userfunction() execution */
// (...) some code

在上面的示例中，您可以替换

memcpy(&msg, pmsg, sizeof(SY_msg));

经过

msg = *pmsg;

Answer 2

我的问题中有错误的假设。用户 function() 的参数是在函数调用之后的堆栈中分配的。也许某种“contextswich”可以解决这个问题。例子：

• 调用用户函数（）；
• “上下文”
• 释放堆
• “上下文”
• 恢复用户函数（）；

但无论如何，都需要汇编代码片段。