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。总结一个简单的例子：

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

SY_msg 的大小为1024个字节。因此，堆栈中有1024个字节。

指向 userfuncion（）的指针是calback_wrapper []中的第一个元素。

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

但是我想要以下内容：

// (...) 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

可以找到“某个地方”的地址吗？我的编译器是gcc。

Answer 1

使您无法做的事情

// (...) 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));

作者

msg = *pmsg;

Answer 2

我的问题有错误的假设。用户function（）的参数在函数调用之后立即分配到堆栈中。也许某种“上下文”可以解决这个问题。示例：

调用userfunction（）;
“上下文三明治”
释放堆
“ contextswich”
恢复userfunction（）;

但是在任何情况下，都要求提供汇编代码段。

如何将参数传递给函数指针

2 个答案: