0

给定

void* hello () {
   cout << "Test.\n";          
}

struct _table_struct {
    void *(*hello) ();
};

我们如何将函数(hello)分配给函数成员指针?

我试过这个(主要):

 _table_struct g_table;
 _table_struct *g_ptr_table = &g_table;

 // trying to get the struct member function pointer to point to the designated function
 (*(g_ptr_table)->hello) =  &hello; // this line does not work

 // trying to activate it
 (*(g_ptr_table)->hello)();
4

1 回答 1

1

在将指针分配为指向对象时,您不会取消引用它。

g_ptr_table->hello = &hello; // note: the & is optional
g_ptr_table->hello(); // dereferencing the function pointer is also optional
于 2015-03-11T01:39:16.940 回答