将指定的函数分配给函数成员指针？

Question

特定

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

和

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

我们如何将函数（hello）分配给函数成员指针？

我尝试了这个（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)();

Answer 1

在分配指向对象的指针时，您无需取消引用。

g_ptr_table->hello = &hello; // note: the & is optional
g_ptr_table->hello(); // dereferencing the function pointer is also optional