分享程序员开发的那些事...
更新时间：2023-02-14 22:43:14
ol>
这是正确的行为。以这种方式考虑： ClassB 的所有实例都具有成员 ClassA :: foo ，但不是所有 ClassA 有成员 ClassB :: foo ;只有那些实际是 ClassB 实例的基类子对象的 ClassA 实例才有它。因此，将 ClassB :: foo 分配到 ClassAFoo ，然后使用 ClassAFoo 结合纯 ClassA 对象将试图调用不存在的函数。
如果从 ClassB 中删除 foo ，表达式 ClassB :: foo acutally是指在 ClassB 中继承的 ClassA :: foo ，因此没有问题。
要详细说明1. further：指向成员的指针实际上与正常指针相反。使用正常指针，可以将 ClassB * 分配到 ClassA * 中（因为类B 也是 ClassA 的实例，但反之亦然。使用成员指针，可以将 ClassA :: * 分配到 ClassB :: * （因为 ClassB 包含 ClassA 的所有成员，但反之不行。
I am using function pointer in my project, facing problem, created a test case to show it... below code fail with below error on MSVC2005 (in simple words i want to access dervied class function through base class function pointer)
error C2440: '=' : cannot convert from 'void (__thiscall ClassB::* )(void)' to 'ClassAFoo'
class ClassA {
public:
virtual void foo()
{
printf("Foo Parent");
}
};
typedef void (ClassA::*ClassAFoo)();
class ClassB : public ClassA {
public:
virtual void foo()
{
printf("Foo Derived");
}
};
int main() {
ClassAFoo fPtr;
fPtr = &ClassB::foo;
}
My questions are
ClassB::foo, this code compile fine, without any further modification, Why is this so, should not fPtr = &ClassB::foo; again result in compile time error?
It's correct behaviour. Think of it this way: all instances of ClassB have the member ClassA::foo, but not all instances of ClassA have the member ClassB::foo; only those instances of ClassA which are actually the base class subobject of a ClassB instance have it. Therefore, assigning ClassB::foo into ClassAFoo and then using ClassAFoo in combination with a "pure" ClassA object would try to call a nonexistent function.
If you remove foo from ClassB, the expression ClassB::foo acutally refers to ClassA::foo which is inherited in ClassB, so there's no problem there.
To elaborate on 1. further: pointers to members actually work the opposite way to normal pointers. With a normal pointer, you can assign ClassB* into ClassA* (because all instances of ClassB are also instances of ClassA), but not vice versa. With member pointers, you can assign ClassA::* into ClassB::* (because ClassB contains all the members of ClassA), but not vice versa.