如何在C ++中将成员变量指针转换为泛型类型 [英] How to cast member variable pointer to generic type in C++

问题描述

我在我的应用程序中有类似的代码：

I have code similar to this in my application:

class A
{
  public: int b;
}

class C
{
  public: int d;
}

void DoThings (void *arg1, MYSTERYTYPE arg2);

A obj_a;
C obj_c;

DoThings(&obj_a, &A::b);
DoThings(&obj_c, &C::d);

问题是MYSTERYTYPE应该是什么？

The question is - What should MYSTERYTYPE be? neither void* nor int work, despite the value &A::b being printed just fine if you output it through a printf.

说明：是，&A;如果你通过printf输出的话，无论是void *还是int工作，尽管值& :: b在C ++下定义。是的，我想获得类成员的偏移量。是的，我很棘手。

Clarifications: Yes, &A::b is defined under C++. Yes, I am trying to get the offset to a class member. Yes, I am being tricky.

编辑：哦，我可以使用offsetof（）。非常感谢。

Oh I can use offsetof(). Thanks anyway.

推荐答案

你有一个数据成员指针指向两个不相关的类。嗯，你找不到一个可以容纳两个指针的常见类型。它只有在函数参数是指向成员的数据成员指针时才会起作用，因为如果基数包含它，它也保证包含成员：​​

You have a data member pointer to two unrelated classes. Well, you can't find a common type that can hold both pointers. It will only work if the function parameter is a data member pointer to a member of the derived, because it's guaranteed to contain the member too, if a base contains it:

struct a { int c; }; struct b : a { }; int main() { int b::*d = &a::c; }

---

strong> Update ：我想我应该写为什么上面的转换从 a :: * 到 b :: * implicitly。毕竟，我们通常有 b * 到 a * ！请考虑：

Update: I think i should write why the above converts from a::* to b::* implicitly. After all, we usually have b* to a* ! Consider:

struct a { };
struct b : a { int c; };
struct e : a { };
int main() { int a::*d = &b::c; e e_; (e_.*d) = 10; /* oops! */ }

如果上述操作有效，上面的不是有效，因为从 b :: * 到 a :: * 不是隐式的。正如你所看到的，我们为b :: c分配了一个指针，然后我们可以使用一个不包含它的类来解引用它。 （ e ）。编译器强制执行这个顺序：

If the above would be valid, you would really much screw up. The above is not valid, because conversion from b::* to a::* is not implicit. As you see, we assigned a pointer to b::c, and then we could dereference it using a class that doesn't contain it at all! (e). The compiler enforces this order:

int main() { int b::*d = &b::c; e e_; (e_.*d) = 10; /* bug! */ }

现在无法编译，因为 e 不是从成员指针指针所属的类 b 派生的。好！然而，以下是非常有效的并且编译，当然（改变的类 a 和 b ）：

It fails to compile now, because e is not derived from b, the class the member pointer pointer belongs to. Good! The following, however, is very valid and compiles, of course (changed classes a and b):

struct a { int c; };
struct b : a { };
struct e : a { };
int main() { int e::*d = &a::c; e e_; (e_.*d) = 10; /* works! */ }

---

为了使它适用于你的情况，你必须使你的函数成为一个模板：

To make it work for your case, you have to make your function a template:

template<typename Class>
void DoThings (int Class::*arg) { /* do something with arg... */ }

b $ b

现在，编译器将自动推导出给定成员指针所属的类。您必须在成员指针旁边传递实例才能使用它：

Now, the compiler will auto-deduce the right class that the given member pointer belongs too. You will have to pass the instance alongside of the member pointer to actually make use of it:

template<typename Class>
void DoThings (Class & t, int Class::*arg) { 
    /* do something with arg... */ 
    (t.*arg) = 10;
}

如果你只想设置一些成员， DoThings，以下足够：

If you just want to set some member you already know at the time you write DoThings, the following suffices:

template<typename Class>
void DoThings (Class & t) {  
    t.c = 10;
}

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