最近，我一直在阅读“inside the c++ object model”。它说基类中使用的填充也应该复制到派生类中，以防你想将基类分配给派生类。于是，我在64位电脑下进行测试:

    class A {
    public:
        int valA;
        char a;
    };
    class B : public A {
    public:
        char b;
    };
    class C : public B {
    public:
        char c;
    };
    int main(){
        std::cout << sizeof(A) << " " << sizeof(B) << " " << sizeof(C) 
        << std::endl;
        C c;
        printf("%p\n%p\n%p\n",&c,&c.b,&c.c);
    }

结果如下:

    8 12 12

    0x7ffd22c5072c

    0x7ffd22c50734

    0x7ffd22c50735

那么为什么 C 和 B 的大小一样呢？虽然貌似B用了A中的3字节padding。

最佳答案

So why is C the same size of B?

因为 B 的尾部填充被 C::b 重用。填充可以重复使用，因为 B 不是 POD(普通旧数据)类(因为它不是标准布局类)。

Although it seems that B used the 3 byte padding in A.

A 的 padding 不能被 B 的其他子对象重用，因为 A 是一个标准的布局类并且可以简单地复制，即A 是一个 POD 类。

will the padding of base class be copied into the derived class?

我想您不是要问复制，而是要问派生类的基类子对象是否具有与单个类型相同的填充。

答案是，从上面可以推导出:填充将是相同的，除了尾部填充可以重新用于其他子对象，除非基类是 POD，在这种情况下它的填充不能被重复使用。

在padding可能被重用的情况下，是否会被标准没有规定，并且编译器之间存在差异。

Please explain or link to a definition of "standard layout types".

当前标准草案:

[basic.types]

... Scalar types, standard-layout class types ([class.prop]), arrays of such types and cv-qualified versions of these types are collectively called standard-layout types.

---

[class.prop] (in older versions of the standard, these may be found under [class] directly)

A class S is a standard-layout class if it:

• (3.1) has no non-static data members of type non-standard-layout class (or array of such types) or reference,

• (3.2) has no virtual functions and no virtual base classes,

• (3.3) has the same access control for all non-static data members,

• (3.4) has no non-standard-layout base classes,

• (3.5) has at most one base class subobject of any given type,

• (3.6) has all non-static data members and bit-fields in the class and its base classes first declared in the same class, and

• (3.7) has no element of the set M(S) of types as a base class, where for any type X, M(X) is defined as follows.107 [ Note: M(X) is the set of the types of all non-base-class subobjects that may be at a zero offset in X. — end note]

  • (3.7.1) If X is a non-union class type with no (possibly inherited) non-static data members, the set M(X) is empty.

  • (3.7.2) If X is a non-union class type with a non-static data member of type X0 that is either of zero size or is the first non-static data member of X (where said member may be an anonymous union), the set M(X) consists of X0 and the elements of M(X0).

  • (3.7.3) If X is a union type, the set M(X) is the union of all M(Ui) and the set containing all Ui, where each Ui is the type of the ith non-static data member of X.

  • (3.7.4) If X is an array type with element type Xe, the set M(X) consists of Xe and the elements of M(Xe).

  • (3.7.5) If X is a non-class, non-array type, the set M(X) is empty.

项目 (3.6) 适用于这种情况。 B 的一些成员没有首先在 B 中声明。特别是，B::A::valA 和 B::A::a 首先在 A 中声明。更友好的描述规则的方式是:类必须有没有直接成员，或者它的祖先都不能有成员。在这种情况下，基类和派生类都有成员，因此它不是标准布局。

关于c++ - 基类的填充会被复制到派生类中吗？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/52818411/