对象的大小是否受访问说明符类型和继承类型影响？ [英] Is size of the object affected by type of access-specifier and type of inheritance?

问题描述

在回答一个问题时，有一个讨论线程在我的答案下面。这表明取决于访问说明符（或可能是继承的类型） private / protected / public sizeof class 对象可能有所不同！

While answering one of the question, there was a discussion thread below my answer. Which suggests that depending on the access specifier (or may be the type of inheritance) private/protected/public the sizeof the class object may vary!

从他们的简短讨论中我仍然不明白，这怎么可能？

I still don't understand from their brief discussion, how is that possible ?

推荐答案

请注意以下C ++ 11的新语言

在C ++ 03 中，有一种语言使之成为可能，即9.2 [class.mem] / 12（强调我的意思）：

In C++03, there is language that makes this possible, 9.2 [class.mem]/12 (emphasis mine):

分配了声明为而没有中间访问说明的（非联盟）类的非静态数据成员，以便以后的成员在类对象中具有更高的地址。 由访问说明符分隔的非静态数据成员的分配顺序不确定（11.1）。实施一致性要求可能会导致两个相邻成员不能彼此立即分配；因此管理虚拟功能（10.3）和虚拟基类（10.1）的空间要求也可能如此。

Nonstatic data members of a (non-union) class declared without an intervening access-specifier are allocated so that later members have higher addresses within a class object. The order of allocation of nonstatic data members separated by an access-specifier is unspecified (11.1). Implementation alignment requirements might cause two adjacent members not to be allocated immediately after each other; so might requirements for space for managing virtual functions (10.3) and virtual base classes (10.1).

因此，给出以下定义：

class Foo
{
    char a; //8 bits
    // a must come before b, so 3 bytes of padding have to go here to satisfy alignment
    int b; //32 bits
    char c; //8 bits
    // 24 bits of padding required to make Foo a multiple of sizeof(int)
};

在具有32位（ int ）对齐方式，不允许编译器重新排序 c 到 b 之前，从而在其中强制插入其他填充在 a 和 b 之间，并且在 c 之后对象的大小（使 sizeof（Foo）== 12 ）。但是，为此：

on a system with 32 bit (int) alignment, the compiler is not allowed to reorder c to come before b, forcing the insertion of additional padding padding in between a and b, and after c to the end of the object (making sizeof(Foo) == 12). However, for this:

class Foo
{
    char a;
public:
    int b;
public:
    char c;
};

a 和（ b 和 c ）由访问说明符分隔，因此编译器可以自由地执行这种重新排序，从而使

a and (b and c) are separated by an access specifier, so the compiler is free to perform such reordering, making

memory-layout Foo
{
    char a; // 8 bits
    char c; // 8 bits
    // 16 bits of padding
    int b; // 32 bits
};

sizeof（Foo）== 8 。

在C ++ 11 中，语言略有变化。 N3485 9.2 [class.mem] / 13说（强调我的意思）：

In C++11, the language changes slightly. N3485 9.2 [class.mem]/13 says (emphasis mine):

（非联盟）类的非静态数据成员具有相同的访问控制（第11条），以便以后的成员在类对象中具有更高的地址。 未指定具有不同访问控制的非静态数据成员的分配顺序（第11条）。实施一致性要求可能会导致两个相邻成员不能彼此立即分配；管理虚拟功能（10.3）和虚拟基类（10.1）的空间要求也可能如此。

Nonstatic data members of a (non-union) class with the same access control (Clause 11) are allocated so that later members have higher addresses within a class object. The order of allocation of non-static data members with different access control is unspecified (Clause 11). Implementation alignment requirements might cause two adjacent members not to be allocated immediately after each other; so might requirements for space for managing virtual functions (10.3) and virtual base classes (10.1).

这意味着在C ++中参照图11，在上面的示例（由3个公众分隔）中，仍然不允许编译器执行重新排序。

This means that in C++11, in the above example (separated by 3 publics), the compiler is still not allowed to perform the reordering. It would have to be something like

class Foo
{
    char a;
public:
    int b;
protected:
    char c;
};

，其中放置 a ， b 和 c 具有不同的访问控制。

, which places a, b, and c with different access control.

请注意， C ++ 11规则，给出如下定义：

Note that under the C++11 rules, given a definition like:

class Foo
{
    char a;
public:
    int b;
protected:
    char c;
public:
    int d;
};

编译器必须在<$之后放置 d c $ c> b ，即使它们被访问说明符隔开。

the compiler must put d after b, even though they are separated by access specifiers.

（那说，我不知道有任何实际利用任何一种标准提供的纬度的实现）

(That said, I'm not aware of any implementation that actually takes advantage of the latitude offered by either standard)

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