亡灵对象([basic.life]/8):为什么允许引用重新绑定(和const修改)?

问题描述

我将undead子句称为C ++规则，即销毁对象后，如果在同一地址创建一个新对象，则有时可以将其视为与旧对象相同的对象.该规则始终存在于C ++中，但在附加条件上有所更改.

I call the undead clause the C++ rule that after the destruction of an object, if a new object is created at the same address, it can sometimes be considered the same object as the old one. That rule always existed in C++ but with some changes on the additional conditions.

这个问题使我阅读了最新的不死条款. 生命周期中的修订条件为:

[basic.life]/8 :

I was made to read the latest undead clause by this question. The revised conditions in Lifetime [basic.life]/8 are:

嗯，嗯.位于不同地址的对象将不是同一对象.

Well, duh. An object at a different address would not be the same object.

再次，du.

它不能是经典的基类(或具有特殊声明使其地址不唯一的成员).再次，嗯.

It cannot a base class, classic (or a member with a special declaration that makes its address not unique). Again, duh.

现在很有趣.被替换的对象不能是:

Now that's interesting. The object being replaced can't be either:

• 完整的const对象
• 完整的const对象的一部分

另一方面，要复活的对象可以是:

On the other hand, the object being resurrected can be:

1. 一个const成员子对象
2. 此类const成员的子对象
3. 一个const对象数组中的元素

const子对象

所以在我看来，所有这些对象x都可以复活:

const成员子对象

Const member subobject

struct CI {
  const int x;
};

CI s = { 1 };
new ((void*)&s.x) int(2);
int r = s.x; // OK, 2

const成员的子对象:

Subobject of const member:

struct T {
  int x;
};

struct CT {
  const T m = { 1 };
};

CT s;
new ((void*)&s.m.x) int (2);
int r = s.m.x;

一个const对象数组中的元素:

Element in an array of const objects:

const int x[1] = { 1 };
new ((void*)&x[0]) int (2);
int r = x[0];

具有const和引用成员的类

似乎也不禁止带有const或引用成员的类类型的对象；恢复的对象仍称为x.

具有const成员的类:

Class with a const member:

struct CIM {
  CIM(int i): m(i) {}
  const int m;
};

CIM x(1);
new ((void*)&x) CIM(2);
int r = x.m; // OK, 2

带有参考成员的课程:

struct CRM {
  CRM (int &r): m(r) {}
  int &m;
};

int i=1,j=2;
CRM x(i);
new ((void*)&x) CRM(j);
int r = x.m; // OK, 2

问题

1. 该条款的解释正确吗?
2. 如果是这样，还有其他禁止这些重写操作的子句吗?
3. 如果是，那是故意的吗?为什么改变了?
4. 这是代码生成器的重大变化吗?所有的编译器真的都支持吗?它们不是基于const成员进行优化，数组的const元素是不可变的并且引用不是可反弹的吗?
5. 奖金问题:这会影响具有足够存储类(当然不是动态创建的对象)和足够的初始化的const对象的ROM能力吗?

注意:我后来添加了奖金，因为讨论中出现了将常量放入ROM的情况.

Note: I added the bonus later because putting constants in ROM came up in the discussion.

推荐答案

如果与对象生存期相关的标准的所有要求都不在[basic-life]中，那将是令人惊讶的.

It would be surprising if all requirement of the standard related to object life-time were not in [basic-life].

在您引用的标准段落中，很少有将完整"形容词无意中添加到名称对象"的情况.

There are few chances that the "complete" adjective has been inadvertedly added to the name "object" in the standard paragraph you cite.

在论文 P0137 中，人们可以读懂这一道理(在下面的@LanguageLawyer评论中引用的论文):

In the paper P0137, one can read this rational (paper cited in @LanguageLawyer comment below):

为了使我们放心，我们可以验证编译器确实遵循字母的标准措辞:它们对完整的const对象执行常量优化，但对非const完整对象的const成员对象不执行常量优化:

To reassure us, we can verify that compilers do follow the standard wording at the letter: they perform constant optimization for complete const objects but not for const member suboject of non const complete objects:

让我们考虑此代码:

struct A{const int m;};

void f(const int& a);

auto g(){
    const int x=12;
    f(x);
    return x;
}

auto h(){
    A a{12};
    f(a.m);
    return a.m;
}

在将x86_64作为目标时，Clang和GCC都会生成此程序集:

Both Clang and GCC generates this assembly when targeting x86_64:

g():                                  # @g()
        push    rax
        mov     dword ptr [rsp + 4], 12
        lea     rdi, [rsp + 4]
        call    f(int const&)
        mov     eax, 12     ;//the return cannot be anything else than 12
        pop     rcx
        ret
h():                                  # @h()
        push    rax
        mov     dword ptr [rsp], 12
        mov     rdi, rsp
        call    f(int const&)
        mov     eax, dword ptr [rsp]  //the content of a.m is returned
        pop     rcx
        ret

返回的值放在寄存器eax中(根据ABI规范:System V x86处理器特定的ABI):

The returned value is placed in register eax (according to the ABI specification: System V x86 processor specific ABI):

• 在函数g中，编译器可以自由假定不能在调用f的过程中更改x，因为x是完整的const对象.因此，值12作为直接值直接放置在eax寄存器中:mov eax, 12.

• In the function g the compiler is free to suppose that x can not be changed accross the call to f because x is a complete const object. So the value 12 is placed directly in the eax register as an immediate value: mov eax, 12.

在函数h中，编译器不能随意假设不能在调用f的过程中更改a.m，因为a.m不是完整const对象的子对象.因此，在调用f之后，必须将a.m的值从内存加载到eax:mov eax, dword ptr [rsp].

In the function h the compiler is not free to suppose that a.m can not be changed accross the call to f because a.m is not a suboject of a complete const object. So after the call to f the value of a.m must be loaded from memory to eax : mov eax, dword ptr [rsp].

这篇关于亡灵对象([basic.life]/8):为什么允许引用重新绑定(和const修改)?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！