确实的sizeof（T）==的sizeof（常量T）和alignof（T）== alignof（常量T） [英] Does sizeof(T) == sizeof(const T) and alignof(T) == alignof(const T)

问题描述

似乎可以合理地假设 T 和常量牛逼将是两种类型，这将是相同的大小和具有相同的取向，但想一些实际系统后，似乎他们可以是不同的。

It seems reasonable to assume that T and const T would be two types that would be the same size and have the same alignment, but after thinking about some real systems, it seems that they could be different.

让我解释一下：

假设你有两种类型的存储系统：RAM和闪存（这是只读）。 RAM为8位寻址，而Flash是只有16位寻址。假设这是 T ：

Suppose you have a system with two types of memory: RAM and Flash (which is read only). The RAM is 8 bit addressable, while the Flash is only 16 bit addressable. Suppose this is T:

struct T
{
  uint8_t x;
  uint16_t y;
};

在字节寻址RAM这个结构是3个字节长....但在双字节寻址闪存（这是一个常量变量将驻留）这结构将必须至少是4个字节长，由于对准问题

In the byte addressable RAM this struct would be 3 bytes long.... but in the double byte addressable Flash (which is where a const variable would reside) this struct would have to be at least 4 bytes long, because of alignment issues.

因此​​，这里是我的问题：

So here is my question:

执行C和C ++标准保证了常量和非常量类型的大小和对齐方式？

Do the c and c++ standards guarantee the sizes and alignment of const and nonconst types?

推荐答案

第3.9.3节：

一型的CV-合格或CV-不合格的版本是不同的
  类型;但是，应当有相同的再presentation和对齐
  要求（3.11）。 53

The cv-qualified or cv-unqualified versions of a type are distinct types; however, they shall have the same representation and alignment requirements (3.11). 53

CV-合格这里指的是常量和挥发性。因此，答案是肯定的。

"cv-qualified" here refers to const and volatile. So the answer is, yes.

常量和挥发性只能指定访问指定对象的局限性/属性。它们不被认为是基类型本身的一部分;因此，它们可以在不影响的类型的属性。

const and volatile only specify the limitations/attributes of access to the specified object. They are not considered to be a part of the base type itself; hence they cannot affect the type's properties.

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