sizeof（T）== sizeof（const T）和alignof（T）== alignof（const T） [英] Does sizeof(T) == sizeof(const T) and alignof(T) == alignof(const T)

问题描述

似乎合理的假设 T 和 const T 将是两种类型的大小相同，

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：

假设你有一个有两种类型内存的系统：RAM和Flash（这是只读的）。 RAM是8位可寻址的，而闪存只有16位可寻址。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;
};

在字节可寻址RAM中，这个结构将是3字节长....但是在双字节可寻址的闪存（这是一个 const 变量将驻留）这个结构将不得不至少4个字节长，因为对齐问题。

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.

这里是我的问题：

c和c ++标准保证 const 和非 const 类型？

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

推荐答案

3：

类型的cv限定或cv非限定版本是不同的
类型;然而，它们应具有相同的代表和一致性
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-qualified这里指 const 和 volatile 。所以答案是，是的。

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

const 和 volatile 仅指定对指定对象的访问的限制/属性。它们不被认为是基本类型本身的一部分;因此它们不会影响类型的属性。

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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