位字段如何与C中的位填充相互作用

问题描述

当存在填充位时，我有两个关于位字段的问题.

I have two questions concerning bit fields when there are padding bits.

说我有一个结构定义为

struct T { 
    unsigned int x: 1; 
    unsigned int y: 1;
};

结构T仅具有实际使用的两位.

Struct T only has two bits actually used.

问题1:这两位始终是基础无符号int的最低有效位吗?还是取决于平台?

Question 1: are these two bits always the least significant bits of the underlying unsigned int? Or it is platform dependent?

问题2:那些未使用的30位是否总是初始化为0? C标准对此有何评论?

Question 2: Are those unused 30 bits always initialized to 0? What does the C standard say about it?

推荐答案

否，它既取决于系统，也取决于编译器.您永远无法假设或知道它们是MSB或LSB.

No, it is both system and compiler dependent. You can never assume or know that they are MSB or LSB.

取决于初始化结构的方式.在本地范围内未初始化的结构可能包含以填充位/字节为单位的垃圾值.使用至少一个初始化程序集进行初始化的结构，即使在填充字节:my_struct = { something };中也保证包含零.

Depends on how you initialize the struct. A struct at local scope which isn't initialized may contain garbage values in padding bits/bytes. A struct that is initialized with at least one initializer set, is guaranteed to contain zero even in padding bytes: my_struct = { something };.

来源

上述原因为何的语言律师细节有些复杂.

The language-lawyer details of why the above works are somewhat complex.

C17 6.7.9/9(重点是我)说:

C17 6.7.9/9 (emphasis mine) says this:

这意味着我们完全不能相信填充位/字节.但是上面的规则(§20强调我的规则)有一个例外:

This means that we cannot trust padding bits/bytes at all.But then there's this exception to the above rule (§20 emphasis mine):

意思是，如果至少有一个初始化程序，则适用以下静态存储初始化规则:

Meaning that if there's at least one initializer, then the following rule of static storage initialization applies:

C17 6.7.9/10(强调我的意思):

C17 6.7.9/10 (emphasis mine):

• 如果是聚合，则根据这些规则初始化(递归)每个成员，并且将任何填充都初始化为零位;

• if it is an aggregate, every member is initialized (recursively) according to these rules, and any padding is initialized to zero bits;

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