能否位字段只能是结构/联合的领域，从来不与QUOT;正常＆QUOT;，＆QUOT;独立＆QUOT;变量？ [英] Can bit-fields only be fields of a structure/union, never "normal", "stand-alone" variables?

问题描述

位字段的字段部分似乎表明，他们只能是结构或联合中的字段。

The field part of bit-fields seems to suggest that they can only be fields inside a structure or union.

位域可以是一个典型的独立的变量，任何集合数据类型一样工会或结构之外如下：

Can a bit-field be a typical "stand-alone" variable, outside any aggregate data-type like union or structure as follows:

   int sum:6;   //Can this work as a declaration statement?
   sum=4;

如果没有，为什么？如果位字段基本上都是打算用更少的内存，我们为什么不能声明任何变量作为位字段，如果我们知道它不会超过大小？

If not, why so? If bit-fields are basically intended to use less memory, why can't we declare any variable as a bit-field if we know it won't exceed that size?

推荐答案

位字段只有结构或联合的一部分，因为那是什么的C的标准允许。这本来是可以决定不同。为什么委员会决定写标准的方式是，你得去问他们。

Bit-fields are only part of structs or unions because that's what the C standard allows. It would have been possible to decide differently. Why the committee decided to write the standard the way it is, you would have to ask them.

要发现在C99标准的信息：

To find the information in the C99 standard:

从表的内容：

6.7声明

6.7.2类型说明符（好吧，这个人是有点晦涩）

6.7.2 Type specifiers (okay, this one is a little bit obscure)

6.7.2.1结构和联合说明符。

6.7.2.1 Structure and union specifiers.

一个位字段是 declarator_opt：恒恩pression 语法的一部分。此语法只允许在这里，和一个后果是，一个人不能在别处声明位字段。

One bit-field is the declarator_opt : constant-expression part of the syntax. This syntax is only allowed here, and a consequence is that one cannot declare bit-fields elsewhere.

语法告诉你有一个位字段的名称可以被省略，如果你好奇的这类信息。第6.7.2.1:3更进一步，禁止名字0长度位字段。

The syntax tells you that the name of a bit-field can optionally be omitted, if you were curious for this sort of information. Clause 6.7.2.1:3 goes further and prohibits names for 0-length bit-fields.

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