问题描述
一个典型的使用位域是定义一个节省空间的变量小于8位。我不明白的是,宣布这些位作为short,int和长,布尔等。例如值
A typical use of bitfield is to declare a space efficient variable smaller than 8 bits. What i don't understand is the value of declaring those bits as short, int , long , bool etc. For example
typedef struct{
int first:3,
short second:3,
char third:3
} somestruct;
在上述情况下,所有的3个变量,即第一,第二和第三个是3位长。什么是首先声明变量为int的价值,第二短,第三为char?
In above case, all 3 variables, i.e. first, second and third are 3 bit long. What is the value of declaring the variable first as int, second as short and third as char?
或者,为什么连一个数据类型要求?我应该能够申报如上
Or, why is even a data type required? I should be able to declare the above as
typedef struct{
first:3,
second:3,
third:3
} modifiedstruct;
该modifiedstruct不承担任何数据类型变量的第一,第二和第三。除$ P $的pting 3位的字符,数字或浮动的责任应的的责任东西的东西。
GCC和G ++在linux上允许上述行为。
Both gcc and g++ on linux allow the above behavior.
推荐答案
事实上,C标准只允许位域的类型为签署int或unsigned int(和_Bool在C99)的。如果你可以抛出一个短,长或CHAR在那里,这是一个编译器扩展。
Actually, the C standard only allows bitfields to be of type signed int or unsigned int (and _Bool in C99). If you can throw a short, long or char in there, that's a compiler extension.
至于为什么,主要原因是符号性。试想一下:
As to why, the main reason is signedness. Consider:
struct {
int s: 3;
unsigned u: 3;
} bf;
bf.s = 7;
bf.u = 7;
这两个位域的都是人。不过,C preserves星座,所以:
Both of these bitfields are all ones. However, C preserves sign, so:
(int)bf.s == -1 // Because signed conversions preserve the sign bit
bf.s >> 1 == -1 // So do right shifts on signed values
,而
(int)bf.u == 7 // Because the source is unsigned and so just a series of bits
bf.u >> 1 == 3 // Unsigned right shifts are just moving bits around as well
有关编译器,允许字符,它可能是同一种思维。字符的默认符号性是实现定义的,所以如果你想有一个位域的符号性,以配合您的编译器的字符的符号性,您可以将其定义为char。
For compilers that allow char, it's probably the same sort of thinking. The default signedness of char is implementation-defined so if you want a bitfield's signedness to match your compiler's char's signedness, you can define it as char.
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