联合的位字段用法 [英] Bit Field Usage with Union
问题描述
我想通过使用Union创建一个带有位字段的简单包.但是,当我尝试将"bit1"设置为1时,所有我的位字段都变为"1".我该如何解决此问题,我想使用联合而不是使用struct来完成位字段部分.
I want to create a simple package with bit fields by using union. But when I tried to set "bit1" to 1, then all off my bit fields became "1". How can I solve this problem, I want to do that bit field part by using union not by using struct.
所以这是我的结构;
struct {
union{
uint8_t bit1 :1 ;
uint8_t bit2 :1 ;
uint8_t bit3 :1 ;
uint8_t bit4 :1 ;
uint8_t bit5 :1 ;
uint8_t bit6 :1 ;
uint8_t bit7 :1 ;
uint8_t bit8 :1 ;
}bits;
uint8_t trial;
}myStruct_t;
int main(int argc, char *argv[]) {
myStruct_t.bits.bit1 = 1;
myStruct_t.bits.bit2 = 0;
printf("%x",myStruct_t.bits);
printf("%x",myStruct_t.bits.bit1);
printf("%x",myStruct_t.bits.bit2);
return 0;
}
输出为:000.
推荐答案
用您的 struct
交换您的 union
.
即我相信您想要一个结构和一个整数的联合,而不是一个联合和一个整数的联合.(我保留了现在具有误导性的名称 myStruct_t
,现在可能应该是 myUnion_t
.)
I.e. I believe you want a union of a struct and an int, not a struct of a union and an int.
(I kept the now misleading name myStruct_t
, which probably should be myUnion_t
now.)
#include <stdio.h>
#include <stdint.h>
union
{
struct
{
uint8_t bit1 :1 ;
uint8_t bit2 :1 ;
uint8_t bit3 :1 ;
uint8_t bit4 :1 ;
uint8_t bit5 :1 ;
uint8_t bit6 :1 ;
uint8_t bit7 :1 ;
uint8_t bit8 :1 ;
}bits;
uint8_t trial;
}myStruct_t;
int main(int argc, char *argv[])
{
myStruct_t.trial=0; // use the encompassing union member for init
myStruct_t.bits.bit1 = 1; // now use the bitwise view to set bits
myStruct_t.bits.bit2 = 0;
printf("%x",myStruct_t.trial);
printf("%x",myStruct_t.bits.bit1);
printf("%x",myStruct_t.bits.bit2);
return 0;
}
我得到的输出是110,您可能会看到12810.除了缺少换行符(这会有所帮助)之外,这还意味着:
The output I get is 110, you might see 12810. Apart from the absent newlines, which would have helped, it means:
- 包括8bit-view在内的8bit-view的值至少为1,您可能会看到128
- 第1位的值为1
- 第2位的值为0
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