[英]Initilizing structures containing bit-fields in C
Im trying to understand a bit more about the workings of bitfields. 我试图进一步了解位域的工作原理。
Given the following code: And assuming int is 32 bits 给定以下代码:并且假设int是32位
#include <stdio.h>
int main()
{
struct byte
{
int one:1;
};
struct byte var = {3};
printf("%d\n", var.one);
printf("%#x\n", var);
return 0;
}
The output I get is: 我得到的输出是:
-1 0x1
However I was expecting to see: 但是我期望看到:
-1 0x3
Since 以来
struct byte var = {3};
is assigning the value 3 to the 4 bytes of int, isn't it? 将值3分配给int的4个字节,不是吗?
From the output I actually get it appears as if the above code line tries to store the value 3 into the 1 bit field hence printing 0x1
as the second output line. 从输出中我实际上得到的结果似乎是上述代码行试图将值3存储到1位字段中,因此将
0x1
打印为第二条输出行。
So my question would be: 所以我的问题是:
How does these initializations and assignments on whole structures work? 这些对整个结构的初始化和分配如何工作?
Also, why are the {}
necessary? 另外,为什么
{}
必需的?
int one:1;
With this, you declare an int with only one bit which is used for the sign
bit. 这样,您就可以声明只有一个位用于
sign
位的整数。 So you see -1
. 所以您看到
-1
。
If you want to store 3 (011), then you need to have 2 (data) +1(sign) bits in total. 如果要存储3(011),则总共需要2(数据)+1(符号)位。 So, it should be:
因此,应为:
struct byte
{
int one:3;
};
Or use an unsigned int. 或使用unsigned int。
struct byte
{
unsigned int one:2;
};
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