如何创建可变大小的位字段数组?
下面的代码是我尝试的,但没有成功。
#include <stdio.h>
int main()
{
int n=4;
struct bite{
unsigned a1:2;
unsigned a2:2;
:
:
unsigned a(n-1):2;
unsigned a(n):2;
}bits;
for(i=1;i<=n;i++)
bits.a[i]=i;
for(i=1;i<=n;i++)
printf("%d ",bits.a[i]);
return 0;
}
最佳答案
无法在运行时定义struct
的成员。
您可以使用char
数组和一些宏来模拟位数组。
#define BitArray(array, bits) \
unsigned char array[bits / 8 + 1]
#define SetBit(array, n) \
do { array[n / 8] |= (1 << (n % 8)) } while (0)
#define GetBit(array, n) \
((array[n / 8] >> (n % 8)) & 1)
int main(void)
{
BitArray(bits, 42); /* Define 42 bits and init to all 0s
(in fact this allocates memory for (42/8 + 1) * 8 bits). */
SetBit(bits, 2); /* Set bit 2. */
int bit2 = GetBit(bits, 2); /* Get bit 2 */
...
类似的2位单词是根据您的代码编写的:
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#define Define2BitWordArray(array, two_bit_words) \
unsigned char array[two_bit_words / 4 + 1]
#define Set2BitWord(array, n, value) \
do { array[n / 4] |= (unsigned char)((value & 0x11) << (2 * (n % 4))); } while (0)
#define Get2BitWord(array, n) \
((array[n / 4] >> (2 * (n % 4))) & 0x11)
int main(void)
{
size_t n = 10;
Define2BitWordArray(bits, n); /* Define 10 two-bits words
(in fact this allocates memory
for (10/4 + 1) * 4 two-bit bits). */
memset(bits, 0, sizeof bits); /* Set all bits to 0. */
for(size_t i = 0; i < n; i++) /* C array's indexes are 0-based. */
{
Set2BitWord(bits, i, i);
}
for(size_t i = 0; i < n; i++)
{
printf("%d ", Get2BitWord(bits, i));
}
}
关于c - C中的位域结构数组,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/46141169/