c - 从 A 中提取某些位并仅替换 B 中某个位置的那些提取位
问题描述
提取和修改位
从打包的 final_value 中的一个位置选择 n 个位并写入任意位置,而不修改 uint16_t test_bit = 0x3048 的原始内容。预期输出 = 0x5048
例如:从 final_val 中选择 010( 3bits from position 17 ) 并写入任意位置( position 11 )
0x3048 = 01 11 0 000 0100 1000;
0x5048 = 01 01 0 000 0100 1000
几个例子:
例子A:
粗体是提取的。我们从 Val_1 中提取位 0 到 7,并仅替换 Val_2 中的位 0 到 7,使位 8 到 15 保持不变。
Val_1 0x1b7 0000 0001 1011 0111
Val_2 0x27b7 0010 0111 1011 0111
示例 B:
从 Val1[从位 8 到 10] 中提取 3 位,并将其替换为 Val_2[从 11 到 13]。
Val_1 0x129 0000 0 001 0010 1001
Val_2 0x4C48 01 00 1 100 0100 1000
到目前为止尝试过:
#include <stdio.h>
#include <stdint.h>
void read_and_write(uint32_t* final_val, uint16_t* write_val, uint8_t start_pos, uint8_t end_pos)
{
uint32_t temp = *final_val;
*write_val = (uint16_t) ((temp >> start_pos) & ((1 << end_pos) - 1)); // store the desired number of bits in write_val
*final_val = (temp >> end_pos); //shift final_val by end_pos since those bits are already written
printf("\n temp %x, write_val %x, final_val %x ", temp, *write_val, *final_val);
}
void main()
{
uint32_t final_val = 0x0; //Stores 20 extracted bits from val1, val2 and val3 into final_val (LSB to MSB in order)
uint16_t ext_val1 = 0x80;
uint8_t ext_val2 = 0x0;
uint8_t ext_val3 = 0x2;
final_val = (ext_val1 | (ext_val2 << 9) | (ext_val3 << 17));
printf ("\n final_val %x", final_val);
uint16_t data_1, data_2, data_3, write_val1, write_val2, write_val3;
// Read first 9 bits of final_val and write only into [0:9] position of existing data_1
uint8_t start_pos = 0;
uint8_t end_pos = 9;
data_1 = 0x80;
read_and_write(&final_val, &write_val1, start_pos, end_pos);
write_val1 = write_val1 | data_1;
// Read next 8 bits of final_val and write only into [0:8] position of existing data_2
start_pos = 0;
end_pos = 8;
data_2 = 0x27b7;
read_and_write(&final_val, &write_val2, start_pos, end_pos);
write_val2 = write_val2 | data_2;
//Read next 3 bits of final_val and write only into[13:11] position of existing data_3
start_pos = 11;
end_pos = 13;
data_3 = 0x3048;
read_and_write(&final_val, &write_val3, start_pos, end_pos);
write_val3 = write_val3 | data_3;
printf ("\n val1 0x%x val2 0x%x val3 0x%x final_val 0x%x", write_val1, write_val2, ext_val3, final_val);
}
有人可以帮忙吗?使用旧方法忽略陈旧的代码。
解决方案
这是一个带有辅助函数和测试框架的简单实现:
((uint32_t)1 << n) - 1
计算设置了低位的二进制数 2 n -1n
。uint32_t mask = (((uint32_t)1 << n) - 1) << pos2;
mask
使用n
从位置开始设置的位进行计算pos2
。(val2 & ~mask)
清除 位置 的位n
,其他位保持不变。pos2
val2
extract(val1, pos1, n) << pos2
将n
提取的位移动val1
到位置pos2
,其他位是0
。- 对这些值进行或运算会计算预期结果,其中复制的
n
位从in的位置开始。val1
pos2
val2
这是代码:
#include <stdio.h>
#include <stdint.h>
uint32_t extract(uint32_t data, int pos, int n) {
return (data >> pos) & (((uint32_t)1 << n) - 1);
}
uint32_t replace(uint32_t val1, uint32_t val2, int pos1, int n, int pos2) {
uint32_t mask = (((uint32_t)1 << n) - 1) << pos2;
return (val2 & ~mask) | (extract(val1, pos1, n) << pos2);
}
void test(uint32_t val1, uint32_t val2, int pos1, int n, int pos2) {
uint32_t res = replace(val1, val2, pos1, n, pos2);
printf("extracting bits [%d-%d] from 0x%04X to bits [%d-%d] in 0x%04X -> 0x%04X\n",
pos1, pos1 + n - 1, val1, pos2, pos2 + n - 1, val2, res);
}
int main() {
test(0x1234, 0x7048, 1, 3, 11);
test(0x01b7, 0x2743, 0, 8, 0);
test(0x0129, 0x7448, 8, 3, 11);
return 0;
}
输出:
extracting bits [1-3] from 0x1234 to bits [11-13] in 0x7048 -> 0x5048
extracting bits [0-7] from 0x01B7 to bits [0-7] in 0x2743 -> 0x27B7
extracting bits [8-10] from 0x0129 to bits [11-13] in 0x7448 -> 0x4C48