从 A 中提取某些位，并仅在特定位置替换 B 中提取的那些位

Question

提取和修改位
从packed final_value中的一个位置取出n个bit，写入任意位置，不修改uint16_t test_bit = 0x3048的原始内容。预期输出 = 0x5048
例如：从final_val取010(3bits from position 17)写入任意位置(position 11)
0x3048 = 0111 0000 0100 1000；
0x5048 = 0101 0000 0100 1000

几个例子：
示例 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，而保留位 8 到 15 不变。
Val_1 0x1b7 0000 0001 1011 0111
Val_2 0x27b7 0010 0111 1011 0111

示例 B：
从 Val1[from bits 8 to 10] 中提取 3 bits 并替换 Val_2[from 11 to 13].
Val_1 0x129 0000 0001 0010 1001
Val_20x4C48 0100 1100 0100 1000
目前尝试过：

#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);
}

有人可以帮忙解决以上问题吗？使用旧方法忽略陈旧代码。

Answer 1

这里有一些函数可以提取特定位的值并将它们替换为从另一个值中提取的值。

uint32_t extract(uint32_t read_data, int n_bits, int pos) 
{ 
    uint32_t mask = (n_bits < (CHAR_BIT * sizeof(mask) - 1)) ? ((1LU << n_bits) - 1) << pos  : -1; 


    return (mask & read_data) >> pos;
} 

uint32_t replace(uint32_t read_val, uint32_t write_val, int read_pos, int write_pos, int n_bits)
{
    uint32_t extracted = extract(read_val, n_bits, read_pos);
    uint32_t mask = (n_bits < (CHAR_BIT * sizeof(mask) - 1)) ? ((1LU << n_bits) - 1) << write_pos  : -1; 
    write_val &= ~mask;
    write_val |= extracted << write_pos;

    return write_val;
}


int main (void)
{
    uint32_t val = 0x1234;
    uint32_t val2 = 0xabcd;

    printf("%x\n", extract(val, 4, 8));
    printf("%x\n", replace(val, val2, 4, 8, 4));
}

Answer 2

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 n 位的二进制数 2ⁿ-1。
uint32_t mask = (((uint32_t)1 << n) - 1) << pos2; 计算 mask，从位置 pos2.

n

(val2 & ~mask) 清除 val2 位置 pos2 的 n 位，其他保持不变。
extract(val1, pos1, n) << pos2将val1中提取的n位移动到位置pos2，其他位为0.
or-ing 这些值计算预期结果，从 val1 中的 pos2 位置开始复制 val1 中的 n 位。

代码如下：

#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

从 A 中提取某些位，并仅在特定位置替换 B 中提取的那些位

Extract certain bits from A and replace only those extracted bits in B at a certain position

c

bit-manipulation

operators

bit

bitwise-operators