从一个寄存器中读取某些位，然后写入另一个寄存器中的某些位

问题描述

我正在寻求实现read_and_write的原型，但无法正确完成。任何帮助表示赞赏。我能够提取位。

在下面的write_val1代码中，我的意图是仅将final_val的前9位写入read_data。 read_data的其余7位应保持不变。 write_val1的预期输出= 0x0080

与write_val2类似，我的意图是仅将final_val中的接下来的8位写入read_data。 read_data的其余8位应保持不变。 write_val2的预期输出= 0x2700

与write_val3类似，我的意图是仅将final_val中的后3位（010）写入read_data的位置[11:13]，而其余read_data保持不变。

预期输出= 0x5448;例如：从中选择010提取的位 write_val3; 0x3048 = 01 11 0 000 0100 1000; 0x5048 = 01 01 0 000 0100 1000

#include <stdio.h>
#include <stdint.h>

 // Extracts n bits from a given position in LSB. 
int bitExtracted(uint16_t read_data,uint8_t n_bits,uint8_t pos) 
{ 
    return (((1 << n_bits) - 1) & (read_data >> (pos - 1))); 
} 

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() 
{
    uint16_t read_data = 0x0; //assume some read value
    uint16_t ext_val1 = bitExtracted(read_data,9,1);  //Read BITS [8:0] from read_data
    uint8_t ext_val2 = bitExtracted(read_data,8,1);   //Read BITS [7:0] from read_data
    uint8_t ext_val3 = bitExtracted(read_data,3,5);   //Read BITS [7:4] from read_data
    uint32_t final_val = 0x0; //Stores 20 extracted bits from val1,val2 and val3 into final_val (LSB to MSB in order)
    uint16_t write_val1,write_val2,write_val3;
    uint8_t start_pos = 0,end_pos =8;
    ext_val1 = 0x80,ext_val2 = 0x0,ext_val3 = 0x2;
    final_val = (ext_val1 | (ext_val2 << 9) | (ext_val3 << 17));
    printf ("\n final_val %x",final_val);
    
    //Read first 9 bits of final_val and write only into [8:0] position of existing read_data
    read_and_write(&final_val,&write_val1,9); 
    read_data = 0x80;
    write_val1 = write_val1 | read_data;
    
    //Read next 8 bits of final_val and write only into [7:0] position of existing read_data
    start_pos = 0;
    end_pos = 7;
    read_data = 0x27b7;
    read_and_write(&final_val,&write_val2,start_pos,end_pos);
    write_val2 = write_val2 | read_data;

    //Read next 3 bits of final_val and write only into[13:11] position of existing read_data
    start_pos = 11;
    end_pos = 13;
    read_data = 0x3048;
    read_and_write(&final_val,&write_val3,end_pos);
    write_val3 = write_val3 | read_data;
    printf ("\n val1 0x%x val2 0x%x val3 0x%x final_val 0x%x",write_val1,ext_val3,final_val);
}

解决方法

如果您希望自己的read_and_write()函数能够更改该函数的前两个输入的值（例如final_val和write_val1），则应将它们作为指向函数的指针。因此，您的函数原型应如下所示：

void read_and_write(uint32_t* final_val,uint16* write_val1,uint8_t start_pos,uint8_t end_pos);

在主代码中调用函数时，需要使用&来获取变量的地址。因此，您可以使用以下方式致电给他们：

read_and_write(&final_val,&write_val1,start_pos,end_pos);

对于读写部分，它与您在bitExtracted函数中所做的非常相似。

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 

*将引用指针并访问存储在特定位置的数据。请注意，由于类型需要匹配，因此您还需要为(uint16_t)进行write_val强制转换。