c-从一个寄存器中读取某些位，并写入另一个寄存器的某些位

Question

我想实现read_and_write的原型，但无法实现。感谢您的帮助。我能提取一些碎片。

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

类似地，对于write_val2，我的意图是只将final_val的后面8位写入read_data。read_data的其余8位应保持不变。write_val2=0x2700 的预期输出

类似地，对于write_val3，我的意图是只将final_val的下3位(010(写入read_data的位置[11:13]，而不影响read_data其余部分。

预期输出=0x5448；例如：从中拾取010提取的位write_val3；0x3048=0111 0000 0100 1000；0x5048=0101 00000100 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, 0, 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, start_pos, 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, write_val2, ext_val3, final_val);
}

Answer 1

如果您希望read_and_write()函数能够更改函数前两个输入的值(例如final_val和write_val1(，那么它们应该作为指向函数的指针传递。因此，你的功能原型应该是这样的：

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

在调用主代码中的函数时，需要使用&获取变量的地址。所以你用来称呼他们

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 

*是为了尊重指针并访问存储在特定位置的数据。请注意，您还需要为write_val进行(uint16_t)强制转换，因为类型需要匹配。