MSB与LSB相结合

Question

我有一个返回1 Byte的函数

uint8_t fun();

函数应该运行9次，所以我得到9字节，我想在这里将最后一个8作为4 short values，但是我不确定我得到的值是否正确：

char array[9];
.............

for ( i = 0; i< 9 ; i++){
array[i] = fun();

}

printf( " 1. Byte %x  a = %d , b=%d c =%d \n" ,
    array[0],   
            *(short*)&(array[1]),
            *(short*)&(array[3]),
            *(short*)&(array[5]),
            *(short*)&(array[7]));

是这样吗？

Answer 1

最好自己显式地将8位值加入到16位值中：

uint8_t  bytes[9];
uint16_t words[4];

words[0] = bytes[1] | (bytes[2] << 8);
words[1] = bytes[3] | (bytes[4] << 8);
words[2] = bytes[5] | (bytes[6] << 8);
words[3] = bytes[7] | (bytes[8] << 8);

顺便说一句，上面的假设是很少的。

Answer 2

你会遇到对齐问题。任何指向短时的指针都可以被看作是指向字符的指针，但是在非8位机器上，逆不能保证。

国际水文学组织，这将更安全：

struct {
    char arr0;
    union {
        char array[8];
        uint16_t sarr[4];
    } u;
} s;

s.arr0 = fun();
for ( i = 0; i< 8 ; i++){
    s.u.array[i] = fun();
}

printf( " 1. Byte %x  a = %d , b=%d c =%d d=%d\n" ,
    s.arr0,   
            s.u.sarr[0],
            s.u.sarr[1],
            s.u.sarr[2],
            s.u.sarr[3]);

但是我想您应该正确地处理机器上的endianness，并且知道如何转换2字符<=> 1的简短工作.

Answer 3

尝试使用struct来安排数据和移位操作以转换为异能。

// The existence of this function is assumed from the question.
extern unsigned char fun(void);

typedef struct
{
    unsigned char Byte;
    short WordA;
    short WordB;
    short WordC;
    short WordD;
}   converted_data;

void ConvertByteArray(converted_data* Dest, unsigned char* Source)
{
    Dest->Byte = Source[0];
    // The following assume that the Source bytes are MSB first.
    // If they are LSB first, you will need to swap the indeces.
    Dest->WordA = (((short)Source[1]) << 8) + Source[2];
    Dest->WordB = (((short)Source[3]) << 8) + Source[4];
    Dest->WordC = (((short)Source[5]) << 8) + Source[6];
    Dest->WordD = (((hshort)Source[7]) << 8) + Source[8];
}

int main(void)
{
    unsigned char array[9];
    converted_data convertedData;

    // Fill the array as per the question.    
    int i;
    for ( i = 0; i< 9 ; i++)
    {
        array[i] = fun();
    }

    // Perform the conversion
    ConvertByteArray(&convertedData, array);

    // Note the use of %h not %d to specify a short in the printf!
    printf( " 1. Byte %x  a = %h , b=%h c =%h d =%h\n",
        (int)convertedData.Byte,  // Cast as int because %x assumes an int.
        convertedData.WordA,
        convertedData.WordB,
        convertedData.WordC,
        convertedData.WordD );

   return 0;
}