Question

我正在编写一个我正在撰写的项目代码。我正在尝试将两个字节的数据组合成一个小端的16位数（即00 02到十六进制的02 00或二进制的0000 0000 0000 0010到0000 0010 0000 0000）。我找到了一个算术解决方案，其中（最高有效位* 256）+最低有效位。但是，每当我尝试这样做时，输出都会得到零。任何人都可以弄清楚我在做错了什么吗？我的想法已经不多了。谢谢！

int main (int argc, char **argv){

int fd;
unsigned char *buffer = malloc(512);
struct fileSystem_info fat_12;

//Check if argv[1] exists
if(argc > 2){
    printf("Error: The file does not exist. \n");
}

//Open the file
fd = open("fat_volume.dat", O_RDONLY, S_IROTH);
if (fd == -1){
    printf("Error: Opening the file was unsuccessful \n");
}

//Read the file
if (read(fd,buffer,512) < 0){
    printf("Error: Read was unsuccessful \n");
}
else{
    printf("Read is successful \n" );
}

//Converts two 8 bit data to one 16 bit data
int converter(int mostSignificant_bit, int leastSignificant_bit){
    return((mostSignificant_bit * 256) + leastSignificant_bit);
}

//Parse data
unsigned char first_byte = buffer[11]; //(raw byte that is read from a hex editor is 00)
unsigned char second_byte = buffer[12]; //(raw byte that is read from  a hex editor is 02)
unsigned char test = converter( (int)second_byte, (int)first_byte);
printf("%hhu \n", first_byte);
//outputs 0 
printf("%hhu \n", second_byte);
//outputs 2 
printf("%hhu \n", test);
//outputs 0

}

Answer 1

所有类型都是char，test需要大于char。请参阅下面的两个^^更改。

unsigned char test = converter( (int)second_byte, (int)first_byte);
         ^^^^ should be int
printf("%hhu \n", first_byte);
//outputs 0 
printf("%hhu \n", second_byte);
//outputs 2 
printf("%hhu \n", test);
         ^^^ should be u

另外，将转换为：

unsigned converter(unsigned mostSignificant_bit, unsigned leastSignificant_bit);

Answer 2

您可以通过移动msb并按位或使用lsb来完成此操作。

uint16_t value = (uint16_t)(msb << 8) | lsb;

将两个8位数据转换为一个16位数据变为零

2 个答案: