Question

假设我有一个存储2个字节的缓冲区：

char *buf=new char[4];
// 00000010 00000000 (.. other stuff ..)

我想要做的是用最重要的字节切换最少的字节，并将该值存储在变量中。试着这样做：

short len=buf[1];
len <<= 8;
len |= buf[0];
    // Result, as expected: 00000000 00000010

它可以处理罚款，除非最重要的字节（buf [0]）是＆gt; = 128，这使得或运算符（|）用1填充短的一半。例如：

Original: 10000110 00000000
Should be: 00000000 10000110
But is: 11111111 10000110

谢谢（哦，我正在用file.read（...，4）读取文件中的字节; - 如果这是相关的话，甚至不知道了）

Answer 1

您的实施显然使用已签名号码的two's complement表示。改为使用无符号值

unsigned char * buf = new unsigned char[2];
...

unsigned short len=buf[1];
len <<= 8;
len |= buf[0];

Answer 2

试试这个：

unsigned short buf;
...
buf = buf << 8 | buf >> 8;

或者只是执行排列：

char buf[2];
char tmp;
...
tmp = buf[0];
buf[0] = buf[1];
buf[1] = tmp;

我希望这有帮助！

Answer 3

 buff = 11001010 01011100;
 //copy this buff, I am writing it directly below  

 copy_of_buff = 11001010 01011100;  
 buff_hi = (buff >> 8);  //Hence after this exec, buff_hi = 00000000 11001010  
 buff_lo = (copy_of_buff << 8); //Hence after this exec, buff_lo = 01011100 00000000

 out_buff = (buff_hi)||(buff_lo); //Hence after this exec, out_buff = 01011100 11001010

按位切换最重要和最不重要的字节

3 个答案: