c++ - C位运算题

Question

谁能帮我理解这段代码是怎么回事。看起来它正在从一个位数组中生成一个整数。我不确定它是怎么做到的。为什么在OxFF上会有按位&操作？难道这只会产生相同的结果吗？

//first take the first  4 bytes read out of the socket into an array and
//make them a 32 bit integer

        long ltemp =0;
        long ltemp2 = 0;
        ltemp  = ltemp | (unsigned char)(analog_val_ptr[0] & 0xff);
        ltemp  = ltemp << 24;
        ltemp2 = ltemp2 | (unsigned char)(analog_val_ptr[1] & 0xff);
        ltemp2 = ltemp2 << 16;
        ltemp = ltemp2 | ltemp;
        ltemp2 =0;
        ltemp2 = ltemp2 | (unsigned char)(analog_val_ptr[2] & 0xff);
        ltemp2 = ltemp2 << 8;
        ltemp = ltemp2 | ltemp;
        ltemp  =  ltemp | (unsigned char)(analog_val_ptr[3] & 0xff);

///then convert that integer into a float, passing

Answer 1

将四个 8 位字节转换为 32 位字节是一种非常冗长的方法。

and0xff 只是确保只使用每个值的低 8 位(0xff == binary 11111111) .

移位(8 的倍数)只是为了让每个字符到正确的位置。

整个事情可以用类似的东西代替:

unsigned long ltemp  = (unsigned char)(analog_val_ptr[0] & 0xff);
ltemp = (ltemp << 8) | (unsigned char)(analog_val_ptr[1] & 0xff);
ltemp = (ltemp << 8) | (unsigned char)(analog_val_ptr[2] & 0xff);
ltemp = (ltemp << 8) | (unsigned char)(analog_val_ptr[3] & 0xff);

或者，或者(假设它们可用)，使用正确的工具来完成工作，特别是 htonl() 和 ntohl()。