连接两个数字的位的代码不起作用

问题描述：

任务是连接两个给定数字的二进制。

The task is to concat the binary of 2 given numbers.

示例：

给出 5 （ 101 ）和 3 （ 011 ），结果为 46 （ concat（101，011）= 101011 ）

Given 5 (101) and 3 (011), the result is 46 (concat(101, 011) = 101011)

到目前为止的代码：

public class Concat {
    public static void main(String[] args) {
      int t = 0;
      int k = 5;
      int x = 3;
      int i = 0;
       while (i < 3) {
           t = x % 2;
            x /= 2;
            k <<= 1;
            k |= t;
           ++i;
       }

      System.out.println(k);
    }

}

但是问题在于代码给出 101110 ，而不是 101011 。

But the problem is that the above code gives 101110, not 101011.

什么是

答

您的问题是您正在向后中输入第二个数字的位。这是因为 x％2 是低位：

Your problem is that you're feeding the bits of the second number in backwards. That's because x%2 is the low order bit:

+---+---+---+       <110
| 1 | 0 | 1 | <-----------------+^
+---+---+---+                   |1
              +---+---+---+     |1
              | 0 | 1 | 1 | ----+0
              +---+---+---+ 011>

Cringe 尽我最大的艺术才能：-)但是，如果您已经知道它是3位宽，只需使用：

Cringe at my awesome artistic abilities :-) However, if you already know that it's 3 bits wide, just use:

public class concat {
    public static void main (String[] args) {
        int t = 0;
        int k = 5;
        int x = 3;

        k <<= 3;
        k |= x;
        // or, in one line: k = (k << 3) | x;

        System.out.println(k);
    }
}

就图形外观而言：

In terms of how that looks graphically:

                  +---+---+---+
                k:| 1 | 0 | 1 |
                  +---+---+---+
                  +---+---+---+
                x:| 0 | 1 | 1 |
                  +---+---+---+

      +---+---+---+---+---+---+
k<<=3:| 1 | 0 | 1 | 0 | 0 | 0 |
      +---+---+---+---+---+---+
                  +---+---+---+
                x:| 0 | 1 | 1 |
                  +---+---+---+

      +---+---+---+---+---+---+
 k|=3:| 1 | 0 | 1 | 0 | 1 | 1 |
      +---+---+---+---+---+---+
                    ^   ^   ^
                  +---+---+---+
                x:| 0 | 1 | 1 |
                  +---+---+---+

没有明显的理由一次这样做。

There's no apparent reason for doing it one bit at a time.

连接两个数字的位的代码不起作用

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