P3. UDP and TCP use 1s complement for their checksums. Suppose you have the following three 8-bit bytes: 01010011, 01100110, 01110100. What is the 1s complement of the sum of these 8-bit bytes? (Note that although UDP and TCP use 16-bit words in computing the checksum, for this problem you are being asked to consider 8-bit sums.) Show all work. Why is it that UDP takes the 1s complement of the sum; that is, why not just use the sum? With the 1s complement scheme, how does the receiver detect errors? Is it possible that a 1-bit error will go undetected? How about a 2-bit error?用中文回答

首先，我们需要将这三个8位字节相加，得到01010011 + 01100110 + 01110100 = 101011001。接下来，我们需要对这个和值取反，即111100110，这就是这三个8位字节的1s补码校验和。

UDP和TCP使用1s补码校验和是因为这种校验方式可以检测出1位比特错误。如果只使用简单的求和，那么可能会出现和值相同但数据不同的情况，这种情况是无法被检测出来的。

接收方如何检测错误呢？接收方会将接收到的数据按照与发送方相同的方法计算1s补码校验和，然后将接收到的校验和与计算出来的校验和进行比较。如果两个校验和不相等，那么就说明数据出现了错误。如果出现了2位比特错误，那么也有可能会被检测出来，但是存在一定的概率检测不出来。

相关问题

Consider two 4-bit binary numbers A and B. Bits of A are a3, a2, a1, and a0, and bits of B are b3, b2, b1, and b0. A is greater than B if a3 is 1 and b3 is 0, but if a3 and b3 are the same, then if a2 is 1 and b2 is 0, we can determine that A is greater than B. This evaluation continues until a0 and b0 are considered. If a0 and b0 are equal then A and B are equal. Using discrete gates and Verilog gate primitives build a 4-bit comparator that generates a 1 on its GT output when its 4-bit input A is greater than its B input, and a 1 on its EQ output when A and B are equal. Use the generate statement and other Verilog iterative logic support structures.

Here is one possible implementation of a 4-bit comparator in Verilog:

module comparator (
  input [3:0] A,
  input [3:0] B,
  output GT,
  output EQ
);

  // Generate the comparison logic for each bit position
  genvar i;
  generate
    for (i = 3; i >= 0; i--) begin
      assign GT = (A[i] > B[i]) ? 1 : 0;
      assign EQ = (A[i] == B[i]) ? 1 : 0;
      if (i > 0) begin
        assign GT = (A[i] == B[i]) ? GT : (!A[i] & B[i]);
      end
    end
  endgenerate

endmodule

This implementation uses a `generate` block to create the comparison logic for each bit position. The first two `assign` statements check if the current bit of `A` is greater than or equal to the current bit of `B`, and set `GT` and `EQ` accordingly. The third `assign` statement checks if the current bits of `A` and `B` are equal, and if they are, it keeps the previous value of `GT`, since the current bit doesn't affect the comparison. If the current bits of `A` and `B` are not equal, then `GT` is set to 1 if `A` has a 1 in that position and `B` has a 0, otherwise it stays the same.

Note that this implementation assumes that `A` and `B` are always 4-bit inputs, and does not handle cases where they are different sizes. It also does not handle signed or two's complement numbers.

SQL 错误 [1292] [22001]: Data truncation: Incorrect datetime value: 'yyyy-MM-dd HH:mm:ss' for column 'etl_time' at row 1

这个错误提示表明你在插入数据时，`etl_time` 这一列的值的格式不正确，导致数据截断错误。一般来说，你需要将时间值转换为 MySQL 支持的正确格式，例如 `yyyy-MM-dd HH:mm:ss`。

你可以尝试更改插入数据的语句，将 `etl_time` 的值改为符合格式的时间值，例如：

INSERT INTO `test_mysql_bushu_detail` (`rank`, `complement_date`, `etl_time`) VALUES ('1', '2022-01-01 00:00:00', '2022-01-01 10:00:00');

如果你仍然遇到问题，请检查你的时间值格式是否正确，并确保时间值在 MySQL 支持的时间范围内。