Answer:
words.hasNext()
Explanation:
Given the code snippet below:
- while (inputFile.hasNextLine()) {
- String word = "";
- String line = inputFile.nextLine();
- Scanner words = new Scanner(line);
- while (words.hasNext()) {
- word = words.next();
- }
- System.out.println(word); }
- }
We have a inputFile Scanner object that can read data from a text file and we presume the inputFile has read several rows of data from the text file. So long as there is another line of input data available, the outer while loop will keep running. In each outer loop, one line of data will be read and assign to line variable (Line 3). Next, there is another Scanner object, words, which will take the current line of data as input. To get the last word of that line, we can use hasNext() method. This method will always return true if there is another tokens in its input. So the inner while loop will keep running so long as there is a token in current line of data and assign the current token to word variable. The word will hold the last token of current line of data upon exit from the inner loop. Then we can print the output (Line 8) which is the last word of the current line of data.
Answer
First part:
The transmitted 8-bit sequence for ASCII character '&' with odd parity will be 00100110. Here leftmost bit is odd parity bit.
Second part:
The invalid bit sequence are option a. 01001000 and d. 11100111
Explanation:
Explanation for first part:
In odd parity, check bit of either 0 or 1 is added to the binary number as leftmost bit for making the number of 1s in binary number odd.
If there are even number of 1s present in the original number then 1 is added as leftmost bit to make total number of 1s odd.
If there are odd number of 1s present in the original number then 0 is added as leftmost bit to keep the total number of 1s odd.
Explanation for second part:
A valid odd parity bit sequence will always have odd number of 1s.
Since in option a and d, total number of 1s are 2 and 6 i.e. even number. Therefore they are invalid odd parity check bit sequences.
And since in option b and c, total number of 1s are 5 and 7 i.e. odd numbers respectively. Therefore they are valid odd parity check bit sequences.
