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Ainat [17]
3 years ago
9

Assume inputFile is a Scanner object used to read data from a text file that contains a number of lines. Each line contains an a

rbitrary number of words, with at least one word per line. Select an expression to complete the following code segment, which prints the last word in each line. while (inputFile.hasNextLine()) { String word = ""; String line = inputFile.nextLine(); Scanner words = new Scanner(line); while (_________________) { word = words.next(); } System.out.println(word); }
Computers and Technology
1 answer:
anzhelika [568]3 years ago
7 0

Answer:

words.hasNext()

Explanation:

Given the code snippet below:

  1.        while (inputFile.hasNextLine()) {
  2.            String word = "";
  3.            String line = inputFile.nextLine();
  4.            Scanner words = new Scanner(line);
  5.            while (words.hasNext()) {
  6.                word = words.next();
  7.            }
  8.            System.out.println(word); }
  9.    }

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.

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aleksandrvk [35]

Explanation:

The below code has been written in C language

void rotateright(int list[], int n)

{

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  int i;

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   int i;

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3 0
3 years ago
Assume a TCP sender is continuously sending 1,090-byte segments. If a TCP receiver advertises a window size of 5,718 bytes, and
Arturiano [62]

Answer:

for the 5 segments, the utilization is 3.8%

Explanation:

Given the data in the question;

segment size = 1090 bytes

Receiver window size = 5,718 bytes

Link transmission rate or Bandwidth = 26 Mbps = 26 × 10⁶ bps

propagation delay = 22.1 ms

so,

Round trip time = 2 × propagation delay = 2 × 22.1 ms = 44.2 ms

we determine the total segments;

Total segments = Receiver window size / sender segment or segment size

we substitute

Total segments = 5718 bytes / 1090 bytes

Total segments = 5.24587 ≈ 5

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Throughput = 1090 bytes / 44.2 ms

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⇒ 5 × 0.76% = 3.8%

Therefore, for the 5 segments, the utilization is 3.8%

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