In the following code segment, assume that the ArrayList numList has been properly declared and initialized to contain the Integer values [1, 2, 2, 3]. The code segment is intended to insert the Integer value val in numList so that numList will remain in ascending order. The code segment does not work as intended in all cases.

int index = 0;
while (val > numList.get(index))
{
index++;
}
numList.add(index, val);

For which of the following values of val will the code segment not work as intended?

a. 0
b. 1
c. 2
d. 3
e. 4

Answer:

e. 4

Explanation:

The code segment will not work as intended if the value of the variable val is 4. This is because the while loop is comparing the value of the variable val to the value of each element within the numList. Since there is no element that is bigger than or equal to the number 4, this means that if the variable val is given the value of 4 it will become an infinite loop. This would cause the program to crash and not work as intended.

The program illustrates the use of while loops.

The code segment will not work as intended when the value of val is 4

Loops are used to perform repeated operations

From the program, the while loop is repeated under the following condition:

val > numList.get(index)

This means that:

The loop is repeated as long as val has a value greater than the index of numList (i.e. 3)

From the list of options, 4 is greater than 3.

Hence, the code segment will not work as intended when the value of val is 4

Read more about loops at:

https://brainly.com/question/19344465

Consider the following method.


public int locate(String str, String oneLetter)

{
int j = 0;
while(j < str.length() && str.substring(j, j+1).compareTo*oneLetter < 0)
{
j++;
}
return j;
}

Which of the following must be true when the while loop terminates?

a. j == str.length()
b. str.substring(j, j+1) >= 0
c. j <= str.length() || str.substring(j, j+1).compareTo(oneLetter) > 0
d. j == str.length() || str.substring(j, j+1).compareTo(oneLetter) >= 0
e. j == str.length() && str.substring(j, j+1).compareTo(oneLetter) >= 0

You can expect to see

During TCP/IP communications between two network hosts, information is encapsulated on the sending host and decapsulated on the receiving host using the OSI model. Match the information format with the appropriate layer of the OSI model.

a. Packets
b. Segments
c. Bits
d. Frames

1. Session Layer
2. Transport Layer
3. Network Layer
4. Data Link Layer
5. Physical Layer