Answer:
b) objects are resuable
Explanation:
In OOP there's code reuse where a method or any other body of code is defined once and called or reused severally.
Answer:
Tab b. CTRL+A c. Alt d. Enter 18. Animated graphics that are displayed on the screen after a set of time when the computer is unattended. a. Screen Saver b. Title Bar c. Scroll Bar d.
Explanation:
Tab b. CTRL+A c. Alt d. Enter 16. Animated graphics that are displayed on the screen after a set of time when the computer is unattended
Computer-generated motion graphics[edit]. Before computers were widely available, motion graphics were costly and time-consuming, limiting their use to high-budget filmmaking and ...
Answer:
public ArrayList onlyBlue(String[] clothes){
ArrayList<String> blueCloths = new ArrayList<>();
for(int i =0; i<clothes.length; i++){
if(clothes[i].equalsIgnoreCase("blue")){
blueCloths.add(clothes[i]);
}
}
return blueCloths;
}
Explanation:
- Create the method to accept an Array object of type String representing colors with a return type of an ArrayList
- Within the method body, create and initialize an Arraylist
- Use a for loop to iterate the Array of cloths.
- Use an if statement within the for loop to check if item equals blue and add to the Arraylist.
- Finally return the arrayList to the caller
Answer:
The program in Python is as follows:
valCount = int(input())
reports = []
for i in range(valCount):
num = int(input())
reports.append(num)
for i in reports:
print(i,"reports.")
Explanation:
This gets input for valCount
valCount = int(input())
This creates an empty list
reports = []
This gets valCount integer from the user
<em>for i in range(valCount):</em>
<em> num = int(input())</em>
<em>Each input is appended to the report list</em>
<em> reports.append(num)</em>
This iterates through the report list
for i in reports:
This prints each element of the report list followed by "reports."
print(i,"reports.")
Answer:
<em>(c) The method call, which worked correctly before the change, will now cause a run-time error because it attempts to access a character at index 7 in a string whose last element is at index 6.</em>
<em />
Explanation:
Given
printAllCharacters method and printAllCharacters("ABCDEFG");
Required
What happens when x < str.length() is changed to x <= str.length()
First, we need to understand that str.length() gets the length of string "ABCDEFG"
There are 7 characters in "ABCDEFG".
So: str.length() = 7
The first character is at index 0 and the last is at index 6
Next, we need to simplify the loop:
for (int x = 0; x< str.length(); x++) means for (int x = 0; x< 7; x++)
The above loop will iterate from the character at the 0 index to the character at the 6th index
while
for (int x = 0; x<=str.length(); x++) means for (int x = 0; x<=7; x++)
The above loop will iterate from the character at the 0 index to the character at the 7th index
Because there is no character at the 7th index, the loop will return an error
Hence: (c) is correct
