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Westkost [7]
3 years ago
15

You are given the following segment of code:Line 1: PROCEDURE printScorePairs()Line 2: {Line 3: count <-- 0Line 4: sum <--

0Line 5: i <-- 1Line 6: scores <-- [73, 85, 100, 90, 64, 55]Line 7: REPEAT UNTIL ( i > LENGTH (scores) )Line 8: {Line 9: sum <-- scores[i] scores[i 1]Line 10: DISPLAY ( sum )Line 11: i <-- i 2Line 12: }Line 13: }The ABC company is thrifty when it comes to purchasing memory (also known as RAM) for its computers. Therefore, memory is of the utmost importance and all programming code needs to be optimized to use the least amount of memory possible. What modifications could be made to reduce the memory requirements without changing the overall functionality of the code?
Computers and Technology
1 answer:
Zolol [24]3 years ago
3 0

Answer:

Move Line 10 to after line 12

Explanation:

Required: Modify the program

From the procedure above, the procedure prints the sum at each loop. Unless it is really necessary, or it is needed to test  the program, it is not a good practice.

To optimize the program, simply remove the line at displays the sum (i.e. line 10) and place it at the end of the loop.

So, the end of the procedure looks like:

<em>Line 10: i <-- i 2 </em>

<em>Line 11: } </em>

<em>Line 12: DISPLAY ( sum )</em>

<em>Line 13: </em>

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4 0
3 years ago
The following code does not work as intended. It is meant to input two test grades and return the average in decimal format:
vredina [299]

Answer:

Type casting error

Explanation:

Logically, the program is correct and it is expected to return the average value. However, test1 and test2 are int values and when you applies any algebraic expression on Int, it will result into Int.

So, in this part (test1 + test2 )/2, two integers are adding and the final result is still an integer and this part will act as an integer even if you multiple or divide with any external number.

For example,

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Now, if you divide it by 2, it will still react as an integer portion. So, 9/2 would result in 4 instead of 4.5. All the calculation is done on the right side before assigning it to a double variable. Therefore, the double variable is still getting the int value and your program is not working correctly.

The solution is to typecast the "(test1 + test2 )/2" portion to double using Double.valueOf(test1 + test2)/2 ;

I have also attached the working code below.

import java.util.Scanner;

public class Main

{

public static void main(String[] args) {

 System.out.println("Hello World");

 Scanner scan = new Scanner (System.in);

 int test1 = scan.nextInt();

       int test2 = scan.nextInt();

       double average = Double.valueOf(test1 + test2)/2 ;

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}

}

4 0
4 years ago
RAM
Alex Ar [27]

Answer:

The answer to this question is given below in the explanation section

Explanation:

The correct answer is RAM.

RAM is used for storing programs and data currently being processed by the CPU.  So, the data in the RAM, can be easily accessible and processed by the CPU more fastly.

While Mass memory and neo volatile memory is not correct options. because these types of memory can stores a large amount of data but CPU fetch data from these memories into RAM. and, RAM can only be used by the CPU when performing operations.

7 0
4 years ago
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