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mojhsa [17]
3 years ago
8

Analyst is investigating proxy logs and found out that one of the internal user visited website storing suspicious java scripts.

After opening one of them he noticed that it's very hard to understand the code and all code differs from typical java script. What is the name of this technique to hide the code and extend analysis time?
Computers and Technology
1 answer:
Snowcat [4.5K]3 years ago
4 0

Answer:

Obfuscation

Explanation:

The fact that the analyst can open the Javascript code means that the code is not encrypted. It simply means that the data the analyst is dealing with here is hidden or scrambled to prevent unauthorized access to sensitive data. This technique is known as Data Obfuscation and is a form of encryption that results in confusing data. Obfuscation hides the meaning by rearranging the operations of some software. As a result, this technique forces the attacker to spend more time investigating the code and looking for encrypted parts.

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QUESTIONS Which of the following use cases are suitable for compute-optimized cloud offering? ОА. None of the listed O B. Highly
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E. All of the listed ​

Explanation:

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3 years ago
How to reply to text from unknown number?
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3 years ago
Define a character variable letterStart. Read the character from the user, print that letter and the next letter in the alphabet
lesantik [10]

Answer:

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Explanation:

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3 0
4 years ago
In this problem, there will be the construction of a program that reads in a sequence of integers from standard input until 0 is
Sedaia [141]

Answer:

Check the explanation

Explanation:

package util;

import java.util.ArrayList;

import java.util.Iterator;

import java.util.List;

import java.util.Scanner;

public class Assignment9 {

public Assignment9() {

List<Integer> numList = new ArrayList<Integer>();

Scanner sc = new Scanner(System.in);

System.out.println("Please enter the numbers , press 0 to stop");

int x = 0;

do {

x = sc.nextInt();

numList.add(x);

} while (x != 0);

int size = numList.size();

Integer[] numArray = new Integer[size];

Iterator<Integer> it = numList.iterator();

int i = 0;

while (it.hasNext()) {

numArray[i] = it.next();

i++;

}

int min = findMin(numArray, 0, 2);

System.out.println("The minimum number is " + min);

int sum = computeNegativeSum(numArray, 0, 2);

System.out.println("The sum of the negative numbers is " + sum);

int sumOdd = computeSumAtOdd(numArray, 0, 3);

System.out

.println("The sum of the numbers at odd indexes is " + sumOdd);

int countEven = computeCountEven(numArray, 0, 3);

System.out.println("The total count of even integers is " + countEven);

}

/**

* This method is used to compute the minimum number

* "at"param NumArray

* "at"param startIndex

* "at"param endIndex

* "at"return

*/

public int findMin(Integer[] NumArray, int startIndex, int endIndex) {

if (startIndex == endIndex)// base case50.

{

return NumArray[startIndex]; // return value is there is only one

// entry

} else if (findMin(NumArray, startIndex, endIndex - 1) < NumArray[endIndex]) {

return findMin(NumArray, startIndex, endIndex - 1);

} else {

return NumArray[endIndex];

}

}

/**

* This method is used to find the sum of negative numbers in the array

* "at"param NumArray

* "at"param startIndex

* "at"param endIndex

* "at"return

*/

public int computeNegativeSum(Integer[] NumArray, int startIndex,

int endIndex) {

if (startIndex == endIndex) {

if (NumArray[startIndex] > 0) {

return 0;

} else {

return NumArray[startIndex];

}

} else if (NumArray[endIndex] < 0) {

return computeNegativeSum(NumArray, startIndex, endIndex - 1)

+ NumArray[endIndex];

} else {

return computeNegativeSum(NumArray, startIndex, endIndex - 1); // if

}

}

/**

* This method is used to find the sum of numbers at odd indexes (1,3, 5,...),

*

* "at"param NumArray

* "at"param startIndex

* "at"param endIndex

* "at"return

*/

public int computeSumAtOdd(Integer[] NumArray, int startIndex, int endIndex) {

if (startIndex == endIndex) {

if (startIndex % 2 == 1) {

return NumArray[startIndex];

} else {

return 0;

}

} else {

if (endIndex % 2 == 1) {

return computeSumAtOdd(NumArray, startIndex, endIndex - 1)

+ NumArray[endIndex];

} else {

return computeSumAtOdd(NumArray, startIndex, endIndex - 1);

}

}

}

/**

* This method is used to find the number of even numbers within the array

*

* "at"param NumArray

* "at"param startIndex

* "at"param endIndex

* "at"return

*/

public int computeCountEven(Integer[] NumArray, int startIndex, int endIndex) {

if (startIndex == endIndex) {

if (NumArray[startIndex] % 2 == 0) {

return NumArray[startIndex];

} else {

return 0;

}

} else if (NumArray[endIndex] % 2 == 0) {

return computeCountEven(NumArray, startIndex, endIndex - 1)

+ NumArray[endIndex];

} else {

return computeCountEven(NumArray, startIndex, endIndex - 1); // if

}

}

public static void main(String args[]) {

Assignment9 assignment9 = new Assignment9();

}

}

Sample Output is :

*************************

Please enter the numbers , press 0 to stop

5

7

3

2

4

-7

-3

0

The minimum number is 3

The sum of the negative numbers is 0

The sum of the numbers at odd indexes is 9

The total count of even integers is 2

8 0
3 years ago
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