Answer:
Implementation file
Explanation:
In popular object oriented programming languages like C/C++, The implementation file (source file) of a class is used to hold the code implementaion of the method(s) of the class, this is helpful for seperating interface and method implementation. When this seperation exists, header files will be used to declare all the methods and fields of the class.
In this way, the implementaion file will hold the actual source code of the methods that are declared in the header file and will have a line to include its associated header file. A major advantage of seperating code in this way is the enhancement of better code organization and code re-use
You can go to jail get in alot of trouble plus you always have a chance of letting a virus into yoru computer
Answer:
see the code snippet below writing in Kotlin Language
Explanation:
fun main(args: Array<String>) {
sumOfNumbers()
}
fun sumOfNumbers(): Int{
var firstNum:Int
var secondNum:Int
println("Enter the value of first +ve Number")
firstNum= Integer.valueOf(readLine())
println("Enter the value of second +ve Number")
secondNum= Integer.valueOf(readLine())
var sum:Int= firstNum+secondNum
println("The sum of $firstNum and $secondNum is $sum")
return sum
}
The term vulnerability describes the issue that Adam discovered.
b. vulnerability
<u>Explanation:</u>
SQL injection attack is an attack in which an external party can execute SQL commands on the database that serves as a back-end for a particular website. The SQL commands can be used to modify the contents of the website, modify the records, delete the records, and retrieve confidential information as well.
As Adam believes that the code has an issue that allows a SQL injection attack, the term that best describes the issue that he discovered is vulnerability. The website is vulnerable since the code does not have a proper procedure to tackle a situation of SQL injection attack.
Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.
public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}
So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.
A quick explanation:
We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.
The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.