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3 years ago

Which of the following is true of a procedure? Check all that apply.

Computers and Technology

2 answers:

kow [346]3 years ago

7 0

Answer:

It can be reused.

It saves programmers time.

Explanation:

1 and 2 is the correct answer

pls mark me as brainleast

Virty [35]3 years ago

5 0

Answer:

It can be reused, It saves programmers time, it is a block of code that performs a single task.

Explanation:

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No production Web application, whether it resides inside or outside of the firewall, should be implemented without:

Alona [7]

Answer:

Stay ahead of threats with machine learning-based WAF protection! Learn more today. Integrate application security and application delivery to protect your network. 100% Privacy. Keep Costs Down. Business Continuity. Maximum Productivity. Types: DDoS Protection

WELCOME

5 0

3 years ago

7.8.1: Function pass by reference: Transforming coordinates. Define a function CoordTransform() that transforms the function's f

Lilit [14]

Answer:

Here is the CoordTransform() function:

void CoordTransform(int xVal, int yVal, int &xValNew, int &yValNew){

xValNew = (xVal + 1) * 2;

yValNew = (yVal + 1) * 2; }

The above method has four parameters xVal and yVal that are used as input parameters and xValNew yValNew as output parameters. This function returns void and transforms its first two input parameters xVal and yVal into two output parameters xValNew and yValNew according to the formula: new = (old + 1) *

Here new variables are xValNew and yValNew and old is represented by xVal and yVal.

Here the variables xValNew and yValNew are passed by reference which means any change made to these variables will be reflected in main(). Whereas variables xVal and yVal are passed by value.

Explanation:

Here is the complete program:

#include <iostream>

using namespace std;

void CoordTransform(int xVal, int yVal, int &xValNew, int &yValNew){

xValNew = (xVal + 1) * 2;

yValNew = (yVal + 1) * 2;}

int main()

{ int xValNew;

int yValNew;

int xValUser;

int yValUser;

cin >> xValUser;

cin >> yValUser;

CoordTransform(xValUser, yValUser, xValNew, yValNew);

cout << "(" << xValUser << ", " << yValUser << ") becomes (" << xValNew << ", " << yValNew << ")" << endl;

return 0; }

The output is given in the attached screenshot

7 0

3 years ago

Suppose you have two arrays of ints, arr1 and arr2, each containing ints that are sorted in ascending order. Write a static meth

telo118 [61]

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.

4 0

4 years ago

If you want to access your home network from your distant garage, a ________ might help boost the signal.

Kryger [21]

Wireless range extender.

6 0

3 years ago

Please help This is a homework in ICT class

Pavlova-9 [17]

Answer:

ejwksjnsskawii28393939393030jedjdjjccjncncnccnnccn

Explanation:

jsskaaaahujpjbhdueiwiwopwpwpwwieuehbfcnnccjxkxooss8ahwhwwnekdodod8djhdmslwoaishwhwkqoq18829293i4644673819191jednsmzljesksowlwlwpwpwnwkpwwwww waaaaayushsjsnmssoss ss sosijw

8 0

3 years ago