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lora16 [44]
3 years ago
7

Match the context details with the pattern that applies.

Computers and Technology
1 answer:
tia_tia [17]3 years ago
6 0

Answer:

1. - Observer

2. - iterator

3. - strategy

4. - composite

5. - decorator

Explanation:

An object contains other objects (elements) and must give access to them for clients, without exposing its internal structure. This is an observation.

The subject-object can generate events as an iterable (iterator) and could use different strategic algorithms to achieve a task. Class objects can be used to aggregate other class objects.

In python, the decorator function mimics the outer function of a nested function, but still maintains the functionality of the second function

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An Operating System is an application just like any other applications we use on our
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Web client software, such as Chrome, Firefox, Internet Explorer, Safari, and Opera, that are used to view Web pages are known as
evablogger [386]

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8 0
3 years ago
Read 2 more answers
Array Challenge Have the function ArrayChallenge (arr) take the array of numbers stored in arr and return the string true if any
vampirchik [111]

Answer:

The code is given as follows,

Explanation:

Code:

#include <stdio.h>

#include <string.h>  

int n; //to store size of array  

char* ArrayChallenge(int arr[]) //function returns string true or false

{

  int i, j;

  int sum = 0;

  for(i = 0; i < n; i++)

  {

      sum = sum + arr[i]; //count sum

  }  

  for(i = 0; i < n; i++)

  {

      for(j = i+1; j < n; j++) //loop for every two elements in array

      {

          if(arr[i]*arr[j] > 2*sum) //check if proudct of two elements > 2 times sum

          {

              printf("\n%d x %d = %d, 2xSum = %d\n", arr[i], arr[j], arr[i]*arr[j], 2*sum);

              return "true"; //If proudct of two elements > 2 times sum. return true

          }

      }

  }

  return "false"; // If proudct of two elements < 2 times sum. return false

}  

int main()

{  

  printf("\nEnter size of array: ");

  scanf("%d", &n); //read size of array

  int A[n]; //array of size n

  printf("\nEnter array elements: ");

  int i;

  for(i = 0; i < n; i++)

  {

      scanf("%d", &A[i]); //read array from stdin

  }

  printf("%s\n",ArrayChallenge(A)); //ccall function and print answer

 

  return 0;

}

6 0
3 years ago
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