Answer:
Digital communication technologies
Explanation:
connecting people across the globe has become easier and faster. Platforms such as Zoom, Room, Microsoft teams, WebEx, and many others are buzzing virtual world spaces to connect and share.
Answer:
The answer is "a1 and a2 is an array of pointers".
Explanation:
In this question, A collection of pointers refers to an array of elements where each pointer array element points to a data array element. In the above-given statement, the two-pointer type array "a1 and a2" is declared that holds the same size "8" elements in the array, and each element points towards the array's first element of the array, therefore, both a1 and a2 are pointer arrays.
Answer:
Option B is the correct answer.
Explanation:
- In the above code, the loop will execute only one time because the loop condition is false and it is the Do-While loop and the property of the Do-while loop is to execute on a single time if the loop condition is false.
- Then the statement "x*=20;" will execute one and gives the result 200 for x variable because this statement means "x=x*20".
- SO the 200 is the answer for the X variable which is described above and it is stated from option B. Hence it is the correct option while the other is not because--
- Option A states that the value is 10 but the value is 200.
- Option C states that this is an infinite loop but the loop is executed one time.
- Option D states that the loop will not be executed but the loop is executed one time
Answer:
Salting is the preservation of food with dry edible salt. It is related to pickling in general and more specifically to brining also known as fermenting (preparing food with brine, that is, salty water) and is one form of curing.
Explanation:
Answer:
answer:
#include <iostream>
#include<list>
using namespace std;
bool Greater(int x) { return x>3; } int main() { list<int>l; /*Declare the list of integers*/ l.push_back(5); l.push_back(6); /*Insert 5 and 6 at the end of list*/ l.push_front(1); l.push_front(2); /*Insert 1 and 2 in front of the list*/ list<int>::iterator it = l.begin(); advance(it, 2); l.insert(it, 4); /*Insert 4 at position 3*/ for(list<int>::iterator i = l.begin();i != l.end();i++) cout<< *i << " "; /*Display the list*/ cout<<endl; l.erase(it); /*Delete the element 4 inserted at position 3*/ for(list<int>::iterator i = l.begin();i != l.end();i++) cout<< *i << " "; /*Display the list*/ cout<<endl;
l.remove_if(Greater); for(list<int>::iterator i = l.begin();i != l.end();i++) cout<< *i << " ";
/*Display the list*/
cout<<endl; return 0;
}
