Answer:
hope this helps. I am also a learner like you. Please cross check my explanation.
Explanation:
#include
#include
using namespace std;
int main()
{
int a[ ] = {0, 0, 0}; //array declared initializing a0=0, a1=0, a3=0
int* p = &a[1]; //pointer p is initialized it will be holding the address of a1 which means when p will be called it will point to whatever is present at the address a1, right now it hold 0.
int* q = &a[0]; //pointer q is initialized it will be holding the address of a0 which means when q will be called it will point to whatever is present at the address a0, right now it hold 0.
q=p; // now q is also pointing towards what p is pointing both holds the same address that is &a[1]
*q=1
; //&a[0] gets overwritten and now pointer q has integer 1......i am not sure abut this one
p = a; //p is now holding address of complete array a
*p=1; // a gets overwritten and now pointer q has integer 1......i am not sure abut this one
int*& r = p; //not sure
int** s = &q; s is a double pointer means it has more capacity of storage than single pointer and is now holding address of q
r = *s + 1; //not sure
s= &r; //explained above
**s = 1; //explained above
return 0;
}
Answer:
The correct answer is C. One reason why a business may want to move entirely online is to focus on a global market.
Explanation:
The fact that a business wishes to move entirely towards the online sales modality implies that there is a desire to expand the possibilities of selling its products beyond the physical place where it has its store.
It is a reality that starting an online business implies that the offered product can be purchased anywhere in the world, thanks to advances in technology and transportation that allow the product to be purchased and delivered in a matter of days, thanks to the advances produced by globalization.
Therefore, the fact that the store goes from physically to online selling makes its potential customers go from being the ones who know the store and live close to it to everyone in the world with access to internet.
Answer: You need a temporary variable to hold the value 3
Explanation:
So, aList[0] is 3 and aList[1] is 19, if it will be as it is you litteraly say to the compiler to change aList[0] to aList[1] at this moment aList[0] is 19 and aList[1] also is 19 and if you try to change aList[1] to aList[0] it will not change its value because they are the same.
You need temp variable to keep one of the values.
In this program, I am using the school-based grading system and the program should accept the subject and the number of students.
Program approach:-
- Using the necessary header file.
- Using the standard I/O namespace function.
- Define the main function.
- Declare the variable.
- Display enter obtain marks in 5 subjects.
- Return the value.
Program:-
//header file
#include<iostream>
//using namespace
using namespace std;
//main method
int main()
{
//declare variable
int j;
float mark, sum=0, a;
//display enter obtain marks in 5 subjects
cout<<"Enter Marks obtained in 5 Subjects: ";
for(j=0; j<5; j++)
{
cin>>mark;
sum = sum+mark;
}
a = sum/5;
//display grade
cout<<"\nGrade = ";
if(a>=91 && a<=100)
//display a1
cout<<"a1";
else if(a>=81 && a<91)
//display a2
cout<<"a2";
else if(a>=71 && a<81)
cout<<"b1";
else if(a>=61 && a<71)
cout<<"b2";
else if(a>=51 && a<61)
//display c1
cout<<"c1";
else if(a>=41 && a<51)
//display c2
cout<<"c2";
else if(a>=33 && a<41)
//display d
cout<<"d";
else if(a>=21 && a<33)
//display e1
cout<<"e1";
else if(a>=0 && a<21)
//display e2
cout<<"e2";
else
//display invalid
cout<<"Invalid!";
cout<<endl;
//return the value
return 0;
}
Learn more grading system
brainly.com/question/24298916
