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Write a class called Adder that stores the sum of all the ints given to it. Your Adder class should allow you to write the following main function (and code like it):

int main()
{
int x,y;
cin >> x >> y;
Adder sum1; // sum1 is initialized to 0 
Adder sum2(x); // sum2 is initialized to x
Adder sum3(y); // sum1 is initialized to y 
cout << "sum1 is " << sum1 << '\n'; // prints "sum1 is 0" 
cout << "sum2 is " << sum2 << '\n'; // prints "sum2 is x"
cout << "sum3 is " << sum3 << '\n'; // prints "sum3 is y" 
sum1 += x; //adds x to sum1; now sum1 is x 
sum2 += -3; // adds -3 to sum2; now sum2 is x-3
sum3 += 2; //adds 2 to sum3; now sum3 is y+2 
if (sum1 == sum2)
cout << "sum1 and sum2 are the same\n";
else cout << "sum1 and sum2 are not the same\n";
if (sum1 == sum3)
cout << "sum1 and sum3 are the same\n";
else cout << "sum1 and sum3 are not the same\n";
}

You should write the functions that are necessary for Adder to be used as in the above program. Combine them all together in your answer to make a perfect program. Use const wherever appropriate, and do not write or use a cast operator. Make sure to include any necessary header files.

Example input:

3 5

Example output:

sum1 is 0
sum2 is 3
sum3 is 5
sum1 and sum2 are not the same
sum1 and sum3 are not the same

 

[Exercise] Coding (C) - asked in C++
ID: 24588 - Available when: Unlimited - Due to: Unlimited

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#include *** ** ******* **



using namespace std;



class Adder{
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}



int main()

{
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* **** ** ** * *** ***** * x * *** * ** y;
*** ** * * * *** * ***** sum1; // sum1 is initialized to 0
* *** * *** * ** *** *** ** sum2(x); // sum2 is initialized to x
* *** ** * * * * sum3(y); // sum1 is initialized to y
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** * *** ** *** ** * * * * * ***** * is " ** * * sum2 * *** * '\n'; // prints * ** ** is x"
*** ** ** ****** * * **** * ** * * * *** * * is " * *** sum3 ** * ** * '\n'; // prints * * * is y"
* * ***** * * * * * ** += x; //adds x to sum1; now sum1 is x
* * * * * ** * *** *** **** += -3; // adds -3 to sum2; now sum2 is x-3
* * ******** * ** ** * * += 2; //adds 2 to sum3; now sum3 is y+2
* ** *** *** * *** * (sum1 == sum2)
* * ** ** * * **** * ** *** * * ** ** * ** * * * * * ** * * * ** and sum2 are the *** ** * * **
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* ** * *** ***** *** 0;

}
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