PLEASE DO NOT DELETE MY QUESTION. IF YOU DO NOT KNOW HOW TO ANSWER IT THEN DO NOT ANSWER IT AND LEAVE IT FOR SOMEONE ELSE TO ANSWER. ALSO DO NOT COPY THE ANSWER FROM CHEGG BECAUSE THAT ANSWER IS WRONG. **1. Given the program below,**```cpp#include using namespace std;int fun(int *k) { *k += 4; return 3 * (*k) - 1;}int main() { int i = 10, j = 10, sum1, sum2; sum1 = (i/2) + fun(&i); sum2 = fun(&j) + (j/2); cout << sum1 << endl; cout << sum2 << endl; return 0;}```**a) Predict the values of sum1 and sum2 if the operands in the expressions for sum1 and sum2 are evaluated left to right.****b) Predict the values of sum1 and sum2 if the operands in the same expressions are evaluated right to left.****c) Predict the values of sum1 and sum2 if functions are evaluated first.****d) Compile with g++ and run the program on pluto or your favorite C++ environment, what are the values for sum1 and sum2?****2. Given the program below,**```cpp#include using namespace std;int a = 5;int fun1() { a = 17; return 3;}int fun(int* i) { *i += 5; return 4;}int main() { a = a + fun1(); cout << "a = " << a << endl; int x = 3; x = x + fun(&x); cout << "x = " << x << endl; return 0;}```**a) What would be the output if you compile with g++ and run the program on pluto?****b) Based on 3 and 4 a), what conclusion can you make on operands evaluation order in C++, left-to-right or right-to-left?**---### Explanation:#### Program 1:1. **`fun(int *k)` Function:** - Adds 4 to the value pointed to by `k`. - Returns `3 * (*k) - 1`.2. **`main` Function:** - Initializes `i` and `j` to 10. - Calculates `sum1` and `sum2` with expressions using `fun`.####

The solution to the given problem is below. ***Note: According to Bartleby policy we are not allowed to answer more than 1 question if multiple are given, request you to upload others as a separate question.****Answer 1) a) Sum1 = 46, sum2 = 48 Given, i = 10, j = 10 sum1 = (i / 2) + fun(&i);sum2 = fun(&j) + (j / 2); Evaluating from left to right sum1 = (10/2) + fun(&i) = 5 + 41 = 46 sum2 = fun(&j) + (j/2) = 41 + (14/2) = 41 + 7 = 48 b) Sum1 = 48, sum2 = 46 Given, i = 10, j = 10 sum1 = (i / 2) + fun(&i);sum2 = fun(&j) + (j / 2); Evaluating from right to left sum1 = (i/2) + fun(&i) = 41 + (14/2) = 41 + 7 = 48 sum2 = fun(&j) + (10/2) = 5 + 41 = 46 c) Sum1 = 48, sum2 = 48 Given, i = 10, j = 10 sum1 = (i / 2) + fun(&i);sum2 = fun(&j) + (j / 2); Evaluating function first sum1 = (i/2) + fun(&i) =41 + (14/2) = 41 + 7 = 48 sum2 = fun(&j) + (j/2) = 41 + (14/2) = 41 + 7 = 48 d) Sum1 = 46, sum2 = 48 Given, i = 10, j = 10 sum1 = (i / 2) + fun(&i);sum2 = fun(&j) + (j / 2); Using C++ compiler sum1 = (i/2) + fun(&i) = 5 + 41 = 46 sum2 = fun(&j) + (j/2) = 41 + (14/2) = 41 + 7 = 48 OUTPUT: SCREENSHOTO OF CODE: