Question

USLUR #include <iostream using namespace std; /** @return a reference to the smaller of the two arguments */ int& maxi(int&x, int& y) { return (x > y) 7X: y; int main() { int a = 10, b = 20; maxi(a,b) = 5; /* Assigns the value 5 to b. / cout << a << " " << b << endl; maxi(a,b) + 6; /* Increases a by 6. a is now 16. =/ cout << a <<"" <b << endl; maxi(a, b) cout << a << return ; / " << b Increments a by 1. */ << endl;

Answer 1

Line by line explanation

1) int a=10, b=20;

assigns 10 and 20 to a and b respectvely.

2) maxi(a,b)=5;

calling maxi(a,b) returns the reference of greater variable, in this case b is greater. Therefore maxi(a,b) returns b. Which is then assigned to 5. Therefore now value of a = 10, b=5.

3) cout<<a<<" "<<b<<endl;

Print the values a and b. 10 5 gets printed.

4) maxi(a,b)+=6;

maxi(a,b) returns a as 10 is greater than 5. Now the expression becomes a = a+6. Therefore new value of a is 10+6 = 16. And b remains 5.

5) cout<<a<<" "<<b<<endl;

Print a and b. 16 5 gets printed.

6) ++maxi(a,b);

Again maxi(a,b) returns reference to a as 16 is greater than 5. Therefore expression now becomes ++a, which is same as a=a+1. Now a become 16+1=17, and b remains 5.

7) cout<<a<<" "<<b<<endl;

print a and b. 17 5 gets printed.

OUTPUT SS:

185 165 175