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output

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Question:

#include <iostream>
using namespace std;
int func (double &, int);
int main ()
{
double a = 19.0;
int x = 0, y = 0;
y = func (a, x);
cout << "a = " << a << ", x = " << x ", y = " << y << endl;
return (0);
}
int func (double &first, int second)
{
int third;
first /= 2.0;
second = first;
first *= 4.0;
third = first;
cout << "Second = " << second << ", and third = " << third << endl;
return (third);
}

Answers:

TDoT

Expert

Karma Points: 809

Date Posted: 7/22/2008 9:43:13 PM Status: Live

Asker's Rating: Helpful

Response:

Second = 9, and third = 38
a = 38, x = 0, y = 38

Hopefully its correct. rate that atleast.thanks

Suicider's Comment:

little explanation would help....

sam_GT

Expert

Karma Points: 811

Date Posted: 7/22/2008 11:50:49 PM Status: Live

Asker's Rating: Lifesaver

Response:

in function func (a, x) at line no. 8, the first parameter is passed by reference but second is passed by value. So the value of "a" will change when it will be changed in function func. But second valued is passed by value, it will remain same in main function.

so value of a will be (19.0 /2 ) * 4 = 9.5 *4 = 38
value of x remain same, so it will be 0
value of y will be return value of function so it will also be 38.

second is integer so it's value is 19.0 /2 = 9
third will be same as first = 38

So the output will be ...

Second = 9, and third = 38
a = 38, x = 0, y = 38

Suicider's Comment:

thanks.............

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