quiestion on function parameters

Hi all,

I'm learning about the different ways of passing arguments for functions and everything (i think!) is clear in Koeings book (esp. p58). However, when I'm trying to pass an argument to modify it and keep its new value outside the function then it runs like I'm passing references. Can you please explain why the program below passes "final" as a refence even though I have put "bouble& fin"

Thank you
Spiros

Code:

---

#include<iomanip>
#include<ios>
#include<iostream>
#include<string>

using std::cin;
using std::cout;
using std::endl;
using std::setprecision;
using std::string;
using std::streamsize;

double grade(double mid, double& fin, double sum_count) {
        fin+=100;
        cout << endl
              << "Final grade (inside function) is: " << fin << endl;
        return 0.2*mid + 0.4*(fin-100) +0.4*sum_count;
}
int main()
{
        cout << "Enter your midterm and final exam grades: ";
        double midterm, final;
        cin >> midterm >> final;
        cout << "Enter your homework grades, followed by an CTRL-D to stop: ";

        int count=0; double sum=0; double x;
        while (cin >> x) {
                ++count;
                sum += x;
        }

        streamsize prec = cout.precision();
        cout << "Your total grade is " << setprecision(3)
                << grade(midterm, final, sum/count) << endl
                << "Also final grade (outside function) is: " << final
                << setprecision(prec) << endl;
        return 0;
}

---

double& fin is a reference.

I had to reread this to understand the question (my dyslexia kicking in on me...). You use the & operator to denote a reference. So you are calling final by reference.

it isn't. you call functions, not variables.

It's being passed by reference because you have a reference as the parameter that you are passing to in your function.

double grade(double mid, double& fin, double sum_count)

thanks for the replies but I'm still getting confused. I'm reading Koeing's book on p58 again and again, but I still get the impression that if you put & to a function parameter then you work on the parameter itself rather than a copy...

Koeing example on

Code:

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const vector<double>&

---

and he says:

Quote:

---

In this type, the & asks the implementation not to copy the argument, and the const promises the program wil not change the parameter

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doesn't that mean hat by putting & we are not using a reference?

I've tried my program without the & but the results are the same; "final" does not change outside the function.

thanks

Ok, what he means by copy and what you think he means by copy seem to vary. He means copy to the stack. You mean copy as in faximaly(sp)?

Think of double& x as saying I don't want to create my own double and call it x (copy), I want the exact one you are handing to me (reference).

Quote:

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doesn't that mean hat by putting & we are not using a reference?

---

On the contrary, the & after the type name means that pass by reference is used.

thank you for your time...

Things are improving now and I understand what we mean when saying "reference". But I'm still confused :(

Koeing on p54 sgives the example:

Code:

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vector<double> homework;
vector<double>& hw = homework;    // hw is a synonym for homework

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and states:

Quote:

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we are saying that hw is another name for homework. From that point on, anything we do to hw is equivalent to doing th same thing to homework, and vice versa.

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Doesn't that mean that if we eg increment an element in "hw" by 50, then we do the same thing to the element in "homework"?

Using my definition again think of it as saying, hw wants your homework. Not hw wants its own homework.

It wants consider itself the same as homework, not make a new homework and copy all its contents to itself.

Quote:

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Doesn't that mean that if we eg increment an element in "hw" by 50, then we do the same thing to the element in "homework"?

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Yes. Of course, try it out yourself to convince yourself that it happens.

Why don't you try things?

Code:

---

#include <iostream>
int main()
{
    int n = 50;
    int& r = n;
    r += 10;
    std::cout << n << ' ' << r << '\n';
}

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Try with and without the & to see the difference.

I have tried it on mine but without success, whereas anon's example works exactly as I expected to. However, when I put & and take it off in my program, then it doesn't make any difference in "final" so I suppose there's another mistake in my program... Any ideas why this happens in my program?

Could you give input/output for an example run?

Code:

---

        cout << "Your total grade is " << setprecision(3)
                << grade(midterm, final, sum/count) << endl
                << "Also final grade (outside function) is: " << final
                << setprecision(prec) << endl;

---

Within this statement, the value of final is used twice and it is changed in the meantime (grade has side-effects). I'm not sure if this is undefined like i = i++; but to be safe you might break it up into two statements:

Code:

---

        cout << "Your total grade is " << setprecision(3)
                << grade(midterm, final, sum/count) << endl;
        cout  << "Also final grade (outside function) is: " << final
                << setprecision(prec) << endl;

---

Anyway, for me it gives the expected output (fin == final) either way.

it works!
I've split the cout statement into two as you suggested and it worked. Thanks so much, it would've taken me donkeys years to figure this out on my own...

cheers!