CODEWITHSUNDEEP
c++referencedo-whileternary-operator
Red One
Red One

Reputation: 101

References, Logical Operators, and Loop Condition

I'm having a problem understanding what I'm doing wrong in my code. What I'm trying to do is write a condition for a ternary operator and a do-while loop to recognize if one of my variables is above 1. Well, it is giving me an error that I don't know how to fix. What puzzles me the most is what I'll give an example of shortly. Here's my overall code. Keep in mind I'm a beginner, so there may be things that make you cringe or parts that could be improved.

#include <iostream>
using namespace std;

void getInfo(int&, int&, double&);

int main() {

   int ordered, stock;
   double charges = 10.00;

   getInfo(ordered, stock, charges);

   system("pause");
   return 0;

}

void getInfo(int& ordered, int& stock, double& charges) {

   do {
       printf("Enter the amount of spools ordered, in stock, and handling charges: ");
       scanf_s("%i %i %lf", &ordered, &stock, &charges);

       printf((&ordered > 1 && &stock > 0 && &charges > 0) ? "All added!\n"
           : "You messed one up. AGAIN!\n");
   } while (&ordered > 1 && &stock > 0 && &charges > 0);

}

Now, the error I'm getting is specifically in the ternary and the while condition. It gives me an error where the > is after ordered for both. Now, if I make it ordered instead of &ordered, the error goes away. Yet, I never get an error for &stock or &charges. I don't know why it's treating &ordered differently. It also doesn't check ordered correctly when I take off the &, for reasons I'm not entirely sure on.

Thank you to whomever is willing to help!

Upvotes: 0

Views: 78

Answers (2)

Julian Sivertsen
Julian Sivertsen

Reputation: 151

The & operator does different things depending on where you put it. If it's in a type declaration (e.g. int& foo), it means that the type is a reference. If however the & is used as an unary operator in an expression it becomes the Address-of operator, and returns a pointer to the object it's used on. So for example int* bar = &spam (assuming spam is an integer) would assign a pointer to spam in in the pointer bar.

Note that reference types behaves identical to the real type. This is perhaps better illustrated with a piece of code:

#include <iostream>
int main() {
    int foo = 12;
    int& bar = foo;  // a reference expects a variable of the same type in the initializer
    bar = 24; // Once the reference has been made the variable behaves indentically to the
              // to the variable it's a reference to.
    std::cout << foo << std::endl;  // outputs 24

    // if you use the & operator on a reference you get the address the variable it is a
    // reference to.
    std::cout << &bar << ' ' << &foo << std::endl; // Outputs two equal addresses.
}

There is also a third meaning of & in C++. As the bitwise and operator. foo & bar would result in the bitwise and of the variable foo and bar.

Upvotes: 1

Sam Varshavchik
Sam Varshavchik

Reputation: 118340

...(&ordered > 1 && &stock > 0 && &charges > 0) ? "All added!\n"

Here, "&ordered" means "the address of the ordered variable. You're obviously not trying to compare the address of ordered, but rather ordered itself. This should be

...(ordered > 1 && stock > 0 && charges > 0) ? "All added!\n"

The same problem is with your while() statement too.

In C++, "&" means two things. In declarations, it's used to declare a reference. In expression, it's the "address of" operator.

Once you declare a reference, like:

int &whatever;

Subsequently, using just whatever refers to the referenced object itself.

       : "You messed one up. AGAIN!\n");

Upvotes: 1

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