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can someone please tell me what this does?

I'm taking a beginner's C++ class, and long story, but for now I have to work without the textbook.

This is from a class point (for a triangle) example.

I have a basic understanding of classes, but what I can't find any help on is this:

double distance (const point& another) const;

My questions: what does the ampersand at the end of point& do, and is "another" a keyword, or is it an arbitrary assignment, and what does it do?

Thank You Thank You, for your time.
0
celere
Asked:
celere
  • 2
1 Solution
 
AxterCommented:
The ampersand means the variable takes a reference, and "another" is a type, not a keyword.
0
 
Peter KwanAnalyst ProgrammerCommented:
& means "reference" in C++. As in your example, the code says that you pass the argument into the function as reference, which means that the argument you passed into the function can be changed and the change inside the function is reflected outside the function.

Let me give another example to illustrate and help you to understand:

void fn(int& x) {
   x=1;
}

void fn1(int x) {
   x=4;
}

int main() {
  int x=9;
  cout << "Before: " << x << endl;
  fn(x);
  cout << "After fn: " << x << endl;
  fn1(x);
  cout << "After fn1: " << x << endl;
  return 0;
}

In this example, the output is:
Before: 9
After fn: 1
After fn1: 1

=========================================================
"another" is not a keyword, it is just a name of the argument. You can replace it with any valid name you like (of course you need to change the name of all the instances with the name "another" inside that function as well).
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MDarlingCommented:
Another is not a type or a keyword, it is a variable of type "point".

I'm guessing that

double distance (const point& another) const;

is supposed to be a member function of the class point.  If so then good old Pythagoras would help you determine the distance from point to point.

class point
{
double x,y;
public:
point(double _x,double _y);
double distance(const point& another) const
{
   double d=...;
   // pythagoras
}
};

Regards,
Mike.
0
 
MDarlingCommented:
Correction: Another is not a type or a keyword, it is a reference to a variable of type "point".
0
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