Denominator for "operator /"

Question

Is there a way to overload operator/() to use the class as the denominator?

Like so: int foo = 5 / object;

This excellent SO post on [operator overloading](http://stackoverflow.com/questions/4421706/operator-overloading) includes good information about implementing binary arithmetic operators. — Blastfurnace, Jul 21 '12 at 02:33

score 4 · Accepted Answer · answered Jul 21 '12 at 02:31

4

Use a free function:

int operator/ (const int, const MyClass &);

If it needs to access private members that you have no interface for, make it a friend inside your class as well:

friend int operator/ (const int, const MyClass &);

answered Jul 21 '12 at 02:31

chris

What exactly is a free function? I've never heard of it before. – Dasaru Jul 21 '12 at 02:36
@Dasaru, One that's not enclosed in a class. In this case, it being a member function constrains it to take a `MyClass *` as the first argument, leaving one open. The order of arguments corresponds to the way you call it, so it's impossible to switch them if it's in the class. – chris Jul 21 '12 at 02:36
@Dasaru: You'll also see them called `non-member` functions. – Blastfurnace Jul 21 '12 at 02:37

mavam · Answer 2 · 2012-07-21T02:38:01.397

3

Use a free function instead of a member function for operator/.

Binary operators typically have the same operand types. Assuming that foo has a non-explicit constructor taking an int, you would have:

struct foo
{
  foo(int i) {};
};

int operator/(foo const& x, foo const& y);

edited Jul 21 '12 at 02:38

answered Jul 21 '12 at 02:30

mavam

score 0 · Answer 3 · answered Jul 21 '12 at 02:31

0

Free function is your friend:

int operator/ (int, const MyClass &);

answered Jul 21 '12 at 02:31

iTayb

5

Where friend does not mean `friend` ;-). – mavam Jul 21 '12 at 02:31
@MatthiasVallentin: Not necessarily -- but `friend` might come into use here too. – Jerry Coffin Jul 21 '12 at 02:32

3 Answers3