CODEWITHSUNDEEP
c++operator-overloading
Dan Dv
Dan Dv

Reputation: 493

Meaning of int variable in suffix increment operator in C++

I have this code:

class LazyStream {
    ostream& output;
    std::list<string> pending;
public:
    //...
    LazyStream operator++(int n = 0){
        LazyStream other(output);
        if (n>0) output << "<#" << n << ">";
        output << pending.pop_front();
        return other;
    }

I do not understand the meaning of getting an int value for the operator++. I though it was just an indication that the operator is a suffix. How can the operator get a number? Can someone give an example?

Thanks

Upvotes: 3

Views: 325

Answers (2)

jrok
jrok

Reputation: 55425

Apparently it's possible to pass that argument if you use function call syntax to call that operator. This code compiles cleanly with gcc and outputs 42:

#include <iostream>

struct Stream {
    Stream operator++(int n)
    {
        std::cout << n;
        return *this;
    }
};

int main()
{
    Stream s;
    s.operator++(42);
}

If I give it default value, it gives a warning (with -pedantic flag) that it cannot have one, though. It sort of makes sense, because if you also defined prefix increment, then the call s.operator++() would be ambiguous. I didn't, however, find anything in the standard explicitly prohibiting the default value.

Upvotes: 2

James Kanze
James Kanze

Reputation: 154007

Well, it's the first time I've seen the int defaulted.

As you point out, a "dummy" int parameter is use to distinguish the post-fix operator from the prefix. Except that it's not really a dummy: when you write: 

myVar ++;

and myVar has a user defined postfix ++, the compiler actually calls it as:

myVar.operator++( 0 );

And there's nothing to stop you from writing:

myVar.operator++( 42 );

(Of course, having to do so, as in this case, sort of defeats the purpose of operator overloading.)

Upvotes: 3

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