c++ overload ostream doesn't work for default construct with empty argument

Question

Here is an example taken from https://msdn.microsoft.com/en-us/library/1z2f6c2k.aspx

#include <iostream>
using namespace std;

class Date
{
    int mo, da, yr;
public:
    Date()
    {
        mo = 10; da = 10; yr = 99;
    }
    Date(int m, int d, int y)
    {
        mo = m; da = d; yr = y;
    }
    friend ostream& operator<<(ostream& os, const Date& dt);
};

ostream& operator<<(ostream& os, const Date& dt)
{
    os << dt.mo << '/' << dt.da << '/' << dt.yr;
    return os;
}

int main()
{
    Date dt0();
    cout << dt0 << endl;

    Date dt(5, 6, 92);
    cout << dt;
}

I'd expect the output to be

10/10/99
5/6/92

But what I get was

1
5/6/92

What or how can I work this out?

score 4 · Accepted Answer · edited May 23 '17 at 12:07

4

Welcome to C++, where you can get tripped up because of the most vexing parse!

dt0 is not an object, it is a function! When you output it, you are actually outputting the address of the function dt0, which always evaluates to true, i.e. 1.

You can replace the () with {}, so not ambiguity occurs:

Date dt0{};

Or just drop the parenthesis altogether:

Date dt0;

edited May 23 '17 at 12:07

Community

1
1

answered Nov 26 '16 at 14:04

Rakete1111

47,013
16
123
162

Thank you for the clarification! Best answer in a few mins! – user3221454 Nov 26 '16 at 14:09
1

Just so you know, that `cout << &dt0 << endl` will print the same output, because both indicate the address of the function. (And in that explicit way, it could reduce some compiler warnings, if you really mean it.) – Dean Seo Nov 26 '16 at 14:31
1

@Dean I already knew that, but thanks anyway :) – Rakete1111 Nov 26 '16 at 14:32
@Rakete1111 I'm sure you do. That comment was actually for the OP. :) – Dean Seo Nov 26 '16 at 14:34
@Dean Oh, then you should have mentioned him with @, OP might never see it :) – Rakete1111 Nov 26 '16 at 14:34

c++ overload ostream doesn't work for default construct with empty argument

1 Answers1