Default value of a protected member c++

Question

Code:

#include <iostream>

using namespace std;

class Example
{
public:
    int _public;

    friend ostream& operator<< (ostream& stream, Example& o);

protected:
    int _protected;

private:
    int _private;
};

ostream& operator<< (ostream& stream, Example& o) {
    stream <<
    "_public=" << o._public << endl <<
    "_protected=" << o._protected << endl <<
    "_private=" << o._private << endl;
    return stream;
}


int main(int argc, char const *argv[])
{
    Example e;
    cout << e << endl;
    return 0;
}

Output

_public=4196960
_protected=0
_private=4196368

Question:

All the three members are uninitialized. But only the public and private members have garbage values in them. Why is the protected member initialized to zero? Is there a reason for that?

g++ version and flags

g++ (Ubuntu 5.4.0-6ubuntu1~16.04.11) 5.4.0 20160609

-std=c++11

`return stream;` is missing in your code example in the `operator<<` function — Wander3r, Feb 13 '19 at 05:27
You are seeing the symptoms of undefined behavior. Explicitly initialize the member variables and you won't have to worry about strange values at all. — R Sahu, Feb 13 '19 at 05:34

P.W · Accepted Answer · 2019-02-13T05:34:22.663

1

First, note that reading uninitialized variables is undefined behavior. You need to define a constructor for your class. And this has nothing to do with the access specifiers. It just happens (in your case) that the protected member is at an address which previously contained the value 0.

Also better not to use variable names starting with underscore. They are reserved. It is allowed to use them in class scope but remember not to use them in global scope.

5.10 Identifiers [lex.name]

3 In addition, some identifiers are reserved for use by C++ implementations and shall not be used otherwise; no diagnostic is required.
...

(3.2) Each identifier that begins with an underscore is reserved to the implementation for use as a name in the global namespace.

edited Feb 13 '19 at 05:34

answered Feb 13 '19 at 05:20

P.W

26,289
6
39
76

2

There is nothing wrong with those identifiers at class scope. The standard reserves them at the global namespace only. – StoryTeller - Unslander Monica Feb 13 '19 at 05:24
I just ran the above code in VS 2017 and got all values as `0`. Ran it several times, but the output hasn't changed. Is it compiler specific? – Wander3r Feb 13 '19 at 05:25
@P.W Yes, I've only used them for this example purpose, and inside the class scope. – Vijay Chavda Feb 13 '19 at 05:29
@SaileshD: Compiler specific means implementation defined behavior. Reading uninitialized variables is not implementation defined but undefined behavior. – P.W Feb 13 '19 at 05:30
1

See also https://stackoverflow.com/questions/2397984/undefined-unspecified-and-implementation-defined-behavior – aschepler Feb 13 '19 at 05:37
@aschepler I was just going to comment that link :) – Vijay Chavda Feb 13 '19 at 05:41

Default value of a protected member c++

Code:

Output

Question:

g++ version and flags

1 Answers1