Recursive variant output compiles but segfault at runtime

Question

So I wrote an object that could handle JSON-like content, of which there is a minimal example:

#include <iostream>
#include <vector>
#include <variant> 

struct Array;
struct Value;

struct Array: public std::vector<Value>{
    using std::vector<Value>::vector;
};

struct Value: public std::variant<bool, int, Array>{
    using std::variant<bool, int, Array>::variant;
};

Now I wanted to overload the output stream operator. Here is the code:

std::ostream& operator <<(std::ostream& os, const Value value);
std::ostream& operator <<(std::ostream& os, const Array array);

std::ostream& operator <<(std::ostream& os, const Array array){
    for (auto &a : array){
        os << a << ", ";
    }
    return os;
}
std::ostream& operator <<(std::ostream& os, const Value value){
    os << value;
    return os;
}

Now, if I try to run it :

int main()
{
    Array a {1,2,3}; // OK
    std::cout << a;  // Segfault
}

Where's the catch ?

EDIT As pointed, there was a recursion loop.

What did you observe, when stepping through your code with a debugger? — Algirdas Preidžius, Jan 10 '19 at 15:35
Be aware of [the risks of inheriting from standard containers](https://stackoverflow.com/questions/2034916/is-it-okay-to-inherit-implementation-from-stl-containers-rather-than-delegate). — François Andrieux, Jan 10 '19 at 15:35
You have a segfault, so you have a stack trace or easy access to one. Look at it. — Useless, Jan 10 '19 at 15:36
... and the `Value`, which may also contain an `Array` (in general). — Useless, Jan 10 '19 at 15:36
In the `operator <<(std::ostream& os, const Value value)` case, what did you expect `os << value` to do? — Botje, Jan 10 '19 at 15:38
@Useless That's something I never did, before, I'm not sure how to do that. — Oreille, Jan 10 '19 at 15:44
Learn to use your debugger - it will also (on any sane system) handle the segfault for you, or load a core file once you have one. — Useless, Jan 10 '19 at 15:49
Clang warns about it, https://wandbox.org/permlink/XFdStZJjMPTQMHka . — Hiroki, Jan 10 '19 at 15:52

eerorika · Answer 1 · 2019-01-10T15:54:22.510

4

You have a stack overflow due to infinite recursion:

std::ostream& operator <<(std::ostream& os, const Value value){
    os << value; // call to self
    return os;
}

This would be a correct way to insert a variant into a stream:

std::visit([&](const auto& value) {
    os << value;
}, static_cast<const Value::variant&>(value));

edited Jan 10 '19 at 15:54

answered Jan 10 '19 at 15:47

eerorika

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The fix (simply deleting that operator) might be mentioned? – Yakk - Adam Nevraumont Jan 10 '19 at 15:51
@Yakk-AdamNevraumont I don't think that would work. I added a fix though. – eerorika Jan 10 '19 at 15:54
I just tried your solution and it works perfectly. Thank you very much! – Oreille Jan 10 '19 at 15:57
@eerorika Ah, yes, `variant` doesn't expose an `operator<<` overload. – Yakk - Adam Nevraumont Jan 10 '19 at 16:09

Recursive variant output compiles but segfault at runtime

1 Answers1