Why is #include <string> preventing a stack overflow error here?

Question

This is my sample code:

#include <iostream>
#include <string>
using namespace std;

class MyClass
{
    string figName;
public:
    MyClass(const string& s)
    {
        figName = s;
    }

    const string& getName() const
    {
        return figName;
    }
};

ostream& operator<<(ostream& ausgabe, const MyClass& f)
{
    ausgabe << f.getName();
    return ausgabe;
}

int main()
{
    MyClass f1("Hello");
    cout << f1;
    return 0;
}

If I comment out #include <string> I don't get any compiler error, I guess because it's kind of included through #include <iostream>. If I "right-click --> Go to Definition" in Microsoft VS they both point to the same line in the xstring file:

typedef basic_string<char, char_traits<char>, allocator<char> >
    string;

But when I run my program, I get an exception error:

0x77846B6E (ntdll.dll) in OperatorString.exe: 0xC00000FD: Stack overflow (Parameter: 0x00000001, 0x01202FC4)

Any idea why I get a runtime error when commenting out #include <string>? I'm using VS 2013 Express.

Answer 1

Indeed, very interesting behavior.

Any idea why I get I runtime error when commenting out #include <string>

With MS VC++ compiler the error happens because if you do not #include <string> you won't have operator<< defined for std::string.

When the compiler tries to compile ausgabe << f.getName(); it looks for an operator<< defined for std::string. Since it was not defined, the compiler looks for alternatives. There is an operator<< defined for MyClass and the compiler tries to use it, and to use it it has to convert std::string to MyClass and this is exactly what happens because MyClass has a non-explicit constructor! So, the compiler ends up creating a new instance of your MyClass and tries to stream it again to your output stream. This results in an endless recursion:

 start:
     operator<<(MyClass) -> 
         MyClass::MyClass(MyClass::getName()) -> 
             operator<<(MyClass) -> ... goto start;

To avoid the error you need to #include <string> to make sure that there is an operator<< defined for std::string. Also you should make your MyClass constructor explicit to avoid this kind of unexpected conversion. Rule of wisdom: make constructors explicit if they take only one argument to avoid implicit conversion:

class MyClass
{
    string figName;
public:
    explicit MyClass(const string& s) // <<-- avoid implicit conversion
    {
        figName = s;
    }

    const string& getName() const
    {
        return figName;
    }
};

It looks like operator<< for std::string gets defined only when <string> is included (with the MS compiler) and for that reason everything compiles, however you get somewhat unexpected behavior as operator<< is getting called recursively for MyClass instead of calling operator<< for std::string.

Does that mean that through #include <iostream> string is only included partly?

No, string is fully included, otherwise you wouldn't be able to use it.