Why lambda returns bool?

Question

I've started learning C++11 and C++14 and i have a question. Why lambda not returns 23?

template<class T>
auto func(T t)
{
    return t;
}

int main(int argc, char *argv[])
{
    QCoreApplication a(argc, argv);

    auto abc = []()->auto { return func(23); };
    qDebug() << abc; // output: true

    return a.exec();
}

Answer 1

As @Bathsheba pointed out. You have a typo and don't call the lambda. Now, it's rather obvious that operator<< for qDebug() is not overloaded on the lambda's closure type. So naturally an implicit conversion sequence has to happen. The only available one, and only because your lambda is capture-less, starts with a conversion to a function pointer.

Now, which overload of operator<< can be used to print a function pointer? On the face of it, two likely candidates:

operator<<(bool t)         // Because it prints true, duh
operator<<(const void *p)  // Because pointers :)

So why the bool overload? Because a function pointer is not implicitly convertible to void*. That conversion is conditionally supported, and must be performed with a cast ([expr.reinterpret.cast]/8):

Converting a function pointer to an object pointer type or vice versa is conditionally-supported.

That leaves us only with [conv.bool]:

A prvalue of arithmetic, unscoped enumeration, pointer, or pointer to member type can be converted to a prvalue of type bool.