On Wednesday, 21 de September de 2011 15.12.21, Thiago Macieira wrote: > There's no way around it. Since we're using the function pointer (actually, > the PMF) to identify the signal, you must have C++ access to the signal > function. That means you must be able to call it. > > Other template-based signal libraries have the same problem. It cannot be > avoided unless you have a way to identify the signal which doesn't include > access to the actual signal implementation. > > One solution to that would be to have an enum of signals, instead of > identifying by an object or by a PMF.
Another solution (also by way of adding an indirection) is to add a function
that emits, which is protected.
template <typename Klass, typename... Args>
void QObject::emitSignal(void (Klass:: *signal)(Args...), Args... args);
Then you'd write:
emitSignal(&Me::textChanged, newText);
instead of:
emit textChanged(newText);
The template magic to make that happen is already present.
However, it would be completely source-incompatible with the existing code,
which is a big no-no. It would require making the signal functions themselves
not do anything -- only serve as identifiers.
--
Thiago Macieira - thiago (AT) macieira.info - thiago (AT) kde.org
Software Architect - Intel Open Source Technology Center
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