On quinta-feira, 1 de novembro de 2012 13.03.56, Stephen Chu wrote: > I find out what QPrivateSignal is for but it's not clear if lambda can > still be used to connect to such private signals. > > If I connect QTimer::timeout() to a slot that takes no argument, it > compiles fine. But when I try to connect to a lambda function without > any argument, the results in the mentioned error.
Can you rephrase and/or give us code samples? "takes no arguments" and "without any argument" are semantically the same in English. > > Is this designed behavior? It makes using lambda function as slot really > hard since there's no mention in documentation which signals are limited > this way. -- Thiago Macieira - thiago.macieira (AT) intel.com Software Architect - Intel Open Source Technology Center
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