Help on PyQt4 QProcess

[Carl Banks]

On Friday, August 19, 2011 12:55:40 PM UTC-7, Edgar Fuentes wrote:
> On Aug 19, 1:56 pm, Phil Thompson 
>  wrote:
> > On Fri, 19 Aug 2011 10:15:20 -0700 (PDT), Edgar Fuentes
> > <fuen... at gmail.com> wrote:
> > > Dear friends,
> >
> > > I need execute an external program from a gui using PyQt4, to avoid
> > > that hang the main thread, i must connect the signal "finished(int)"
> > > of a QProcess to work properly.
> >
> > > for example, why this program don't work?
> >
> > >    from PyQt4.QtCore import QProcess
> > >    pro = QProcess() # create QProcess object
> > >    pro.connect(pro, SIGNAL('started()'), lambda
> > > x="started":print(x))        # connect
> > >    pro.connect(pro, SIGNAL("finished(int)"), lambda
> > > x="finished":print(x))
> > >    pro.start('python',['hello.py'])        # star hello.py program
> > > (contain print("hello world!"))
> > >    timeout = -1
> > >    pro.waitForFinished(timeout)
> > >    print(pro.readAllStandardOutput().data())
> >
> > > output:
> >
> > >    started
> > >    0
> > >    b'hello world!\n'
> >
> > > see that not emit the signal finished(int)
> >
> > Yes it is, and your lambda slot is printing "0" which is the return code
> > of the process.
> >
> > Phil
> 
> Ok, but the output should be:
> 
>     started
>     b'hello world!\n'
>     finished
> 
> no?.
> 
> thanks Phil

Two issues.  First of all, your slot for the finished function does not have the correct prototype, and it's accidentally not throwing an exception because of your unnecessary use of default arguments.  Anyway, to fix that, try this:

pro.connect(pro, SIGNAL("finished(int)"), lambda v, x="finished":print(x))

Notice that it adds an argument to the lambda (v) that accepts the int argument of the signal.  If you don't have that argument there, the int argument goes into x, which is why Python prints 0 instead of "finished".

Second, processess run asynchrously, and because of line-buffering, IO can output asynchronously, and so there's no guarantee what order output occurs.  You might try calling the python subprocess with the '-u' switch to force unbuffered IO, which might be enough to force synchronous output (depending on how signal/slot and subprocess semantics are implemented).


Carl Banks