Hi All,
I am facing a strange issue on Mac, where wxExecute called with wxEXEC_SYNC, but it returns immediately with the return code -1 and the app that was called continues its execution.
It is like, I have called the app asynchronously. But actually, I am using the argument wxEXEC_SYNC.
This is happening only on MAC. The call is made from a bundle to an application.
What could be the issue in this case .?
There are actually a series of apps in this case
app1 -> calls app2 -> calls app3 -> calls app4 -> loads bundle -> calls app5
The issue happens in the phase loads bundle -> calls app5 in the above sequence.
Let me know if you need anything
The code is as follows
long pid = wxExecute(launcherpath, wxEXEC_SYNC)
So, if I put a message box before and after the call
wxMessageBox("Before calling");
long pid = wxExecute(launcherpath, wxEXEC_SYNC)
wxMessageBox("After calling");
And in the Oninit of the app that I call, I have put another message box
wxMessageBox("Oninit");
When I execute this, I get the message
Before calling
After calling
parallely Oninit is called...
Thanks in advance!
Thanks,
Karan
Mac only -wxExecute called with wxEXEC_SYNC returns immediately
Re: Mac only -wxExecute called with wxEXEC_SYNC returns immediately
Hi,
What version of wx?
What version of OSX?
What is an exact configure line you used?
I presume it is reproducible in the exec sample?
Thankj you.
What version of wx?
What version of OSX?
What is an exact configure line you used?
I presume it is reproducible in the exec sample?
Thankj you.
Re: Mac only -wxExecute called with wxEXEC_SYNC returns immediately
Although 7 years old, this seems relevant:
http://trac.wxwidgets.org/ticket/13033#comment:3
http://trac.wxwidgets.org/ticket/13033#comment:3
Use the source, Luke!