Solved connect cause exception triggered (Beginner)
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when you click the button, plugin in the lambda will be a dangling pointer. In debug mode some compilers set dangling pointers to NULL and this triggers your error.
Bottom line: you are misusing the lambdaabout the delete question, when you call
layout->addWidget();
the layout will take care of deleting the widgets see http://doc.qt.io/qt-5/qlayout.html#addItem -
it's strange i init them in heap store and acces them with a static list:
in mainwindow.cpp's constructorfor(Plugin* p : PluginCollection::get_plugins()) { build_tab_from_plugin(p); }
plugincollection.h
namespace PluginCollection { QList<Plugin *> get_plugins() { static QList<Plugin*> list {new PastebinPlugin() }; return list; } }
should static list stay in memory (as well as allocated obj) until application closing?
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@DumaPlusPlus You're passing references to local variables [&] which only exist while MainWindow::build_tab_from_plugin is being executed. Since everything you use in the lambda are pointers you can pass by value [=]. Is plugin a valid pointer?
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EDIT: I WAS WRONG
no matter how you pass thos variables [&] or [=]. when the lambda gets called all those pointers will be junk so it won't work
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EDIT: I WAS WRONG
when You call
build_tab_from_plugin()
all the pointers are valid and fine, then the function terminates and all pointers go out of scope. when the button is pressed, the code goes into the lambda (imagine agoto
), the environment of build_tab_from_plugin is recreated but you don't know what the pointers point to right now. the lambda won't save the value of any of the pointers in its body uppon declaration -
thanks to all.
@jsulm resolves my problem but now i'm bit confused when do i connect SIGNAL with lambda this isn't stored somewhere? and reference used in lambda isn't an alias to a pointer (something that stay in memoery untile delete operator is called) ? -
@VRonin said:
the lambda won't save the value of any of the pointers in its body uppon declaration
Sure about this? As far as I know C++11 (which isn't so well) a pointer is an automatic storage variable and is copy-captured by
[=]
as any other auto-storage variable.the code goes into the lambda (imagine a goto), the environment of build_tab_from_plugin is recreated
This is very wrong way of thinking about a lambda, sorry for saying. A lambda is a typical functor (and is implemented as such).
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@DumaPlusPlus If you use references then then "point" to the variable. In your case they point to local variables. These local variables disappear as soon as the method finishes, so the "pointers" to them are not valid anymore - because they do not exist anymore.
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@VRonin Sorry, but you're wrong. You can easily test this: using [&] will crash, using [=] works just fine.
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@jsulm I'm really confused now as I tested it and this code works 100% fine, even with [&]. how can it be?!
#include <QCoreApplication> #include<QDebug> #include <QTimer> int main(int argc, char *argv[]) { QCoreApplication appl(argc,argv); QTimer mainTimer; mainTimer.setSingleShot(true); { QString* myString=nullptr; myString=new QString("A Message"); QObject::connect(&mainTimer,&QTimer::timeout,[&](){qDebug() << *myString;}); QObject::connect(&mainTimer,&QTimer::timeout,[=](){qDebug() << *myString;}); // Memory leak! } mainTimer.start(100); return appl.exec(); }
EDIT:
Using MSVC2013 on Windows -
@VRonin said:
QObject::connect(&mainTimer,&QTimer::timeout,[&](){qDebug() << *myString;});
What about:
QObject::connect(&mainTimer,&QTimer::timeout,[&](){myString = nullptr;});
You're capturing
QString*
, so you'd end up with:QString*&
but you don't modify the string pointer, rather you dereference the object it's pointing to, so you'd try to output aQString &
withQDebug
. Also probably your compiler somewhat lax. :)PS.
Well that's really disturbing ... I don't get any errors either. The memory will silently be overwritten. (g++ on Linux) -
second test:
#include <QCoreApplication> #include <QDebug> #include <QTimer> #include <QPointer> int main(int argc, char *argv[]) { QCoreApplication appl(argc,argv); QTimer mainTimer; mainTimer.setSingleShot(true); { QPointer<QObject> mybj; mybj =new QObject(); mybj->setObjectName("A Message"); QObject::connect(&mainTimer,&QTimer::timeout,[&](){qDebug() << mybj->objectName(); mybj->setObjectName("Foo"); qDebug() << mybj->objectName();}); QObject::connect(&mainTimer,&QTimer::timeout,[=](){qDebug() << mybj->objectName(); mybj->setObjectName("Bar"); qDebug() << mybj->objectName();}); // Memory leak! } { QString testString("Occuppy Stack"); } mainTimer.start(100); return appl.exec(); }
Still working correctly. Notice how the first output of the [=] lambda is Foo. HOW?!
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@VRonin said:
second test:
QObject::connect(&mainTimer,&QTimer::timeout,[&](){qDebug() << mybj->objectName(); mybj->setObjectName("Foo"); qDebug() << mybj->objectName();}); QObject::connect(&mainTimer,&QTimer::timeout,[=](){qDebug() << mybj->objectName(); mybj->setObjectName("Bar"); qDebug() << mybj->objectName();});
Still working correctly. Notice how the first output of the [=] lambda is Foo. HOW?!
should be that?
passing by value copy the pointer so you point to objname with modified name...right? -
@DumaPlusPlus The first lambda should never be executed at the creation of the second one. then when the timer times out I was expecting the first to crash or operate on invalid memory and the second to print
A Message Bar
While it looks like the lambda with [&] behaves exactly as the one with [=]
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If I haven't missed anything. my g++ doesn't open a new stack frame when it sees:
{ }
. So in your examples (which I used) all is flat, thusmyString
is inmain()
's stack frame, which means it doesn't go out of scope (i.e. it's not pop-ed from the stack), which ultimately means that the lambda capture is valid.Here's what I have for
main()
from yourQString
test case:# Sets up main()'s stack 0x400fc2 55 push %rbp 0x400fc3 <+0x0001> 48 89 e5 mov %rsp,%rbp ... 0x400fcf <+0x000d> 48 83 ec 78 sub $0x78,%rsp ... # mainTimer.setSingleShot(true); 0x401010 <+0x004e> 48 8d 45 80 lea -0x80(%rbp),%rax 0x401014 <+0x0052> be 01 00 00 00 mov $0x1,%esi 0x401019 <+0x0057> 48 89 c7 mov %rax,%rdi 0x40101c <+0x005a> e8 59 08 00 00 callq 0x40187a <QTimer::setSingleShot(bool)> # QString * myString = nullptr; 0x401021 <+0x005f> 48 c7 85 78 ff ff ff 00 00 00 00 movq $0x0,-0x88(%rbp) # No stack frame was opened as one'd expect from a block ... # main()'s stack's being unwound 0x40113c <+0x017a> 48 83 c4 78 add $0x78,%rsp ... 0x401149 <+0x0187> 5d pop %rbp # And that was all folks, thanks for playing 0x40114a <+0x0188> c3 retq
As for the lambda, it doesn't make any checks. It just stores the captured address (the reference) and ultimately dereferences it when it's executed:
... # qDebug() << *myString; 0x400ec3 <+0x000d> 48 8b 45 b8 mov -0x48(%rbp),%rax # Load QString *& from the base pointer 0x400ec7 <+0x0011> 48 8b 00 mov (%rax),%rax # Dereference once (strip &) 0x400eca <+0x0014> 48 8b 18 mov (%rax),%rbx # Dereference second time i.e. (*myString)
So I hope this explains how and why.
Kind regards.
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Thanks @kshegunov now it make sense, it was just the compiler (I use MSVC btw) optimizing.
This behaves as expected.#include <QCoreApplication> #include<QDebug> #include <QTimer> void makeConnections(QTimer* mainTimer ){ QString* myString=nullptr; myString=new QString("A Message"); QObject::connect(mainTimer,&QTimer::timeout,[&](){qDebug() << *myString;}); QObject::connect(mainTimer,&QTimer::timeout,[=](){qDebug() << *myString;}); } int main(int argc, char *argv[]) { QCoreApplication appl(argc,argv); QTimer mainTimer; mainTimer.setSingleShot(true); makeConnections(&mainTimer); mainTimer.start(100); return appl.exec(); }
I marked my previous post where I was wrong and the final answer to the topic is use [=] in the lambda instead of [&]
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Thanks @kshegunov now it make sense, it was just the compiler (I use MSVC btw) optimizing.
No problem. Yes the compiler was an issue apparently, although that's some strange optimization made. Especially if you take into account we're running in debug mode, two different compilers no less. But even in release mode I'd venture to say one doesn't expect a block to just be ignored ... at least I don't.
This behaves as expected.
Meaning it crashes at the appropriate place? :)
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@kshegunov said:
Meaning it crashes at the appropriate place? :)
Even a crash sometimes is expected behaviour ;)
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wow very good thread i wish i will be professional like you