How are parameters referenced when using QTimer::singleShot?

mirro

void fun(QString& str)
{

}
QString param;
QTimer::singleShot(200,this,[&,param](){}
};
//error:qualifiers dropped in binding reference of type "QString &" to initializer of type "const QString"

Chris Kawa

Did you meant to write QTimer::singleShot(200,this,[&,param](){ fun(param); }); i.e. actually call the fun function from the lambda?

If so, lambdas are immutable by default, which means param becomes const when copied by the capture group. You can't then use it as a non const parameter to fun. Couple of ways to fix this:

If you don't want to modify str then make it a const parameter i.e. void fun(const QString& str).
If you want to modify the original parameter capture it by reference, not by value i.e.
[&](){ fun(param); } or explicitly [&param](){ fun(param); }
If you want to capture param by copy and modify that copy then you have to make the lambda mutable
[=]() mutable { fun(param); } or explicitly [param]() mutable { fun(param); }
There's usually no reason to do that, but if you really want to you can.

JonB

@mirro
Assuming something close to this is your actual code (since above won't compile), I can only guess there is an issue/confusion about passing both & ("by-reference") and param ("not-by-reference"); do you really need this?

Otherwise can you show actual code you are compiling.

Chris Kawa

Did you meant to write QTimer::singleShot(200,this,[&,param](){ fun(param); }); i.e. actually call the fun function from the lambda?

If so, lambdas are immutable by default, which means param becomes const when copied by the capture group. You can't then use it as a non const parameter to fun. Couple of ways to fix this:

If you don't want to modify str then make it a const parameter i.e. void fun(const QString& str).
If you want to modify the original parameter capture it by reference, not by value i.e.
[&](){ fun(param); } or explicitly [&param](){ fun(param); }
If you want to capture param by copy and modify that copy then you have to make the lambda mutable
[=]() mutable { fun(param); } or explicitly [param]() mutable { fun(param); }
There's usually no reason to do that, but if you really want to you can.