Lambda function referenced pointer destroyed detection

Question

I create a lambda function to run code in a different thread or simply to run it a bit later, but it can happen that an object kept by the lambda function is deleted in the mean time.

How can I detect that and not run the function in that case?

for instance

class A
{
public:
    A(){}
    virtual void test(){std::cout << m;}
    int m;
};
int main()
{
    A* a = new A();
    std::function<void ()> function = [=]()->void
    {
        //if( pointer to 'a' still valid )
        {
            a->test();
        }
    };
    delete a;
    //or if( pointer to 'a' still valid )
    function();

    system("pause");
    return 0;
}

or the detection could be done before executing the lambda function too.

An other idea is to have an object "Runnable" keep the lambda function and register it to the one that can be deleted. Then in the destructor I would notify the Runnable and prevent the execution.

Would that be a good way to do it ?

You should be setting a=NULL after you invoke delete. Then you can do if (a != NULL) DO SOMETHING. — Tuxdude
– Tuxdude, Commented Mar 5, 2013 at 21:37
using a shared ptr would keep the object alive until the execution, but I need that object to be deleted when required. Setting the variable to null doesnt change the value inside the lambda function — Michael Ferris
– Michael Ferris, Commented Mar 5, 2013 at 21:43
you can also use std::weak_ptr to have a pointer that can be checked if you have other pointers to this object and convert to smart pointer when needed (and fails graceully, if an object was deleted in the mean time, you need to use it with smart_ptr's, not raw pointers though. weak_ptr — Marcin Deptuła
– Marcin Deptuła, Commented Mar 5, 2013 at 21:43

zzk · Accepted Answer · 2013-03-05 21:46:00Z

2

You cannot test whether the object pointed by the pointer has been deleted or not..

If it has been deleted, your test() would just have undefined behavior.

answered Mar 5, 2013 at 21:46

zzk

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Yakk - Adam Nevraumont · Accepted Answer · 2013-03-06 15:00:32Z

0

Here is a solution:

std::shared_ptr<A> a(new A());
std::weak_ptr<A> weak_a(a);
std::function<void ()> function = [weak_a]()->void
{
    if( std::shared_ptr<A> a = weak_a.lock() )
    {
        // to get the A* from a, do a.get()
        // operator-> and *a work however:
        a->test();
    }
};
a.reset(); // instead of delete

The use of the weak_ptr is optional -- if you instead made it a shared_ptr copied to the lambda, the lifetime of a would be extended by the lifetime of the lambda.

This does require that the code that uses a outside of the lambda be shared_ptr compliant.

answered Mar 6, 2013 at 15:00

Yakk - Adam Nevraumont

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