Question

From my experience it seems that either:

• A lambda expression created inside a function call is destroyed just after the invocation
• Calling a function that expects a std::function creates a temporary object (std::function) out of the lambda, and that object is destroyed after invocation

This behavior can be observed with the following snippet of code:

const function<void()>* pointer;

void a(const function<void()> & f)
{
    pointer = &f;
}

void b()
{
    (*pointer)();
}

int main()
{
    int value = 1;
    std::cout << &value << std::endl;

    // 1: this works    
    function<void()> f = [&] () { std::cout << &value << std::endl; };
    a(f);

    // 2: this doesn't
    a([&] () { std::cout << &value << std::endl; });

    /* modify the stack*/
    char data[1024];
    for (int i = 0; i < 1024; i++)
        data[i] = i % 4; 

    b();

    return 0;
}

What exactly s actually happening in the second case?
Is there a correct way to call a() without creating an explicit std::function object?

Edit::
This both versions (1 and 2) compile just right but result in different outputs:

Version 1:

0x7fffa70148c8
0x7fffa70148c8

Version 2:

0x7fffa70148c8
0

Answer 1

If you create a temporary, it will be gone at the end of the line. This means storing a pointer to it is a bad idea, as you correctly stated.

If you want to store a pointer to a std::function (or anything else really), you need to make sure it’s lifetime doesn’t end before you stop using the pointer. This means that you really do need a named object of type std::function.

As to what is happening in the second case: You create a temporary lambda to be passed to the function. Since the function expects a std::function, a temporary std::function will be created from the lambda. Both of those will be destroyed at the end of the line. Therefore you now have a pointer to an already destroyed temporary, which means that trying to use the pointed to object will bring you firmly into undefined behaviour territory.