c++ - Capturing a static variable by reference in a C++11 lambda

Question

Main question

I'm trying to compile the following code with GCC 4.7.2:

#include <iostream>

int foo() {
    static int bar;
    return [&bar] () { return bar++; } (); // lambda capturing by reference
}

int main (int argc, char* argv[]) {
    std::cout << foo() << std::endl;
    return 0;
}

And it seems that is not going well, as the output is this one:

$p2.cpp: In function ‘int foo()’:
$p2.cpp:6:14: warning: capture of variable ‘bar’ with non-automatic storage duration [enabled by default]
$p2.cpp:4:16: note: ‘int bar’ declared here

So, my first question would be:

Is this a failure of GCC, or the code is not legit C++11? Is this fixed in any recent version of GCC?

Using the trick in a shared_ptr factory

I consider to build an artifact based on this principle but using a non-literal static variable. This artifact is meant to be a factory of shared_ptr< T > objects, which avoid the creation of new T objects when you just need a duplicate shared_ptr container for the same instance.

This artifact would look like:

std::shared_ptr<Foo> create(std::string name) {
    static std::unordered_map<std::string,std::weak_ptr<Foo>> registry;

    if (auto it = registry.find(name) != registry.end())
        return registry[name].lock();

    auto b = std::shared_ptr<Foo>(
        new Foo(name), 
        [&registry] (Foo* p) {
            registry.erase(p->getName());
            delete p;
        });

    registry.emplace(name,b);
    return b;
}

As far as I know, if the GCC issue discussed before is not a problem in terms of C++11 conformance, this artifact shouldn't be an issue either. The only thing to take care by using this hack, is to not set the resulting shared_ptr< T > object to any global object which could be destructed after the static variable.

Am I right about this?

Answer 1

Why are you even trying to capture bar? It's static. You don't need to capture it at all. Only automatic variables need capturing. Clang throws a hard error on your code, not just a warning. And if you simply remove the &bar from your lambda capture, then the code works perfectly.

#include <iostream>

int foo() {
    static int bar;
    return [] () { return bar++; } (); // lambda capturing by reference
}

int main (int argc, char* argv[]) {
    std::cout << foo() << std::endl;
    std::cout << foo() << std::endl;
    std::cout << foo() << std::endl;
    return 0;
}

prints

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