这是程序：

#include <iostream>
#include <functional>

struct Foo
{
    inline Foo() {
        std::cout << "foo constructor" << std::endl;
    }

    inline ~Foo() {
        std::cout << "foo destructor" << std::endl;
    }
};

void NotBroken()
{
    Foo foo;
    std::function<void(Foo&)> callback;
    callback = [](auto&) { // <- notice the auto&
        std::cout << "executed callback" << std::endl;
    };
    callback(foo);
}

void TotallyBroken()
{
    Foo foo;
    std::function<void(Foo&)> callback;
    callback = [](auto) { // <- notice the auto without &
        std::cout << "executed callback" << std::endl;
    };
    callback(foo);
}

int main()
{
    NotBroken();
    std::cout << std::endl;
    TotallyBroken();
}

以下是输出：

foo constructor
executed callback
foo destructor

foo constructor
executed callback
foo destructor
foo destructor

从字面上看，为什么没有&的auto关键字不应该仍然检测到我正在将lambda传递到期望引用的std：：函数中？我用MSVC，Clang和GCC测试了这一点，它们都有这种行为。
额外的析构函数调用发生在lambda执行结束时