Is std::bind exactly equivalent to a function object If I create a similar one ?

<Though the subject line may be a bit misleading, bear with me a little>

I was reading up on pthread `s and then found out std::thread. It ...seems wonderful.

So, are the following versions of the snippet exactly equivalent?

Code:

---

class X
{
    public:
    std::string foo;
    void operator()(){std::cout<<foo<<std::endl;}
};
int main()
{
    X bar;
    cin>>bar.foo;
    thread t(bar);
    t.join();
    return 0;
}

---

Code:

---

void foo(std::string bar)
{
    std::cout<<bar<<std::endl;
}
int main()
{
    string bar;
    cin>>bar;
    thread t(bind(foo,bar));
    t.join();
    return 0;
}

---

Or is there some hidden difference I should know about ?

[Also, I remember seeing a thread here in which someone posted about a bug in gcc's implementation of std::thread. I can't find it now. Anyone remember what it was about?]

They will work the same, yes.
Note that boost::bind stores a copy of bar unless you wrap it in std::ref. Import if you pass a reference to a function which you want it to modify.

Quote:

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Originally Posted by Elysia

They will work the same, yes.
Note that boost::bind stores a copy of bar unless you wrap it in std::ref. Import if you pass a reference to a function which you want it to modify.

---

By "work" do you mean they give the same output? (They do, as I made these snippets with that in mind.)
Or as I read here that bind creates f-objects(not just variables) as an intermediate? (Though I suspect that it is implementation defined. )

Yes, they will give the same output.
boost::bind creates a functor object which it returns, and when invoked, calls the desired function with all the bound parameters, so it is an itermediate step.