[VikingCoder]

You're also consuming the variable name i, which everyone else is probably using.

Speaking of which, this means you can't have one of your locks scoped inside another one.

[cjensen]

Solvable by creating a Preprocessor macro which creates a unique and complex-named identifier.

  #define LOCKED(lk) for(int UNIQUE_ID=0; UNIQUE_ID<1 && !lk_lock(lk); lk_unlock(lk), UNIQUE_ID++)

Use __LINE__ to help make the id unique.

[VikingCoder]

Then you still can't forget and do this:

    lock(lk_var) lock(lk_var2)
    {
        lk_var = lk_var2;
    }

Or this:

    lock(lk_var) { lock(lk_var2 { lk_var = lk_var2; } }

Using the pre-processor to solve a regular, every day, hum-drum, solved-a-billion-times RAII problem is bad news.

[JoeAltmaier]

SUre you could. The{} make it work.

[VikingCoder]

EDIT: I sit corrected. My post is wrong.

No, you can't, and no they don't.

    lock(lk_var) {
        lock(lk_var2) {
            var = var2;
        }
    }

That expands to:

    for(int i=0; (i < 1) && !lk_lock(lk_var); lk_unlock(lk_var), i++)
    {
        for(int i=0; (i < 1) && !lk_lock(lk_var2); lk_unlock(lk_var2), i++)
        {
            var = var2;
        }
    }

You end up with an i scoped inside another i.

[plorkyeran]

C++ has variable shadowing, so that works correctly.

    for (int i = 0; i < 5; ++i) {
        for (int i = 0; i < 5; ++i) {
            printf("hello\n");
        }
    }

prints hello 25 times.

[VikingCoder]

Oh right. My brain was stuck in another language. Wow, I forgot how much I'd forgotten about C++.

[JoeAltmaier]

Actually that's just C (name nesting)