Can I get the size of a struct field w/o creating an instance of the struct?

Question

It's trivial to get the size of a struct's field in C++ if you have an instance of the struct. E.g. (uncompiled):

typedef struct Foo {
    int bar;
    bool baz;
} Foo;

// ...

Foo s;
StoreInSomething(s.bar, sizeof(s.bar)); // easy as pie

Now I can still do something like this, but with the interface I'm implementing (I get a BOOL that indicates what the state of a specific bit in a bitfield should be), I'd be creating the struct solely to get the size of the data member. Is there a way to indicate to the compiler that it should use the size of a struct's field without creating an instance of the struct? It would be the philosophical equivalent of:

SetBit(bool val) {
    StoreInSomething(
        BITFIELD_POSITION_CONSTANT, // position of bit being set
        val,                        // true = 1, false = 0
        sizeof(Foo::bar));          // This is, of course, illegal.  (The method I've been told I must use req's the size of the target field.)
}

Creating the struct on the stack should be fast and cheap, but I suspect I'll get dinged for it in a code review, so I'm looking for a better way that doesn't introduce an add'l maintenance burden (such as #defines for sizes).

Answer 1

You can use an expression such as:

sizeof Foo().bar

As the argument of sizeof isn't evaluated, only its type, no temporary is actually created.

---

If Foo wasn't default constructible (unlike your example), you'd have to use a different expression such as one involving a pointer. (Thanks to Mike Seymour)

sizeof ((Foo*)0)->bar