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

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

...

sizeof(Foo::BarType)

You can use sizeof of with a pointer to the structure. Consider something like the following:

#include <stdio.h>

typedef struct Foo {
    char         cbar;
    short        sbar;
    int          bar;
    bool         baz;
    long long    llbar;
} Foo;




int main (void)
{
    struct Foo    *p_foo = 0;

    printf("Size of cbar: %d\n", sizeof(p_foo->cbar));
    printf("Size of sbar: %d\n", sizeof(p_foo->sbar));
    printf("Size of bar: %d\n", sizeof(p_foo->bar));
    printf("Size of baz: %d\n", sizeof(p_foo->baz));
    printf("Size of llbar: %d\n", sizeof(p_foo->llbar));
}

Which gives results such as:

163> size.exe
Size of cbar: 1
Size of sbar: 2
Size of bar: 4
Size of baz: 1
Size of llbar: 8

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