Does source incompatibility always imply binary incompatibility?

Question

Any examples demonstrating where source compatibility is broken yet binary compatibility is maintained is welcome.

Answer 1

Old version:

struct inner {
  int bar;
}

struct foo {
  struct inner i;
};

void quux(struct foo *p);

New version:

struct inner2 {
  int bar;
};

struct foo {
  struct inner2 i;
};

void quux(struct foo *p);

Broken code:

struct foo x;
struct inner *i = &x.i;
i->bar = 42;
quux(&x);

Since the only difference is the name of the struct, and the inner struct's type name is erased during compilation, there's no binary incompatibility.