Forward declaration vs include in Google c++ style guide

Question

Google c++ style has the following. I don't understand why forward declaration will call f(void*).

It can be difficult to determine whether a forward declaration or a full #include is needed. Replacing an #include with a forward declaration can silently change the meaning of code:

 // b.h: struct B {}; struct D : B {}; // good_user.cc: #include "b.h" void f(B*); void f(void*); void test(D* x) { f(x); } // calls f(B*)

If the #include was replaced with forward decls for B and D, test() would call f(void*).

If you just do struct D; then the inheritance is not known. — Some programmer dude
– Some programmer dude, Commented May 15, 2018 at 15:19

François Andrieux · Accepted Answer · 2018-05-15 15:21:00Z

Consider the two cases individually. By replacing the #include with the header's content, we first have :

struct B {}; struct D : B {}; void f(B*) {} void f(void*) {} void test(D* x) { f(x); }

There are two possible overloads. Since D inherits from B then D* is implicitly convertible to B*. Out of B* and void* the first is a better match so that overload is chosen. But in the case of a forward declared B and D :

struct B; struct D; void f(B*) {} void f(void*) {} void test(D* x) { f(x); }

D is not known to inherit from B so there is no implicit conversion possible from D* to B*. The only matching overload is f(void*).

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