CODEWITHSUNDEEP
c++cgcc
kjh
kjh

Reputation: 3411

Why don't functions marked as "hidden" throw an error when called from external modules in gcc 4?

I'm using gcc. 4.4.7. If I mark a function as hidden in module A.h:

//module A.h
__attribute__ ((visibility ("hidden"))) void foo() { ... }

I can obviously still call foo from within module A.h. If I have a second module B.h that references A::foo:

//module B.h
#include "A.h"
foo();

Why isn't an error thrown? If A::foo is technically undefined in module B?

edit removed "nothing happens when foo is called in module B". had some code set up wrong

edit2 I'm looking at this tutorial for setting up local references in a header file so that they can't be called from other modules. Maybe I'm missing something?

edit3 I am compling with -fvisibility=hidden

Upvotes: 0

Views: 97

Answers (2)

Deduplicator
Deduplicator

Reputation: 45674

Quote from the GCC Manual:

visibility ("visibility_type")

This attribute affects the linkage of the declaration to which it is attached. There are four supported visibility_type values: default, hidden, protected or internal visibility.

[...]

hidden

Hidden visibility indicates that the entity declared has a new form of linkage, which we call “hidden linkage”. Two declarations of an object with hidden linkage refer to the same object if they are in the same shared object.

Two compilation units which are linked together are in the same shared object.

Upvotes: 3

samsun96
samsun96

Reputation: 357

Are you compiling this code with the -fvisibility=hidden flag? If so, then try

#pragma GCC visibility push(hidden)

Upvotes: 0

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