C++: warning: '…' declared with greater visibility than the type of its field '…::'

Question

I\'m getting these two warnings (with GCC 4.2 on MacOSX):

/Users/az/Programmierung/openlierox/build/Xcode/../../src/main.cpp:154:0 /Users/az/Programmierung/openliero

Answer 1

It's because you forgot to declare the inheritance as public.

    struct MainLockDetector : public Action {
         bool wait(Uint32 time) { /* ... */ }
         int handle() { /* ... */ }
    };

This causes the "Action" members to be private. But, you've just overridden an Action private member as public (public default in a struct), which could break encapsulation, hence the warning.

Answer 2

To fix this problem, try one of below.

1. Use #pragma GCC visibility push() statement like this.

   #pragma GCC visibility push(hidden)
   struct MainLockDetector : Action {
        bool wait(Uint32 time) { /* ... */ }
        int handle() { /* ... */ }
   };
   #pragma GCC visibility pop
   

2. Use __attribute__ ((visibility("hidden"))) like this.

   struct __attribute__ ((visibility("hidden"))) MainLockDetector : Action {
        bool wait(Uint32 time) { /* ... */ }
        int handle() { /* ... */ }
   };
   

3. Add the command line option -fvisibility=default.

For more details, refer http://gcc.gnu.org/wiki/Visibility.

Answer 3

There seems to be a problem with shared libraries. I assume you are writing a shared library. Look at this explanation. Try to add the command line option -fvisibility-inlines-hidden. The problem seems to be, that gcc tries to export some symbols of MainLockDetector (visible to linking executables and libraries), whereas Action is not exported (invisible to linking executables and libraries). So the visibility of MainLockDetector really is higher than the visibility of Action.