c++ - Can I use C++14 in a library meant for C++11 client applications? -

if i'm building library assume "clients" of library may using c++11, can compile library using c++14 internals? there api/abi/link compatibility issues versus c++11? safe implement , build library c++14 long avoid new features in public api, , if so, must avoid? or inherently incompatible mix c++11 , c++14 within final software project?

it's cross-platform library, btw, i'll need build on linux, osx, , windows.

if i'm building library assume "clients" of library may using c++11, can compile library using c++14 internals?

yes, in general should possible.

i gcc's implementation of filesystem ts. <experimental/filesystem> header written in pure c++11 , can included c++11 clients, implementation in libstdc++fs.a written in c++14.

are there api/abi/link compatibility issues versus c++11?

there no changes between c++11 , c++14 require implementations break link-time compatibility. doesn't mean implementations won't break it, aren't required to.

for gcc believe c++11 , c++14 entirely api , abi compatible, except constexpr , sized-deallocation issues mentioned below.

is safe implement , build library c++14 long avoid new features in public api, , if so, must avoid?

it depends on compiler, in theory it's possible.

obviously avoid c++14 language features aren't valid in c++11 (such function return type deduction, or generic lambdas auto parameters, or variable templates) , c++14 library entities, std::make_unique, std::integer_sequence, orstd::shared_timed_mutex`.

a list of changes in c++14 can found in sd-6.

one thing watch out meaning of non-static constexpr member function changed between c++11 , c++14. in c++11 member function const:

struct x {   constexpr int foo(); }; 

in c++14 non-const. compatible both c++11 , c++14 should explicitly qualify const:

struct x {   constexpr int foo() const; }; 

that means same thing in both c++11 , c++14.

another caveat in c++11 , c++14 operator means different:

void operator delete(void*, std::size_t); 

if c++11 client code defines function library compiled in c++14 could end calling instead of usual operator delete(void*) , presumably wrong thing. probably uncommon , not problem in real code, it's possible. g++ , clang allow compile c++14 code -fno-sized-deallocation disable new feature, c++14 library code never call version of operator delete.