C++ why the type of parameter of atomic_load is pointer instead of reference?

Question

I agree the answer in When to use references vs. pointers.
But, I am wondering why C++ defines atomic_load as

template<class T>
T atomic_load(const std::atomic<T> *obj) noexcept;
                                   ^

instead of

template<class T>
T atomic_load(const std::atomic<T> &obj) noexcept;
                                   ^

Can anyone help me?

Most likely for uniformity with the [equivalent C interface](http://en.cppreference.com/w/c/atomic/atomic_load). — WhiZTiM, Sep 19 '17 at 12:07
@WhiZTiM But, std::atomic is a class template, why does it require to consistent with C? — Caesar, Sep 19 '17 at 12:08
It is not strictly required, it is a convenience for the programmer who doesn't have to remember yet another pointless difference. You just need to remember that `atomic_load` takes a pointer, regardless of what base type exactly you are passing. — Matteo Italia, Sep 19 '17 at 12:12

score 10 · Answer 1 · edited Sep 19 '17 at 12:52

The reason why we have these free function templates at all is source compatibility with C11:

#ifdef __cplusplus
#include <atomic>
#define _Atomic(X) std::atomic<X>
#else
#include <stdatomic.h>
#endif

_Atomic(int) c;

int get_c(void) { 
    return atomic_load(&c); 
}

C doesn't have references.

If you don't need that, then c.load() or the implicit conversion to T will work just fine. Just forget that the free function ever existed.

(This is also why the memory_order version of the free function template is called atomic_load_explicit: _Generic-powered macros in C can handle varying argument types, but not varying arity.)

C++ why the type of parameter of atomic_load is pointer instead of reference?

1 Answers1