c++ - Evaluation order concerning operator -> and () -

consider code snippet:

std::shared_ptr<...> p = ...; p->f(std::move(p)); 

according cppref, operator -> , () of same precedence , both left-to-right associative. presume p-> evaluated before std::move(), , snippet should fine. vc15 opposite , crashes program. vc15 bug or have got wrong here?

you're invoking undefined behaviour, because order of evaluation of arguments in function call not specified standard, , with...

p->f(std::move(p)); 

you have two arguments being passed. first of all, this, , secondly, std::move(p). now, there's not-so-obvious problem. you're both reading and writing same shared pointer. can lead anything, otherwise known undefined behaviour. let's not mention you're using std::move() on such pointer...

*this pointing *p, first (explicit) argument holds pointer itself, *p not valid! let's suppose didn't crashed program (somehow...). then, can lead object being destroyed f left. but... when shared pointer (assuming single reference left) , object destroyed, function has not still left. again, problems...

that example, , have happened in middle of anything. don't read and write in same expression , you'll okay :).

btw: we know vs standards-complaint, anyway, has nothing order , precedence of operator resolution.

edit: okay if f() takes universal reference (a.k.a: forwarding reference, a.k.a collapsed-rvalue-reference-black-magic-that-somehow-got-into-the-standard) or rvalue reference. so, example...

void f(std::shared_pointer<...> &&ptr) 

everything go beatiful , perfect, _{as long don't bad things ptr...}

but, if f takes object of type std::remove_reference_t<decltype(*p)>>, shared pointer moved on argument , above stuff happens.