C++: Is it safe to pass an argument like unique_ptr::get() to function

Question

Is is safe to pass a function parameter like getAName(getA().get())? getA() return a object unique_ptr<A>.

I test with the whole code below on VS 2010, it works. But I would like to make sure if it's c++ standard, is it safe with other c++ compilers?

#include "stdafx.h"
#include <memory>
#include <iostream>

using namespace std;

class A
{
public:
    A(){ cout<<"A()"<<endl;}
    ~A(){ cout<<"~A()"<<endl;}

    string name() { return "A"; }
};

std::unique_ptr<A> getA()
{
    return std::unique_ptr<A>(new A());;
}

void getAName(A* a)
{
    if(a)
    {
        cout << a->name().c_str() << endl;
    }
}

int _tmain(int argc, _TCHAR* argv[])
{
    getAName(getA().get());
    return 0;
}

The outputs in console are:

A()
A
~()

Is it necessary to make code as below for safe for all compilers?

unique_ptr<A> a = getA();
getAName(a.get());

Answer 1

It is safe. getA() returns a temporary std::unique_ptr, which will be destroyed after the full expression, which constains the invocation of getAName(). So inside the body of getAName() the pointer passed remains valid.

All temporary objects are destroyed as the last step in evaluating the full-expression that (lexically) contains the point where they were created, ...

Note that if the pointer passed in is stored somewhere (e.g. a global variable) and then used later (i.e. after the invocation of getAName()), the object pointed by the pointer has been destroyed by the temporary std::unique_ptr and the pointer becomes dangled; then deference on it would be UB. If this is the case, as you showed, you might need a named variable for it.