在c ++中将指针解引用到引用中

Question

In the following code snippet, are there any caveats that need to be considered?

void do_stuff()
{
    std::unique_ptr<my_type> my(new my());

    my_type& mine = *my;

    //use mine 
}

It gives me a weird feeling to 'bring out' the value of my by dereferencing the implicit pointer, as it feels like it would create a temporary (but I'm quite sure that's not the case)

Answer 1

The code is fine with the usual caveats when holding a reference (or pointer) to an object: you have to make sure that the object outlives the uses of the pointer or reference.

In this particular case, make sure that the object inside the unique_ptr does not get deleted by a call to reset() or delete my.release(); which I think are the only cases where the lifetime of that reference might outlive the lifetime of the object itself.

Side note: I am assuming that you have a good reason to use dynamic allocation for that object. If you are not sure, consider not using pointers/new in the first place.

I am also assuming that the reference is not there for performance , if it is, you might want to consider as access through the reference or the smart pointer will be translated to exactly the same binary.

Answer 2

除了没有多大意义（至少在代码片段中）代码是正常的，如果在构造my和引用的赋值之间确实没有任何内容。

Answer 3

Make sure that the pointer outlives the (use of the) reference.

One simple way to do that is to use const :

void doStuff()
{
    auto const p    = std::unique_ptr<MyType>( new MyType );
    MyType&    m    = *p;

    // use m
}

Of course, if there is not any special reason to prefer dynamic allocation (such as size, or special allocator) then just use automatic allocation for the MyType object, ie declare it directly as a local variable without using the word static .