std::unique_ptr<int> p1(new int);
std::unique_ptr<int> p2(new int);
p2=p1;
It seems here that p1 is no longer "unique" since p2 refer to it also
It is legal c++ ? Does unique_ptr have copy_semantics ? If no, and if it has only move semantics, is p1 set to NULL after assign it to p2 ?
EDIT:
ok so the correct version is
p2=std::move(p1)
According to that, after this assign, p1 is not valid ? And the difference with auto_ptr is here? it is more safe to explictly specfiy transfer of ownership than implicitly as it is the case with auto_ptr I guess
std::unique_ptr is non-assignable and non-copyable. You need to use std::move();
so
p1 = std::move(p2);
Have a look here for more info.
Here is an article I wrote which answers your questions. I originally wrote this article to present an emulation of unique_ptr. However you can ignore the first few paragraphs dealing with the emulation and just start reading at "Basic Examples".
http://howardhinnant.github.io/unique_ptr03.html
Edit:
I had trouble distilling the above linked article down to something small enough to make a practical answer in this format. However here is my best shot:
The reason: Safety in generic code. One can not really make copies of either
auto_ptr
orunique_ptr
. Consider:template <class T> void foo(T t) { T copy_of_t = t; // line 4 assert(copy_of_t == t); }
It is not unusual at all for generic code to look like
foo
above. Theassert
is probably not actually there, but the assumption that theassert
would hold often is there ... implicitly . Indeed, a popular implementation ofstd::sort
had exactly this logic in 1996, which is exactly what prompted the secondauto_ptr
redesign (which helped, but didn't completely fix the problem).
按照此 , p2=p1
是编译错误。
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