What happens to auto_ptr as a vector element

Because of the device limitations, I am now testing under Windows, and all of the following code has been run in vs2013

Here is Example 1:

#include <stdio.h> #include <iostream> #include <memory> #include <vector>using namespace std; Class D{public:d (): D (1) {}~d () {printf ("D destruction\n");  } int D;}; void Autoptrinvec () {vector<auto_ptr<d> > auto_vec;for (int i = 0; i < i++) Auto_vec.push_back (auto_ptr& Lt;d> (new D)); for (int i = 0; i <; i++) printf ("Auto_vec[%d]->d:%d\n", I, auto_vec[i]->d);} int _tmain (int argc, _tchar* argv[]) {Autoptrinvec (); GetChar (); return 0;}

The result of the operation is:

auto_vec[0]->d:1auto_vec[1]->d:1auto_vec[2]->d:1auto_vec[3]->d:1auto_vec[4]->d:1auto_vec[5]- >D:1AUTO_VEC[6]->D:1AUTO_VEC[7]->D:1AUTO_VEC[8]->D:1AUTO_VEC[9]->D:1D DestructionD DestructionD Destructiond destructiond destructiond Destructiond destructiond Destructiond destructiond Destruction

Results analysis: From the above, auto_ptr as a vector element in the above example does not appear error, and does not like the online part of the article said that the compilation will be wrong (compilation error I think may be the result of different compiler).

So, does the above example indicate that auto_ptr can be an element of a vector? The answer is no, see the following example:

#include <stdio.h> #include <iostream> #include <memory> #include <vector>using namespace std;  Class D{public:d (): D (1) {}~d () {printf ("D destruction\n"); }int D;}; void Funauto (auto_ptr<d> D) {printf ("do nothing\n");} void Autoptrinvec () {vector<auto_ptr<d> > auto_vec;for (int i = 0; i < 3; i++) Auto_vec.push_back (AUTO_PTR&L T;d> (new D)); for (int i = 0; i < 3; i++) printf ("Auto_vec[%d].get ():%p\n", I, Auto_vec[i].get ()); for (int i = 0; I & Lt 3; i++) Funauto (Auto_vec[i]), for (int i = 0; i < 3; i++) printf ("Auto_vec[%d].get ():%p\n", I, Auto_vec[i].get ());p rintf ("\ n "); Auto_vec.clear (); for (int i = 0; i < 3; i++) Auto_vec.push_back (auto_ptr<d> (new D)), for (int i = 0; i < 3; i++) printf ("Auto_vec[%d].get ():%p\n", I, Auto_vec[i].get ()); for (int i = 0; i < 3; i++) auto_ptr<d> temp = Auto_v ec[i];for (int i = 0; i < 3; i++) printf ("Auto_vec[%d].get ():%p\n", I, Auto_vec[i].get ());} int _tmain (int argc, _tchar* argv[]) {AutoptrinVec (); GetChar (); return 0;} 

The results of the operation are as follows:

Auto_vec[0].get (): 0041c930auto_vec[1].get (): 0041c9b0auto_vec[2].get (): 0041c970do nothingd Destructiondo NothingD Destructiondo nothingd destructionauto_vec[0].get (): 00000000auto_vec[1].get (): 00000000auto_vec[2].get (): 00000000auto_vec[0].get (): 0041c930auto_vec[1].get (): 0041c970auto_vec[2].get (): 0041c9b0d DestructionD Destructiond destructionauto_vec[0].get (): 00000000auto_vec[1].get (): 00000000auto_vec[2].get (): 00000000

Based on the result analysis: The auto_ptr element in the vector is set to NULL for the value of the member variable _myptr in the original auto_ptr in the process of passing the value (calling the copy constructor of the Auto_ptr Class) and assigning (calling the assignment operator of the Auto_ptr class). In this case the original auto_ptr is invalid (about how the whole process of failure can go directly to the Auto_ptr source code, here is not analyzed), can no longer be accessed, the vector of auto_ptr elements in these two processes will be invalidated.

the C + + standard says: "STL elements must have copy construction and assignable ...", it is clear from the above process that auto_ptr as a vector element obviously can not achieve this condition, so in any case it is best not to put Auto_ PTR, as an element of a vector, may silently fail in the process of invoking a function such as swap, sort, and so on.

What happens to auto_ptr as a vector element