Re: shared_ptr throw error: pointer being freed was not allocated

From:
Marcel Mueller <news.5.maazl@spamgourmet.org>
Newsgroups:
comp.lang.c++
Date:
Fri, 25 Jul 2014 18:51:22 +0200
Message-ID:
<lqu1uc$8gv$1@news.albasani.net>
On 25.07.14 17.51, TZhang wrote:

Thanks Marcel. I was doing this because recently I was working on a code base that is implemented in the following way:

#include<memory>
#include<vector>
using namespace std;

shared_ptr<int> tmpfunc1(int* p)
{
     return shared_ptr<int>(p);
}

void tmpfunc2(int * px)
{
     shared_ptr<int> ptr1 = tmpfunc1(px);
}

class A
{
public:
     vector<int*> ptrvec;
     vector<shared_ptr<int> > shrptrvec;
     A() {};
     ~A() {};
};

int main()
{
     A a;
     for (int i = 0; i< 100; ++i)
     {
         shared_ptr<int> tmpptr(new int);
         a.ptrvec.push_back(tmpptr.get());
         a.shrptrvec.push_back(tmpptr);
     }
     tmpfunc2(a.ptrvec[0]); // <-- BUG !!!
}

The code will still crash though the integers were allocated on heap. ennnnn


This code is also broken. You duplicate the ownership of the 0-th
element. In the last line by invoking the constructor shared_ptr(int*)
again. The conversion from the raw pointer to shared_ptr must be done
exactly once per object.

tmpfunc2 always deletes the object passed to it. But shrptrvec also
deletes all of it's elements at the end of main. The 0-th element is
already deleted so you have undefined behavior.

I repeat myself. Do not use raw pointers (int*) if you deal with
shared_ptr. It is by far too easy to write buggy code this way.

Marcel

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