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

From:
TZhang <tingnan.zhang@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Fri, 25 Jul 2014 08:51:06 -0700 (PDT)
Message-ID:
<8e02f845-69e5-4729-a72e-4ccf858db418@googlegroups.com>
On Friday, July 25, 2014 4:18:38 AM UTC-4, TZhang wrote:

#include <memory>

using namespace std;

shared_ptr<int> tmpfunc1(int* p)

{

    return shared_ptr<int>(p);

}

void tmpfunc2(int * px)

{

    shared_ptr<int> ptr1 = tmpfunc1(px);

}

int main()

{

    int x = 1;

    tmpfunc2(&x);

    return 0;

}

The following code would abort before tmpfunc2() returns (and when ptr1 is destroyed). Can anybody explain what is happening? I think the temporary object returned from tmpfunc1 is causing the problem, but cannot figure out why.

What would be a best practice to return shared_ptr?

Thanks!


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]);
}

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

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