How does the below code can access deleted object.I know that explicitly calling a constructor will create temporary object and that object will be deleted immediately

Question

1.00/5 (1 vote)

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How does this statement is able to run successfully

C++

obj = b[0];
cout<<obj.a<<"   "<<obj.b<<"\n";
obj = b[1];
cout<<obj.a<<"   "<<obj.b<<"\n";

//code

Thanks in advance

C++

#include<vector>
#include<iostream>
#include"conio.h"
using namespace std;
class B{
public:
	int a ;
	int b;
public:
	B(){}
	B(int c,int d):a(c),b(d){cout<<"contructor\n";}
	~B(){cout<<"destructor\n";}
};

int main()
{
std::vector<B> b;
B obj;
b.push_back(B(10,20));
b.push_back(B(30,40));
obj = b[0];
cout<<obj.a<<"   "<<obj.b<<"\n";
obj = b[1];
cout<<obj.a<<"   "<<obj.b<<"\n";
getch();
}

Output:

contructor
destructor
contructor
destructor
destructor
10   20
30   40

Posted 17-Jul-15 7:05am

krish0969

Updated 18-Jul-15 4:08am

Frankie-C

v6

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Comments

PIEBALDconsult 17-Jul-15 14:08pm

Because it C/C++ and someday you may find that you have blown your leg off.

nv3 17-Jul-15 17:36pm

The line "std::vector b;" tells me that you are not showing us the exact code of your example. 'b' is not a type. Hence that statement would not compile. I assume you meant to write: "std::vector b;".

And what is so unusual about the output? Can you elaborate a little more what you actual question is.

[no name] 17-Jul-15 20:52pm

Not a question yet.

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User 59241 · Accepted Answer · 2015-07-17T20:14:00

Your interpretation is wrong. Calling push_back() makes a copy of the stack object to insert in the vector.
This creates an object with automatic storage duration:

C++

b.push_back(B(10,20));

which immediately goes out of scope. The destructor is called on the stack object but a new copy exists in the vector.
http://www.cplusplus.com/reference/vector/vector/[^]