Re: destruction of already destructed pointer variable when copying
an object - abort error
On Sep 10, 11:34 pm, suresh <suresh.amritap...@gmail.com> wrote:
Hi,
Kindly consider the code segment below: I have a function object
containing a pointer variable which is passed to min_element
algorithm. My problem is the pointer variable is deleted twice and I
do not know how to fix this issue.
class FO{
public:
set<int>::size_type size(){return s->size()}
bool operator()(int a, int b);
~FO();
private:
set<int> * s;
};
FO::FO(){
s = new set<int>;
}
FO::~FO(){
delete s;
}
bool FO::operator()(int a,int b){
s->insert(a);
return (a<b);
}
int main(){
vector<int> v;
//vector populated
FO fo;
min_element(v.begin(),v.end(),fo);
}
The variable 's' is defined as a pointer bcz min_element algorithm
takes a copy of its function object argument. Now inside the
min_element algorithm, the copy of 'fo' is deleted which results in
freeing of the memory associated with 's'. But in the main, the
original object fo is destructed and then also the same memory is
freed and this gives a abort error.
How to solve this kind of a problem?
thanks
suresh
solution no. 89 to your question:
/*
* func_obj.cpp
*
* Created on: Sep 11, 2010
* Author: Gill
*/
#include <iostream>
#include <iterator>
#include <set>
#include <vector>
#include <algorithm>
#include <tr1/functional>
#include <cstdlib>
struct func_obj : public std::binary_function< int, int, bool > {
typedef std::set< int >::const_iterator set_citerator_t;
public:
bool operator( )( int a, int b ) {
s_.insert( a );
return ( a < b );
}
std::set< int >::size_type size( ) {
return s_.size( );
}
set_citerator_t begin( ) const {
return s_.begin( );
}
set_citerator_t end( ) const {
return s_.end( );
}
private:
std::set< int > s_;
};
int main( ) {
std::vector< int > v;
//vector populated
std::generate_n( back_inserter( v ), 10, std::rand );
func_obj fo;
std::min_element( v.begin( ), v.end( ), std::tr1::ref( fo ) );
std::copy( fo.begin( ),
fo.end( ),
std::ostream_iterator< int >( std::cout, " " ) );
std::cout << std::endl;
return 0;
}