CODEWITHSUNDEEP
c++listiterator
skypower
skypower

Reputation: 159

List iterator erase "Debug assertion failed"

I have 2 std::lists. I want to delete all the items from list 1 and insert it in the 2nd and vice versa. My code is not working (getting access violation and "list iterator not dereferencable")

for ( std::list<Item *>::iterator it = list1.begin(); it != list1.end(); ++it ) {
        it = list1.erase( it ); 
        list2.push_back( *it ); 
    }
it = list1.begin();
it = list2.erase( it ); // because the last element is not deleted in the above loop
list2.push_back( *it );

Symmetrical code for the 2nd method. I manage to transfer the items between the 2 lists one time, but the next I get the error.

Any help?

Upvotes: 1

Views: 324

Answers (2)

qPCR4vir
qPCR4vir

Reputation: 3571

Of course you have to use list::swap. But your code show you have some misunderstanding. 

for ( std::list<Item *>::iterator it = list1.begin(); it != list1.end(); ++it ) {
   it = list1.erase( it ); // this effectively erase and destroy *it, 
                 // and erase() return an iterator to the NEXT element. 
                 // Now it=it+1
   list2.push_back( *it ); // you copy the NEXT element!!  
   // here is where we efectively get the ++it of the 'for'. 
   // When erase was used when ‘it’ was at end()-1, erase return end()
   // The attempt to do it=end()+1 is an error probably detected by an assertion. 
}

If list1 have originally even number of elements, for example 0,1,2,3,4,5,6,7,8,9 the iterator end() will point to the inexistent 10, with you don’t need (cant) erase. This ‘for’ will delete the even elements (0,2,4,6,8) , and copy to list2 the odd (1,3,5,7,9). But if originally list1 have odd elements, for example 0,1,2,3,4,5,6,7,8 the last deleted is 8, erase return an iterator to the inexistent 9 = end(), and the ‘for’ make an attempt to increment it with don’t pass the assertion.

Upvotes: 0

juanchopanza
juanchopanza

Reputation: 227468

This is easily and efficiently done with std::list's swap member function:

list1.swap(list2);

This has constant time complexity.

Upvotes: 3

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