Does std::vector.clear() do delete (free memory) on each element?


Consider this code:

#include <vector>

void Example()
    std::vector<TCHAR*> list;
    TCHAR* pLine = new TCHAR[20];
    list.clear();    // is delete called here?
    // is delete pLine; necessary?

Does list.clear() call delete on each element? I.e. do I have to free the memory before / after list.clear()?

2/27/2009 9:26:37 AM

Accepted Answer

No (you need to do the delete yourself at the end as you suggest in your example as the destruction of the bald pointer doesnt do anything). But you can use a boost [or other RAII-based idiom] smart pointer to make it Do The Right Thing (auto_ptr would not work correctly in a container as it has incompatible behaviour under copying etc.), but be sure you understand the pitfalls of such smart pointers before use. (As Benoit mentions, in this case, basic_string is what you're really looking for here.)

Having said that there's a need to understand the pitfalls of smart pointers, having them take care of the memory management implicitly so you dont have to do it explicitly is far less error-prone.

EDIT: Substantially revised to encompass the elements Benoit brought into his far more thorough answer, thanks to strong prodding from the Earwicker and James Matta - thanks for pushing me to do the due diligence on this!

2/27/2009 10:48:08 PM

std::vector does call the destructor of every element it contains when clear() is called. In your particular case, it destroys the pointer but the objects remain.

Smart pointers are the right way to go, but be careful. auto_ptr cannot be used in std containers. boost::scoped_ptr can't either. boost::shared_ptr can, but it won't work in your case because you don't have a pointer to an object, you are actually using an array. So the solution to your problem is to use boost::shared_array.

But i suggest you use std::basic_string instead, where you won't have to deal with memory management, while still getting the benefits of working with a string.

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