Can you mix free and constructor in C++?

Question

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Is there any danger in calling free() or delete instead of delete[]?

I was reading this question:
http://stackoverflow.com/questions/184537/in-what-cases-do-i-use-malloc-vs-new/

Someone raised that one reason to use malloc was if you were going to use free.

I was wondering: Is it valid to mix a free call and a constructor initialization in C++?

i.e.

Can I say:

my_type *ptr = new my_type;
free(my_type);

Is that somehow invalid or worse than:

my_type *ptr = new my_type;
delete my_type;

other than the fact that it's not c++ish?

Likewise, could you do the opposite? Can you say

my_type *ptr = (my_type *)malloc(sizeof(my_type));
delete my_type;

Please merge if this is a duplicate, I searched but didn't see a question along this lines exactly about malloc/delete/new/free asked.

Answer 1

No it is invalid. There is no guarantee that new will use malloc or delete will use free.

Moreover, using free instead of delete will skip my_type's destructor. If my_type itself is holding some resources, those will be leaked. Similarly, malloc will skip the constructor so the variable may be in an invalid state.

Answer 2

Is it valid to mix a free call and a constructor initialization in C++?

No it is not.

malloc,calloc,realloc -> free

new -> delete

new[] -> delete[]

Answer 3

No, this is not valid. malloc (and other C allocation functions) must be matched with free, new must be matched with delete, and new [] must be matched with delete []. While your compiler may not necessarily do anything differently if there is no destructor code, this is not something you should rely on.

The main difference is that new/delete call the constructor and destructor of an object; malloc and free just treat it as raw, untyped memory.

Answer 4

1. free() does not call the destructor. It just deallocates the memory, no questions asked.

2. new/delete is not guaranteed to use malloc() as its memory allocator; free might not even know about the memory you're throwing at it