double free or corruption when deleting reference

Question

I have a class with some basic constructor. The code in main

 C1 g = *new C1(2);
 delete &g;

leads to error:

double free or corruption

Isn't this code equivalent to

C1 *g = new C1(2);
delete g;

What is that thing, that I don't understand.?

You meant to write `C1& g = *new C1(2)`, because otherwise there is no reference. But of course it's a horrible idea to use references this way. — Sebastian Redl, Jun 05 '15 at 15:06

score 1 · Accepted Answer · answered Jun 05 '15 at 09:24

Assuming this is C++, your first snippet is equivalent to

C1 *p = new C1(2);
C1 g = *p;
delete &g;

p is a pointer and lies on the stack.
The location p points to lies on the heap.
g is a C1 and lies on the stack.
&g (the value passed to delete) therefore returns a memory address from somewhere in the stack segment.

Now, if I'm not mistaken:
Variables on the stack are allocated and deallocated automatically, so while your delete &g is not itself causing the error, it is the automatic deallocation at the end of the function that triggers it.
Also, g is a copy of the value pointed to by p, and not a reference to it or something.
Plus p does not get deallocated at all.

Now, in your second snippet:

C1 *g = new C1(2);
delete g;

g is a pointer and lies on the stack.
The location pointed to by g lies on the heap. g (the value passed to delete) therefore returns a memory address from somewhere in the heap segment, and everyone is happy.

double free or corruption when deleting reference

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