C++ deleting a pointer

Question

On this page, it's written that

One reason is that the operand of delete need not be an lvalue. Consider:

delete p+1; 
delete f(x); 

Here, the implementation of delete does not have a pointer to which it can assign zero.

Adding a number to a pointer shifts it forward in memory by those many number of sizeof(*p) units.

So, what is the difference between delete p and delete p+1, and why would making the pointer 0 only be a problem with delete p+1?

Answer 1

The reason is that neither of those are lvalues so setting them to NULL (or any other value) is simply not possible.

Answer 2

p and p+1 point to different places, as you rightly said sizeof(*p) units apart. You can't delete something that hasn't been allocated, but for example:

A* p = new A();
p++;
delete p-1;

would delete the original allocation. Deleting p+1 when p+1 wasn't allocated originally is undefined; glib flips out with:

*** glibc detected *** free(): invalid pointer: 0x0804b009 ***

The implementation can't zero p+1 because p+1 isn't a variable to modify. The paragraph was saying that

delete p;

could be translated to

free(p);
p = 0;

That doesn't make sense with p+1

Answer 3

You can't do p + 1 = 0. For the same reason, if you do delete p + 1 then delete cannot zero out its operand (p+1), which is what the question on Stroustrup's FAQ is about.

The likelihood that you'd ever write delete p+1 in a program is quite low, but that's beside the point...

Answer 4

The section is talking about why delete() doesn't set the value of pointer variable to zero.

I'm not sure I can explain it better than the article. p != p+1.