问题
After reading many posts about this, I want to clarify the next point:
A* a = new A();
A* b = a;
delete a;
A* c = a; //illegal - I know it (in c++ 11)
A* d = b; //I suppose it's legal, is it true?
So the question is about using the value of copy of deleted pointer.
I've read, that in c++ 11 reading the value of a leads to undefined behaviour - but what about reading the value of b?
Trying to read the value of the pointer (note: this is different to dereferencing it) causes implementation-defined behaviour since C++14, which may include generating a runtime fault. (In C++11 it was undefined behaviour) What happens to the pointer itself after delete?
回答1:
Both:
A* c = a;
A* d = b;
are undefined in C++11 and implementation defined in C++14. This is because a and b are both "invalid pointer values" (as they point to deallocated storage space), and "using an invalid pointer value" is either undefined or implementation defined, depending on the C++ version. ("Using" includes "copying the value of").
The relevant section ([basic.stc.dynamic.deallocation]/4) in C++11 reads (emphasis added):
If the argument given to a deallocation function in the standard library is a pointer that is not the null pointer value (4.10), the deallocation function shall deallocate the storage referenced by the pointer, rendering invalid all pointers referring to any part of the deallocated storage. The effect of using an invalid pointer value (including passing it to a deallocation function) is undefined.
with a non-normative note stating:
On some implementations, it causes a system-generated runtime
In C++14 the same section reads:
If the argument given to a deallocation function in the standard library is a pointer that is not the null pointer value (4.10), the deallocation function shall deallocate the storage referenced by the pointer, rendering invalid all pointers referring to any part of the deallocated storage. Indirection through an invalid pointer value and passing an invalid pointer value to a deallocation function have undefined behavior. Any other use of an invalid pointer value has implementation-defined behavior.
with a non-normative note stating:
Some implementations might define that copying an invalid pointer value causes a system-generated runtime fault
回答2:
These 2 lines do not have any difference (meaning legality for C++):
A* c = a; //illegal - I know it (in c++ 11)
A* d = b; //I suppose it's legal, is it true?
Your mistake (and it is pretty common) to think if you call delete on a it makes it any different than b. You should remember that when you call delete on a pointer you pass argument by value, so memory, where a points to after delete is not usable anymore, but that call does not make a any different than b in your example.
回答3:
You should not use the pointer after delete. My below example with acessing a is based on implementation-defined behaviour.
(thanks to for M.M and Mankarse for pointing this)
I feel that it is not the variable a (or b, c, d) that is important here, but that the value (=the memory address of a deallocated block) which in some implementations can trigger a runtime fault when used in some 'pointer context'.
This value may be an rvalue/expression, not necessarily the value stored in a variable - so I do not believe the value of a ever changes (I am using the loose 'pointer context' to distinguish from using the same value, i.e. the same set of bits, in non-pointer related expressions - which will not cause a runtime fault).
------------My original post is below.---------------
Well, you are almost there with your experiment. Just add some cout's like here:
class A {};
A* a = new A();
A* b = a;
std::cout << a << std::endl; // <--- added here
delete a;
std::cout << a << std::endl; // <--- added here. Note 'a' can still be used!
A* c = a;
A* d = b;
Calling delete a does not do anything to the variable a. This is just a library call. The library that manages dynamic memory allocation keeps a list of allocated memory blocks and uses the value passed by variable a to mark one of the previously allocated blocks as freed.
While it is true what Mankarse cites from C++ documentation, about: "rendering invalid all pointers referring to any part of the deallocated storage" - note that the value of variable a remains untouched (you did not pass it by reference, but by value !).
So to sum up and to try to answer your question:
Variable a still exists in the scope after delete. The variable a still contains the same value, which is the address of the beginning of the memory block allocated (and now already deallocated) for an object of class A. This value of a technically can be used - you can e.g. print it like in my above example – however it is hard to find a more reasonable use for it than printing/logging the past...
What you should not do is trying to de-reference this value (which you also keep in variables b, c, and d) – as this value is not a valid memory pointer any longer.
You should never rely on the object being in the deallocated storage (while it is quite probable that it will remain there for some while, as C++ does not require to clear the storage freed after use) - you have no guarantees and no safe way to check this).
来源：https://stackoverflow.com/questions/44182049/pointers-in-c-after-delete