This question already has answers here:
C++ deleting a pointer when there are 2 pointers pointing to the same memory locations
(4 answers)
Closed 3 years ago.
Class example
{
};
int main()
{
Example* pointer1 = new example();
Example* pointer2;
pointer2 = pointer1;
delete pointer1;
}
Should I delete pointer2? I think it's in the stack and I don't need to delete.
Short answer: No.
pointer1 and pointer2 are both pointers which exist on the stack. new Example allocates on the heap a new object of type Example, and its memory address is stored in pointer1. When you do delete pointer1, you are freeing the memory allocated on the heap. Since both pointer1 and pointer2 are both referencing the same memory location at the point of the delete call, it does not need to also be deleted, in fact, this would be Undefined Behaviour, and could cause heap corruption or simply your program to crash.
At the end of this, both pointer1 and pointer2 are still pointing to the same block of memory, neither are actually nullptr.
Deleting a pointer is telling the operating system that the memory at the location of that pointer is not needed anymore by the program. Remember that a pointer is just an integer that points to place in your RAM. By doing pointer2 = pointer1, you're only copying the integer and you're not moving any memory around. Therefore, by deleting the first pointer, because the second pointer points to the same location, you don't need to delete it.
Related
I know this is pretty common question, but still new for me!
I don't understand concept of dangling pointer, was googling around, and writing test methods to find one.
I just wonder is this a dangling pointer? As whatever example I found was returning something, here I'm trying something similar!
Thanks!
void foo(const std::string name)
{
// will it be Dangling pointer?!, with comments/Answer
// it could be if in new_foo, I store name into Global.
// Why?! And what is safe then?
new_foo(name.c_str());
}
void new_foo(const char* name)
{
// print name or do something with name...
}
A dangling pointer is a pointer that points to invalid data or to data which is not valid anymore, for example:
Class *object = new Class();
Class *object2 = object;
delete object;
object = nullptr;
// now object2 points to something which is not valid anymore
This can occur even in stack allocated objects:
Object *method() {
Object object;
return &object;
}
Object *object2 = method();
// object2 points to an object which has been removed from stack after exiting the function
The pointer returned by c_str may become invalid if the string is modified afterwards or destroyed. In your example you don't seem to modify it, but since it's not clear what you are going to do with const char *name it's impossible to know it your code is inherently safe or not.
For example, if you store the pointer somewhere and then the corresponding string is destroyed, the pointer becomes invalid. If you use const char *name just in the scope of new_foo (for example, for printing purposes) then the pointer will remain valid.
A dangling pointer is a (non-NULL) pointer which points to unallocated (already freed) memory area.
The above example should be correct given that the string is not modified through new_foo.
Taken from here. Although, even if this is for C, it is the same for C++.
Dangling Pointer
When a pointer is pointing at the memory address of a variable but after some time that variable is deleted from that memory location while the pointer is still pointing to it, then such a pointer is known as a dangling pointer and this problem is known as the dangling pointer problem.
Initially
Later
Example
#include<stdio.h>
int *call();
int main() {
int *ptr;
ptr = call();
fflush(stdin);
printf("%d", *ptr);
return 0;
}
int * call() {
int x=25;
++x;
return &x;
}
Its output will be garbage because the variable x is a local variable. Its scope and lifetime are within the function call hence after returning the address of x variable x becomes dead and the pointer is still pointing to that location.
As a matter of style, I explain a dangling pointer as "a pointer which still exists, even though the object it pointed to no longer exists".
In your case, the pointer name exists for a shorter period that the object that it points to. So it's never dangling.
Inside common C++ classes, pointers dangle for a very short period, inside destructors. That's because the delete statement is before the last } of the destructor, while the pointer itself ceases to exist at the last }. If you don't want to worry about this, use e.g. unique_ptr<T>. The T* pointer will dangle for a very short time inside the unique_ptr::~unique_ptr destructor, which is perfectly safe.
Dangling pointers is a situation where you have valid pointers in the stack, but it is pointing to invalid memory. You might end up in this situation when you deallocate the heap memory before the pointers in stack deallocated.
This is a security issue. Because when you deallocate a memory, we are informing Operating System, that we no longer need this section of memory. So OS will mark that piece of memory as ready to allocate and allocate to other applications when they request for memory.
Usually, in C++, memory allocated and deallocated through a general pattern. Constructor in a class gets invoked when a class initialised and this is the right place to allocate memory in heap.Destructor will be invoked when the class instance goes out of scope, and this is the right place to deallocate memory from heap. Assume we already created a class that does allocation and deallocation of memory in constructor and destructor respectively.
int main() {
SomeClass pointer1 = SomeClass();
SomeClass pointer2 = pointer1;
}
In the above example code, there are two variables declared but both holding the same value. When the constructor invoked, it allocates a heap memory. Then we are declaring one more variable and assigning the same value. In C++ usually, when you assign a value of complex type, it does a shallow copy (unless you explicitly implemented copy constructor) instead of deep copy. That means the only pointer gets copied in Stack, but not the heap memory. Actually it is not recommended to copy heap memory for performance reasons. Now the final memory layout looks like that we have two pointers pointing to the same heap memory.
Now when the function is done with execution, local variables goes out of scope and it invokes destructor. First, pointer2 invokes destructor that deallocates the heap memory. At this point, pointer1 becomes dangling pointer. It points to a memory that is already deallocated.
From this example, we understood that the primary cause of dangling pointer is having multiple owners for the same resource. Because when one pointer deallocates memory other pointers became dangling pointers.
//Declaring two pointer variables to int
int * ptr1;
int * ptr2;
// Allocating dynamic memory in the heap
ptr1 = new int;
ptr2 = ptr1; // Having both pointers to point same dynamic memory location
//deleting the dynamic memory location
delete ptr1;
ptr1 = nullptr;
//ptr2 is still pointing the already deleted memory location
//We call ptr2 is a dangling pointer
Dangling Pointer and dangling pointer problem
If any pointer is pointing the memory address of any variable but after some variable has deleted from that memory location while pointer is still pointing such memory location.
That pointer is called as dangling pointer and the problem that arises at that time is called as dangling pointer problem.
Here are some examples: Dangling Pointer and dangling pointer problem
This question already has answers here:
What's the difference between delete-ing a pointer and setting it to nullptr? [duplicate]
(5 answers)
Closed 6 years ago.
Suppose I have a pointer to MyClass:
MyClass *myPointer = new MyClass();
What is the difference between delete myPointer; and myPointer = NULL;?
Thanks
delete myPointer de-allocates memory, but leaves value of myPointer variable, that it is pointing to some garbage address.
myPointer = NULL only sets value of myPointer to null-pointer, possibly leaking memory, if you don't delete other pointer, which points to same address as myPointer
In ideal world you should use both those strings, like this:
delete myPointer; // free memory
myPointer = nullptr; // setting value to zero, signalizing, that this is empty pointer
But overall, use std::unique_ptr, which is modern approach to memory management.
delete myPointer frees the allocated space but lets you with a dangling pointer (one that points to something not allocated).
myPointer = NULL set your pointer to a value that is used to represent the concept of (pointing to nothing) but gives you a memory leak in return as you didn't deallocate a memory which is now "lost". Memory leak may not be too harmful if you don't abuse, but is always considered as a kind of programming error.
You may use the following idiom to prevent future problems:
delete myPointer;
myPointer = NULL;
Briefly, delete is used to deallocate memory for an object previously allocated with the new keyword. The object's destructor is called before the object's memory is deallocated (if the object has a destructor).
Pointers that dereference the deallocated memory (usually) have not a NULL value after calling delete, but any operation on them cause errors.
Setting a pointer to NULL implies that it does not dereference anything, but the memory allocated for your object still persist.
Sometimes could be useful to set pointers to NULL after deleting object they dereference, so that it's possible to check if they are still valid (dereference a consistent memory area) or not.
I know this is pretty common question, but still new for me!
I don't understand concept of dangling pointer, was googling around, and writing test methods to find one.
I just wonder is this a dangling pointer? As whatever example I found was returning something, here I'm trying something similar!
Thanks!
void foo(const std::string name)
{
// will it be Dangling pointer?!, with comments/Answer
// it could be if in new_foo, I store name into Global.
// Why?! And what is safe then?
new_foo(name.c_str());
}
void new_foo(const char* name)
{
// print name or do something with name...
}
A dangling pointer is a pointer that points to invalid data or to data which is not valid anymore, for example:
Class *object = new Class();
Class *object2 = object;
delete object;
object = nullptr;
// now object2 points to something which is not valid anymore
This can occur even in stack allocated objects:
Object *method() {
Object object;
return &object;
}
Object *object2 = method();
// object2 points to an object which has been removed from stack after exiting the function
The pointer returned by c_str may become invalid if the string is modified afterwards or destroyed. In your example you don't seem to modify it, but since it's not clear what you are going to do with const char *name it's impossible to know it your code is inherently safe or not.
For example, if you store the pointer somewhere and then the corresponding string is destroyed, the pointer becomes invalid. If you use const char *name just in the scope of new_foo (for example, for printing purposes) then the pointer will remain valid.
A dangling pointer is a (non-NULL) pointer which points to unallocated (already freed) memory area.
The above example should be correct given that the string is not modified through new_foo.
Taken from here. Although, even if this is for C, it is the same for C++.
Dangling Pointer
When a pointer is pointing at the memory address of a variable but after some time that variable is deleted from that memory location while the pointer is still pointing to it, then such a pointer is known as a dangling pointer and this problem is known as the dangling pointer problem.
Initially
Later
Example
#include<stdio.h>
int *call();
int main() {
int *ptr;
ptr = call();
fflush(stdin);
printf("%d", *ptr);
return 0;
}
int * call() {
int x=25;
++x;
return &x;
}
Its output will be garbage because the variable x is a local variable. Its scope and lifetime are within the function call hence after returning the address of x variable x becomes dead and the pointer is still pointing to that location.
As a matter of style, I explain a dangling pointer as "a pointer which still exists, even though the object it pointed to no longer exists".
In your case, the pointer name exists for a shorter period that the object that it points to. So it's never dangling.
Inside common C++ classes, pointers dangle for a very short period, inside destructors. That's because the delete statement is before the last } of the destructor, while the pointer itself ceases to exist at the last }. If you don't want to worry about this, use e.g. unique_ptr<T>. The T* pointer will dangle for a very short time inside the unique_ptr::~unique_ptr destructor, which is perfectly safe.
Dangling pointers is a situation where you have valid pointers in the stack, but it is pointing to invalid memory. You might end up in this situation when you deallocate the heap memory before the pointers in stack deallocated.
This is a security issue. Because when you deallocate a memory, we are informing Operating System, that we no longer need this section of memory. So OS will mark that piece of memory as ready to allocate and allocate to other applications when they request for memory.
Usually, in C++, memory allocated and deallocated through a general pattern. Constructor in a class gets invoked when a class initialised and this is the right place to allocate memory in heap.Destructor will be invoked when the class instance goes out of scope, and this is the right place to deallocate memory from heap. Assume we already created a class that does allocation and deallocation of memory in constructor and destructor respectively.
int main() {
SomeClass pointer1 = SomeClass();
SomeClass pointer2 = pointer1;
}
In the above example code, there are two variables declared but both holding the same value. When the constructor invoked, it allocates a heap memory. Then we are declaring one more variable and assigning the same value. In C++ usually, when you assign a value of complex type, it does a shallow copy (unless you explicitly implemented copy constructor) instead of deep copy. That means the only pointer gets copied in Stack, but not the heap memory. Actually it is not recommended to copy heap memory for performance reasons. Now the final memory layout looks like that we have two pointers pointing to the same heap memory.
Now when the function is done with execution, local variables goes out of scope and it invokes destructor. First, pointer2 invokes destructor that deallocates the heap memory. At this point, pointer1 becomes dangling pointer. It points to a memory that is already deallocated.
From this example, we understood that the primary cause of dangling pointer is having multiple owners for the same resource. Because when one pointer deallocates memory other pointers became dangling pointers.
//Declaring two pointer variables to int
int * ptr1;
int * ptr2;
// Allocating dynamic memory in the heap
ptr1 = new int;
ptr2 = ptr1; // Having both pointers to point same dynamic memory location
//deleting the dynamic memory location
delete ptr1;
ptr1 = nullptr;
//ptr2 is still pointing the already deleted memory location
//We call ptr2 is a dangling pointer
Dangling Pointer and dangling pointer problem
If any pointer is pointing the memory address of any variable but after some variable has deleted from that memory location while pointer is still pointing such memory location.
That pointer is called as dangling pointer and the problem that arises at that time is called as dangling pointer problem.
Here are some examples: Dangling Pointer and dangling pointer problem
This question already has answers here:
C++ delete - It deletes my objects but I can still access the data?
(13 answers)
Closed 9 years ago.
I'm facing an issue. I wrote the following programme:
void main()
{
int *ptr;
ptr=new int;
cout<<ptr<<endl;
cout<<&ptr<<endl;
delete ptr;
// ptr= NULL;
cout<<ptr<<endl;
cout<<&ptr<<endl;
}
The output of the programme is:
0x00431810
0x0018FF44
0x00431810
0x0018FF44
"&ptr" would definitely be the same because it is ptr's own address. But 1st and third line of output clearly shows that "ptr" is pointing to the same memory location even after the "delete" is called. Why? If this is the case why do we have "delete" and not just rely on NULLING the pointer. I know that more than one pointers can point to a memory location but yeah "delete" is useless. No ? That's so confusing. please help. I've an exam tomorrow.
delete ptr does not change the value of the pointer, but it does free the memory.
Note that dereferencing ptr after freeing the memory would trigger undefined behaviour.
When you call new, you are asking the computer to allocate some memory on the heap and returns a pointer to where that is on the heap.
Calling delete does the opposite in that it tells the computer to free the memory the pointer is pointing to from the heap. This means that it can be reused by your program at a later stage.
However, it does not change the value of the pointer. It is good practise to set the pointers value to NULL. This is because, where the pointer is now pointing to could contain anything (the computer is free to reuse this memory for something else). Were you to use this pointer, it is undefined what will happen.
If you just change the pointer to NULL, then all that does it forget where you allocated the memory. Because C++ is not a managed language, it won't detect you don't have any pointer to that memory, so the memory will become lost as you can't access it (you don't know where it is) but the computer can't reuse it (it doesn't know you are done with it).
If you use a managed language like C# or Java, when there are no pointers to something on the heap, the computer frees it (known as Garbage Collection) but this comes with performance overheads so C++ leaves it up to the programmer.
pointer doesn't know if the memory is valid or corrupted. You can make pointer to point at any address that you can address (that will fit into pointer). For example on my machine pointer has 8 bytes size so I can do
int main(int argc, char** argv) {
int *ptr;
ptr=new int;
cout<<ptr<<endl;
cout<<&ptr<<endl;
delete ptr;
//ptr= NULL;
cout<<ptr<<endl;
cout<<&ptr<<endl;
ptr = (int*) 0xffffffff;
cout<<ptr<<endl; //prints 0xffffffff
ptr = (int*) 0xffffffffffffffff;
cout<<ptr<<endl; //prints 0xffffffffffffffff
ptr = (int*) 0xffffffffffffffff1;
cout<<ptr<<endl; //prints 0xfffffffffffffff1 truncated
ptr = (int*) 0xffffffffffffffff321;
cout<<ptr<<endl; //prints 0xfffffffffffff321 truncated
return 0;
}
When you are using new it will allocate memory at some address on the heap and return you a pointer to this address.
void* operator new (std::size_t size) throw (std::bad_alloc);
Pointer and memory are two distinguished things. In particular you can even specify an address which you want to use for allocation by hand(using placement new syntax):
int* ptr = new (0xff1256) int;
//correct if the memory at 0xff1256 is preallocated
//because placement new only creates object at address
//doesn't allocate memory
"ptr" is pointing to the same memory location even after the "delete"
is called. Why?
In the same way delete ptr only deallocates memory, leaving the pointer intact (this is a good practice though to assign NULL to a pointer immediately after call to delete). It only deallocate the memory, it is not interested in your pointer but in the address where to free memory. Allocating memory or deallocating it is done by operating system (i.e.the memory is being marked as allocated).
From the documentation,
Deallocates the memory block pointed by ptr (if not null), releasing
the storage space previously allocated to it by a call to operator new
and rendering that pointer location invalid.
That means, delete marks the location to be reused by any other process for any other purpose. It never says that, it will initialize the memory block to any random value, so that the earlier stored data can never be accessed again. So, there is always the possibility of having your data even after the call to delete. But, again, it is just one of the possibility of Undefined Behavior. Thats' why coders, make the pointer NULL after deallocating it.
I know this is pretty common question, but still new for me!
I don't understand concept of dangling pointer, was googling around, and writing test methods to find one.
I just wonder is this a dangling pointer? As whatever example I found was returning something, here I'm trying something similar!
Thanks!
void foo(const std::string name)
{
// will it be Dangling pointer?!, with comments/Answer
// it could be if in new_foo, I store name into Global.
// Why?! And what is safe then?
new_foo(name.c_str());
}
void new_foo(const char* name)
{
// print name or do something with name...
}
A dangling pointer is a pointer that points to invalid data or to data which is not valid anymore, for example:
Class *object = new Class();
Class *object2 = object;
delete object;
object = nullptr;
// now object2 points to something which is not valid anymore
This can occur even in stack allocated objects:
Object *method() {
Object object;
return &object;
}
Object *object2 = method();
// object2 points to an object which has been removed from stack after exiting the function
The pointer returned by c_str may become invalid if the string is modified afterwards or destroyed. In your example you don't seem to modify it, but since it's not clear what you are going to do with const char *name it's impossible to know it your code is inherently safe or not.
For example, if you store the pointer somewhere and then the corresponding string is destroyed, the pointer becomes invalid. If you use const char *name just in the scope of new_foo (for example, for printing purposes) then the pointer will remain valid.
A dangling pointer is a (non-NULL) pointer which points to unallocated (already freed) memory area.
The above example should be correct given that the string is not modified through new_foo.
Taken from here. Although, even if this is for C, it is the same for C++.
Dangling Pointer
When a pointer is pointing at the memory address of a variable but after some time that variable is deleted from that memory location while the pointer is still pointing to it, then such a pointer is known as a dangling pointer and this problem is known as the dangling pointer problem.
Initially
Later
Example
#include<stdio.h>
int *call();
int main() {
int *ptr;
ptr = call();
fflush(stdin);
printf("%d", *ptr);
return 0;
}
int * call() {
int x=25;
++x;
return &x;
}
Its output will be garbage because the variable x is a local variable. Its scope and lifetime are within the function call hence after returning the address of x variable x becomes dead and the pointer is still pointing to that location.
As a matter of style, I explain a dangling pointer as "a pointer which still exists, even though the object it pointed to no longer exists".
In your case, the pointer name exists for a shorter period that the object that it points to. So it's never dangling.
Inside common C++ classes, pointers dangle for a very short period, inside destructors. That's because the delete statement is before the last } of the destructor, while the pointer itself ceases to exist at the last }. If you don't want to worry about this, use e.g. unique_ptr<T>. The T* pointer will dangle for a very short time inside the unique_ptr::~unique_ptr destructor, which is perfectly safe.
Dangling pointers is a situation where you have valid pointers in the stack, but it is pointing to invalid memory. You might end up in this situation when you deallocate the heap memory before the pointers in stack deallocated.
This is a security issue. Because when you deallocate a memory, we are informing Operating System, that we no longer need this section of memory. So OS will mark that piece of memory as ready to allocate and allocate to other applications when they request for memory.
Usually, in C++, memory allocated and deallocated through a general pattern. Constructor in a class gets invoked when a class initialised and this is the right place to allocate memory in heap.Destructor will be invoked when the class instance goes out of scope, and this is the right place to deallocate memory from heap. Assume we already created a class that does allocation and deallocation of memory in constructor and destructor respectively.
int main() {
SomeClass pointer1 = SomeClass();
SomeClass pointer2 = pointer1;
}
In the above example code, there are two variables declared but both holding the same value. When the constructor invoked, it allocates a heap memory. Then we are declaring one more variable and assigning the same value. In C++ usually, when you assign a value of complex type, it does a shallow copy (unless you explicitly implemented copy constructor) instead of deep copy. That means the only pointer gets copied in Stack, but not the heap memory. Actually it is not recommended to copy heap memory for performance reasons. Now the final memory layout looks like that we have two pointers pointing to the same heap memory.
Now when the function is done with execution, local variables goes out of scope and it invokes destructor. First, pointer2 invokes destructor that deallocates the heap memory. At this point, pointer1 becomes dangling pointer. It points to a memory that is already deallocated.
From this example, we understood that the primary cause of dangling pointer is having multiple owners for the same resource. Because when one pointer deallocates memory other pointers became dangling pointers.
//Declaring two pointer variables to int
int * ptr1;
int * ptr2;
// Allocating dynamic memory in the heap
ptr1 = new int;
ptr2 = ptr1; // Having both pointers to point same dynamic memory location
//deleting the dynamic memory location
delete ptr1;
ptr1 = nullptr;
//ptr2 is still pointing the already deleted memory location
//We call ptr2 is a dangling pointer
Dangling Pointer and dangling pointer problem
If any pointer is pointing the memory address of any variable but after some variable has deleted from that memory location while pointer is still pointing such memory location.
That pointer is called as dangling pointer and the problem that arises at that time is called as dangling pointer problem.
Here are some examples: Dangling Pointer and dangling pointer problem