Deleting dynamically allocated variables setting pointer to 0 [duplicate] - c++

This question already has answers here:
Is it good practice to NULL a pointer after deleting it?
(18 answers)
Closed 5 years ago.
I can't understand the end of this code (array = 0;):
#include <iostream>
int main()
{
std::cout << "Enter a positive integer: ";
int length;
std::cin >> length;
int *array = new int[length];
std::cout << "I just allocated an array of integers of length " << length << '\n';
array[0] = 5; // set element 0 to value 5
delete[] array; // use array delete to deallocate array
array = 0; // use nullptr instead of 0 in C++11
return 0;
}
At the end, a dynamically allocated array is deleted (returned to OS) and then assigned a value of 0.
Why is this done? After array has been returned to the OS, there is no need to assign it a value of 0, right?
Code from: http://www.learncpp.com/cpp-tutorial/6-9a-dynamically-allocating-arrays/

After array has been returned to the OS, there is no need to assign it a value of 0, right?
You're right it is not needed because the memory is freed (deallocated) by the operator delete. But think of a case where you may use the pointer in another place in your code (functions, loops, etc.) after you use delete[] on it.
The array variable still holds the address of the old allocation after the delete[] statement was called (dangling pointer). If you would access that address you would get undefined bahaviour (UB) because the memory is no longer yours, in most of the cases your program would crash.
To avoid that you do a null pointer check like:
if (array != nullptr)
{
/* access array */
...
}
which is checking the pointer against the address 0 which represents an invalid address.
To make that check possible you set the pointer to nullptr or NULL if C++11 is not available. The nullptr keyword introduces type safety because it acts like a pointer type and should be preferred over the C-like NULL. In pre C++11 NULL is defined as integer 0, since C++11 it's an alias to nullptr.
To define your own nullptr to use it for pre C++11 compiler look here: How to define our own nullptr in c++98?
An interesting fact about delete or delete[] is that it is safe to use it on a nullptr. It is written at point 2 on cppreference.com or at this SO answer.
operator delete, operator delete[]
2)
[...] The behavior of the standard library implementation of this function is undefined unless ptr is a null pointer or is a pointer previously obtained from the standard library implementation of operator new[](size_t) or operator new[](size_t, std::nothrow_t).

We are setting pointers to NULL (0) to avoid dangling pointers (pointer is still pointing to same memory which is no longer yours). In case of local variables it is not that useful if function doesnt continue after delete (so its obvious pointer won't be reused). In case of global/member poitners its good practice to avoid bugs.
Accessing already deleted pointer may lead to overwriting/reading random memory (it might be more dangerous than crash) and causes undefined behavior, while accessing NULL pointer will immediatelly crash.
Since c++11 you should use nullptr because its defined as pointer type while NULL is more int type and improves type safety + resolves ambiguous situations.
In case of double deleting pointer, its safe to use delete on nullptr and nothing happens but if you delete already deleted non-null pointer, it will cause undefined behavior and most likely program will crash.
In c++ you should avoid using pure pointers since there are STL containers (which frees their resources them self (RAII)) for this usage or smart pointers.
std::vector<int> array{1,2,3,4,5};

This is done so that the pointer is set to NULL (whether in C++, we prefer nullptr, since NULL and 0 may be different things).
This tactic eliminates the possibility of a dangling pointer, because the array may have been deleted, but that doesn't mean that it's set to NULL.
If we don't do that, we run the risk of checking whether the pointer is NULL or not (latter in our code), we will see that it's not NULL, wrongly believe that the pointer is OK to be accessed, and cause Undefined Behavior.

You assign to a value commonly known as "invalid address", i.e. NULL, 0 or the pointer type nullptr, because otherwise there is no way you can know whether you pointer points to an invalid address. In other words when you delete[] your array your pointer "doesn't know" that it is pointing to a no longer usable memory address.

Related

Use of a deleted pointer address

(*) As far as I know the Standard allows an implementation to modify the operand of the delete operator, however most implementations do not do that.
int* ptr = new int(0);
delete ptr; //delete is allowed to modify ptr, for example set it to 0
std::cout << ptr; // UB?
Acknowledging (*), is the reading of ptr (in the form of printing it) well-defined?
If delete does modify ptr, is it allowed to set a trap value, which would make reading ptr UB?
In C++14 this is implementation-defined behaviour, [basic.stc.dynamic.deallocation]/4:
If the argument given to a deallocation function in the standard library is a pointer that is not the null pointer value, 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.
There is a footnote:
Some implementations might define that copying an invalid pointer value causes a system-generated runtime fault.
This changed since C++11 where the bolded text said "undefined behaviour" and there was no footnote.
So to answer your question, delete ptr; is allowed to set a trap value that would cause a runtime fault for std::cout << ptr. The compiler documentation must specify the behaviour. This is a narrower restriction than UB in which case any unstable behaviour would be permissible.
In this example, std::cout << ptr is NOT undefined behavior by default, because ptr is not being dereferenced at all, so it doesn't matter what its value is actually set to.
By default, the STL does not define an operator<< for int* pointers. It defines:
an operator<< for (signed|unsignd) char*, used for printing null-terminated text.
a generic operator<< for void*, which simply prints the memory address itself that the pointer is set to, not the data that is being pointed at.
Since int* is implicitly convertible to void*, calling std::cin << ptr is actually calling operator<<(std::cin, (void*)ptr), and so prints the memory address as-is that ptr holds.
The code would have undefined behavior only if your app defines its own operator<< for int* and then tries to dereference the pointer after it has been deleted.

What is the difference between pointers and pointers with allocated memory?

sstruct *temp1;
sstruct *p;
sstruct *temp2=new sstruct;
If I try to do something like this
temp1=p;
temp2=p;
is there any diffrence?
The pointer p is not explicitly initialized. So if this is a local variable it has an indeterminate value. If, OTOH., it's in namespace scope, then it's zero-initialized.
Assuming that it's an indeterminate value the subsequent assignment temp1=p has Undefined Behavior. Anything might happen. These things that can happen include (1) nothing, and (2) a crash.
If nothing happens, then the second assignment, temp2=p, loses the original value of temp2, which was the pointer to a dynamically allocated object. So this then is a memory leak. On top of the Undefined Behavior.
In original Pascal pointers always pointed to dynamically allocated objects. There was no way to take the address of a variable. This was a somewhat higher level view of pointers, and the Pascal pointers were single object pointers, not pointers to arrays.
In original C, which C++ was based on, pointers were more like low level memory addresses. A C or C++ pointer value can be obtained by using the address operator, &, on some object. Or you can get a pointer as an implicit conversion of an expression referring to an array (you then get a pointer to the array's first item).
There are three main notations for denoting a nullvalue of pointer type.
int* p = 0; // Core language, original C.
int* p = nullptr; // Core language, C++11 and later.
int* p = NULL; // Library, `<stddef.h>`.
I recommend using the newer nullptr notation.
It's more readable and it fares better when used with so called perfect forwarding (whose main imperfection is that it doesn't deal correctly with 0, forwarding it as an int value).
First of all, since p is uninitialized you invoke undefined behavior by reading from it (if the code is in function scope). But for the following code
sstruct *temp1;
sstruct *p = nullptr;
sstruct *temp2=new sstruct;
temp1=p;
temp2=p;
both assignments are no different (sans the destination). temp1 and temp2 are left with nullptr value and the struct you previously allocated for temp2 has been leaked.

Can a unique_ptr be used with a negative index without leaking memory?

I read Are negative array indexes allowed in C? and found it interesting that negative values can be used for the index of an array. I tried it again with the c++11 unique_ptr and it works there as well! Of course the deleter must be replaced with something which can delete the original array. Here is what it looks like:
#include <iostream>
#include <memory>
int main()
{
const int min = -23; // the smaller valid index
const int max = -21; // the highest valid index
const auto deleter = [min](char* p)
{
delete [](p+min);
};
std::unique_ptr<char[],decltype(deleter)> up(new char[max-min+1] - min, deleter);
// this works as expected
up[-23] = 'h'; up[-22] = 'i'; up[-21] = 0;
std::cout << (up.get()-23) << '\n'; // outputs:hi
}
I'm wondering if there is a very, very small chance that there is a memory leak. The address of the memory created on the heap (new char[max-min+1]) could overflow when adding 23 to it and become a null pointer. Subtracting 23 still yields the array's original address, but the unique_ptr may recognize it as a null pointer. The unique_ptr may not delete it because it's null.
So, is there a chance that the previous code will leak memory or does the smart pointer behave in a way which makes it safe?
Note: I wouldn't actually use this in actual code; I'm just interested in how it would behave.
Edit: icepack brings up an interesting point, namely that there are only two valid pointer values that are allowed in pointer arithmetic:
§5.7 [expr.add] p5
If both the pointer operand and the result point to elements of the same array object, or one past the last element of the array object, the evaluation shall not produce an overflow; otherwise, the behavior is undefined.
As such, the new char[N] - min of your code already invokes UB.
Now, on most implementations, this will not cause problems. The destructor of std::unique_ptr, however, will (pre-edit answer from here on out):
§20.7.1.2.2 [unique.ptr.single.dtor] p2
Effects: If get() == nullptr there are no effects. Otherwise get_deleter()(get()).
So yes, there is a chance that you will leak memory here if it indeed maps to whatever value represents the null pointer value (most likely 0, but not necessarily). And yes, I know this is the one for single objects, but the array one behaves exactly the same:
§20.7.1.3 [unique.ptr.runtime] p2
Descriptions are provided below only for member functions that have behavior different from the primary template.
And there is no description for the destructor.
new char[max-min+1] doesn't allocate memory on the stack but rather on heap - that's how standard operator new behaves. The expression max-min+1 is evaluated by the compiler and results in 3, so eventually this expression is equal to allocating 3 bytes on the heap. No problem here.
However, subtracting min results in pointer which is 23 bytes beyond the beginning of the allocated memory returned by new and since in new you allocated only 3 bytes, this will definitely point to a location not owned by you --> anything following will result in undefined behavior.

Is setting a pointer to zero equivalent to setting it to NULL?

I searched in the web but couldn't find a reliable answer.
And what would
someclass* ptr = 1;
char* charptr = 2;
or assigning them to any other integer mean?
What happens if I don't initialize pointers (for native data pointers as well as class pointers) before using them?
Setting a pointer to 0 is equivalent to setting it to NULL. However, this is only true for a constant expression 0, i.e. for compile-time zero value. Trying to set a pointer to a run-time zero value is not guaranteed to produce a null pointer
int *pi = 0; // Initializes a null pointer
char *pc = 2 - 2; // Initializes a null pointer
short *ps = sizeof *pc - 1; // Initializes a null pointer
int x = 0;
double *pd = (double *) x;
// Implementation-defined, not guaranteed to produce a null pointer
You can explore the matter in greater detail C FAQ
To answer the second part of your question:
It is illegal to assign any other integer (besides literal/constant 0) to a pointer. Neither of your initializations (or assignments) will compile. It is illegal in both C and C++, although C compilers are historically more permitting in this regard, responding with a warning instead of refusing to compile the code.
Also, in C++ language there's no difference of how pointers to class types are treated. Pointer to class types are still considered scalar types and behave in this regard the same way as any pointer to a fundamental type does.
Yes, in C++ NULL is defined to be 0. Setting a pointer to some other small integer value would mean it pointed to a -- likely illegal -- portion of the computer's memory (but it wouldn't be considered a NULL pointer).
If you don't initialize a global pointer it will be set to zero (NULL) for you just before the program starts. If you don't initialize pointer variables declared on the stack (i.e. as within functions or methods) they will have garbage in them. Likewise, the contents of any dynamically allocated pointer or any contained in an object as a data member also will have no predefined value. Some compilers have extensions that allow requests for dynamically allocated memory to be initially zeroed, like C's calloc() function does for raw memory.

What does "dereferencing" a pointer mean?

Please include an example with the explanation.
Reviewing the basic terminology
It's usually good enough - unless you're programming assembly - to envisage a pointer containing a numeric memory address, with 1 referring to the second byte in the process's memory, 2 the third, 3 the fourth and so on....
What happened to 0 and the first byte? Well, we'll get to that later - see null pointers below.
For a more accurate definition of what pointers store, and how memory and addresses relate, see "More about memory addresses, and why you probably don't need to know" at the end of this answer.
When you want to access the data/value in the memory that the pointer points to - the contents of the address with that numerical index - then you dereference the pointer.
Different computer languages have different notations to tell the compiler or interpreter that you're now interested in the pointed-to object's (current) value - I focus below on C and C++.
A pointer scenario
Consider in C, given a pointer such as p below...
const char* p = "abc";
...four bytes with the numerical values used to encode the letters 'a', 'b', 'c', and a 0 byte to denote the end of the textual data, are stored somewhere in memory and the numerical address of that data is stored in p. This way C encodes text in memory is known as ASCIIZ.
For example, if the string literal happened to be at address 0x1000 and p a 32-bit pointer at 0x2000, the memory content would be:
Memory Address (hex) Variable name Contents
1000 'a' == 97 (ASCII)
1001 'b' == 98
1002 'c' == 99
1003 0
...
2000-2003 p 1000 hex
Note that there is no variable name/identifier for address 0x1000, but we can indirectly refer to the string literal using a pointer storing its address: p.
Dereferencing the pointer
To refer to the characters p points to, we dereference p using one of these notations (again, for C):
assert(*p == 'a'); // The first character at address p will be 'a'
assert(p[1] == 'b'); // p[1] actually dereferences a pointer created by adding
// p and 1 times the size of the things to which p points:
// In this case they're char which are 1 byte in C...
assert(*(p + 1) == 'b'); // Another notation for p[1]
You can also move pointers through the pointed-to data, dereferencing them as you go:
++p; // Increment p so it's now 0x1001
assert(*p == 'b'); // p == 0x1001 which is where the 'b' is...
If you have some data that can be written to, then you can do things like this:
int x = 2;
int* p_x = &x; // Put the address of the x variable into the pointer p_x
*p_x = 4; // Change the memory at the address in p_x to be 4
assert(x == 4); // Check x is now 4
Above, you must have known at compile time that you would need a variable called x, and the code asks the compiler to arrange where it should be stored, ensuring the address will be available via &x.
Dereferencing and accessing a structure data member
In C, if you have a variable that is a pointer to a structure with data members, you can access those members using the -> dereferencing operator:
typedef struct X { int i_; double d_; } X;
X x;
X* p = &x;
p->d_ = 3.14159; // Dereference and access data member x.d_
(*p).d_ *= -1; // Another equivalent notation for accessing x.d_
Multi-byte data types
To use a pointer, a computer program also needs some insight into the type of data that is being pointed at - if that data type needs more than one byte to represent, then the pointer normally points to the lowest-numbered byte in the data.
So, looking at a slightly more complex example:
double sizes[] = { 10.3, 13.4, 11.2, 19.4 };
double* p = sizes;
assert(p[0] == 10.3); // Knows to look at all the bytes in the first double value
assert(p[1] == 13.4); // Actually looks at bytes from address p + 1 * sizeof(double)
// (sizeof(double) is almost always eight bytes)
++p; // Advance p by sizeof(double)
assert(*p == 13.4); // The double at memory beginning at address p has value 13.4
*(p + 2) = 29.8; // Change sizes[3] from 19.4 to 29.8
// Note earlier ++p and + 2 here => sizes[3]
Pointers to dynamically allocated memory
Sometimes you don't know how much memory you'll need until your program is running and sees what data is thrown at it... then you can dynamically allocate memory using malloc. It is common practice to store the address in a pointer...
int* p = (int*)malloc(sizeof(int)); // Get some memory somewhere...
*p = 10; // Dereference the pointer to the memory, then write a value in
fn(*p); // Call a function, passing it the value at address p
(*p) += 3; // Change the value, adding 3 to it
free(p); // Release the memory back to the heap allocation library
In C++, memory allocation is normally done with the new operator, and deallocation with delete:
int* p = new int(10); // Memory for one int with initial value 10
delete p;
p = new int[10]; // Memory for ten ints with unspecified initial value
delete[] p;
p = new int[10](); // Memory for ten ints that are value initialised (to 0)
delete[] p;
See also C++ smart pointers below.
Losing and leaking addresses
Often a pointer may be the only indication of where some data or buffer exists in memory. If ongoing use of that data/buffer is needed, or the ability to call free() or delete to avoid leaking the memory, then the programmer must operate on a copy of the pointer...
const char* p = asprintf("name: %s", name); // Common but non-Standard printf-on-heap
// Replace non-printable characters with underscores....
for (const char* q = p; *q; ++q)
if (!isprint(*q))
*q = '_';
printf("%s\n", p); // Only q was modified
free(p);
...or carefully orchestrate reversal of any changes...
const size_t n = ...;
p += n;
...
p -= n; // Restore earlier value...
free(p);
C++ smart pointers
In C++, it's best practice to use smart pointer objects to store and manage the pointers, automatically deallocating them when the smart pointers' destructors run. Since C++11 the Standard Library provides two, unique_ptr for when there's a single owner for an allocated object...
{
std::unique_ptr<T> p{new T(42, "meaning")};
call_a_function(p);
// The function above might throw, so delete here is unreliable, but...
} // p's destructor's guaranteed to run "here", calling delete
...and shared_ptr for share ownership (using reference counting)...
{
auto p = std::make_shared<T>(3.14, "pi");
number_storage1.may_add(p); // Might copy p into its container
number_storage2.may_add(p); // Might copy p into its container } // p's destructor will only delete the T if neither may_add copied it
Null pointers
In C, NULL and 0 - and additionally in C++ nullptr - can be used to indicate that a pointer doesn't currently hold the memory address of a variable, and shouldn't be dereferenced or used in pointer arithmetic. For example:
const char* p_filename = NULL; // Or "= 0", or "= nullptr" in C++
int c;
while ((c = getopt(argc, argv, "f:")) != -1)
switch (c) {
case f: p_filename = optarg; break;
}
if (p_filename) // Only NULL converts to false
... // Only get here if -f flag specified
In C and C++, just as inbuilt numeric types don't necessarily default to 0, nor bools to false, pointers are not always set to NULL. All these are set to 0/false/NULL when they're static variables or (C++ only) direct or indirect member variables of static objects or their bases, or undergo zero initialisation (e.g. new T(); and new T(x, y, z); perform zero-initialisation on T's members including pointers, whereas new T; does not).
Further, when you assign 0, NULL and nullptr to a pointer the bits in the pointer are not necessarily all reset: the pointer may not contain "0" at the hardware level, or refer to address 0 in your virtual address space. The compiler is allowed to store something else there if it has reason to, but whatever it does - if you come along and compare the pointer to 0, NULL, nullptr or another pointer that was assigned any of those, the comparison must work as expected. So, below the source code at the compiler level, "NULL" is potentially a bit "magical" in the C and C++ languages...
More about memory addresses, and why you probably don't need to know
More strictly, initialised pointers store a bit-pattern identifying either NULL or a (often virtual) memory address.
The simple case is where this is a numeric offset into the process's entire virtual address space; in more complex cases the pointer may be relative to some specific memory area, which the CPU may select based on CPU "segment" registers or some manner of segment id encoded in the bit-pattern, and/or looking in different places depending on the machine code instructions using the address.
For example, an int* properly initialised to point to an int variable might - after casting to a float* - access memory in "GPU" memory quite distinct from the memory where the int variable is, then once cast to and used as a function pointer it might point into further distinct memory holding machine opcodes for the program (with the numeric value of the int* effectively a random, invalid pointer within these other memory regions).
3GL programming languages like C and C++ tend to hide this complexity, such that:
If the compiler gives you a pointer to a variable or function, you can dereference it freely (as long as the variable's not destructed/deallocated meanwhile) and it's the compiler's problem whether e.g. a particular CPU segment register needs to be restored beforehand, or a distinct machine code instruction used
If you get a pointer to an element in an array, you can use pointer arithmetic to move anywhere else in the array, or even to form an address one-past-the-end of the array that's legal to compare with other pointers to elements in the array (or that have similarly been moved by pointer arithmetic to the same one-past-the-end value); again in C and C++, it's up to the compiler to ensure this "just works"
Specific OS functions, e.g. shared memory mapping, may give you pointers, and they'll "just work" within the range of addresses that makes sense for them
Attempts to move legal pointers beyond these boundaries, or to cast arbitrary numbers to pointers, or use pointers cast to unrelated types, typically have undefined behaviour, so should be avoided in higher level libraries and applications, but code for OSes, device drivers, etc. may need to rely on behaviour left undefined by the C or C++ Standard, that is nevertheless well defined by their specific implementation or hardware.
Dereferencing a pointer means getting the value that is stored in the memory location pointed by the pointer. The operator * is used to do this, and is called the dereferencing operator.
int a = 10;
int* ptr = &a;
printf("%d", *ptr); // With *ptr I'm dereferencing the pointer.
// Which means, I am asking the value pointed at by the pointer.
// ptr is pointing to the location in memory of the variable a.
// In a's location, we have 10. So, dereferencing gives this value.
// Since we have indirect control over a's location, we can modify its content using the pointer. This is an indirect way to access a.
*ptr = 20; // Now a's content is no longer 10, and has been modified to 20.
In simple words, dereferencing means accessing the value from a certain memory location against which that pointer is pointing.
A pointer is a "reference" to a value.. much like a library call number is a reference to a book. "Dereferencing" the call number is physically going through and retrieving that book.
int a=4 ;
int *pA = &a ;
printf( "The REFERENCE/call number for the variable `a` is %p\n", pA ) ;
// The * causes pA to DEREFERENCE... `a` via "callnumber" `pA`.
printf( "%d\n", *pA ) ; // prints 4..
If the book isn't there, the librarian starts shouting, shuts the library down, and a couple of people are set to investigate the cause of a person going to find a book that isn't there.
Code and explanation from Pointer Basics:
The dereference operation starts at
the pointer and follows its arrow over
to access its pointee. The goal may be
to look at the pointee state or to
change the pointee state. The
dereference operation on a pointer
only works if the pointer has a
pointee -- the pointee must be
allocated and the pointer must be set
to point to it. The most common error
in pointer code is forgetting to set
up the pointee. The most common
runtime crash because of that error in
the code is a failed dereference
operation. In Java the incorrect
dereference will be flagged politely
by the runtime system. In compiled
languages such as C, C++, and Pascal,
the incorrect dereference will
sometimes crash, and other times
corrupt memory in some subtle, random
way. Pointer bugs in compiled
languages can be difficult to track
down for this reason.
void main() {
int* x; // Allocate the pointer x
x = malloc(sizeof(int)); // Allocate an int pointee,
// and set x to point to it
*x = 42; // Dereference x to store 42 in its pointee
}
I think all the previous answers are wrong, as they
state that dereferencing means accessing the actual value.
Wikipedia gives the correct definition instead:
https://en.wikipedia.org/wiki/Dereference_operator
It operates on a pointer variable, and returns an l-value equivalent to the value at the pointer address. This is called "dereferencing" the pointer.
That said, we can dereference the pointer without ever
accessing the value it points to. For example:
char *p = NULL;
*p;
We dereferenced the NULL pointer without accessing its
value. Or we could do:
p1 = &(*p);
sz = sizeof(*p);
Again, dereferencing, but never accessing the value. Such code will NOT crash:
The crash happens when you actually access the data by an
invalid pointer. However, unfortunately, according the the
standard, dereferencing an invalid pointer is an undefined
behaviour (with a few exceptions), even if you don't try to
touch the actual data.
So in short: dereferencing the pointer means applying the
dereference operator to it. That operator just returns an
l-value for your future use.