From what I'd learned, a dynamically allocated variable needs to be deleted using the delete operator and will not automatically delete at the end of a scope, like in the case of static variables.
Therefore, in the following example when the loop runs for the 2nd and 3rd time, shouldn't "int *p=new int;" be cosnidered as multiple initialization since the dynamically allocated memory 'p' hasn't been deleted?
#include<iostream>
using namespace std;
void main()
{
int i = 2;
while (i > -1)
{
int *p=new int;
*p = 5;
cout << *p;
--i;
}
}
Using Visual Studio 2015, the above program gives no error. According to my understanding this does not make sense.
I'm assuming there is something wrong with my understanding of dynamically allocated variables. Can anyone please clarify?
Your code shows what is called a "memory leak". The memory allocated for each loop iteration is lost when p goes out of scope without deleteing the memory first. This (usually) does not lead to compiler warnings or runtime errors, as it can be quite complicated for the compiler to find this kind of error. Some static code analysers might be able to detect this though.
What you might notice in the case of a memory leak is that your program uses up more and more memory the longer it runs, meaning that memory leaks are especially problematic in systems with low RAM and for programs that are long running, e.h. system services that are supposed to run for several days.
There are special tools to find memory leaks, e.g. valgrind for Linux or built-in tools in the debug runtime for Visual Studio.
You can't dynamically allocate variables, only objects.
p is not a dynamically allocated object, but *p - the object created by new - is.
delete p would not delete p but the object it points to.
Scope is a syntactic compile-time property that applies to names.
The variable p, having a name, has a scope.
The object it points to is unnamed, so the concept of scope doesn't even apply to it.
At runtime, both p and the object it points to have a lifetime.
The lifetime of p coincides with its scope, since it's an automatic variable. Every iteration of the loop has its own variable, all with the same name - there is no multiple initialization because the variables are distinct.
The lifetime of the object that p points to extends until its address is passed to delete.
Since you never do that, every object you created with new "leaks".
p is just a static variable containing the address to the dynamic data, the pointer itself gets lost at the end of the scope, while the dynamic data persist without having a pointer to access it (memory leak).
The data pointed to by p is dynamic memory (in the heap) and won't get freed after each iteration, however, the content of variable p itself (which is the address of the dynamic data, if you go to p in memory you find it a simple number in stack, which is just the address of the dynamic data in the heap not the data itself) is just a static address in the stack that gets lost at the end of the scope (an iteration).
Related
Suppose that you have the following function:
void doSomething(){
int *data = new int[100];
}
Why will this produce a memory leak? Since I can not access this variable outside the function, why doesn't the compiler call delete by itself every time a call to this function ends?
Why will this produce a memory leak?
Because you're responsible for deleting anything you create with new.
Why doesn't the compiler call delete by itself every time a call to this function ends?
Often, the compiler can't tell whether or not you still have a pointer to the allocated object. For example:
void doSomething(){
int *data = new int[100];
doSomethingElse(data);
}
Does doSomethingElse just use the pointer during the function call (in which case we still want to delete the array here)? Does it store a copy of the pointer for use later (in which case we don't want to delete it yet)? The compiler has no way of knowing; it's up to you to tell it. Rather than making a complicated, error-prone rule (like "you must delete it unless you can figure out that the compiler must know that there are no other references to it"), the rule is kept simple: you must delete it.
Luckily, we can do better than juggling a raw pointer and trying to delete it at the right time. The principle of RAII allows objects to take ownership of allocated resources, and automatically release them when their destructor is called as they go out of scope. Containers allow dynamic objects to be maintained within a single scope, and copied if needed; smart pointers allow ownership to be moved or shared between scopes. In this case, a simple container will give us a dynamic array:
void doSomething(){
std::vector<int> data(100);
} // automatically deallocated
Of course, for a small fixed-size array like this, you might as well just make it automatic:
void doSomething(){
int data[100];
} // all automatic variables are deallocated
The whole point of dynamic allocation like this is that you manually control the lifetime of the allocated object. It is needed and appropriate a lot less often than many people seem to think. In fact, I cannot think of a valid use of new[].
It is indeed a good idea to let the compiler handle the lifetime of objects. This is called RAII. In this particular case, you should use std::vector<int> data(100). Now the memory gets freed automatically as soon as data goes out of scope in any way.
Actually you can access this variable outside of your doSomething() function, you just need to know address which this pointer holds.
In picture below, rectangle is one memory cell, on top of rectangle is memory cell address, inside of rectangle is memory cell value.
For example, in picture above, if you know that allocated array starts from address 0x200, then outside of your function you can do next:
int *data = (int *)0x200;
std::cout << data[0];
So compiler can't delete this memory for you, but please pay attention to compiler warning:
main.cpp: In function ‘void doSomething()’:
main.cpp:3:10: warning: unused variable ‘data’ [-Wunused-variable]
int *data = new int[100];
When you do new, OS allocates memory in RAM for you, so you need to make OS know, when you don't need this memory anymore, doing delete, so it's only you, who knows when execute delete, not a compiler.
This is memory allocated on the heap, and is therefore available outside the function to anything which has its address. The compiler can't be expected to know what you intend to do with heap-allocated memory, and it will not deduce at compile time the need to free the memory by (for example) tracking whether anything has an active reference to the memory. There is also no automatic run-time garbage collection as in Java. In C++ the responsibility is on the programmer to free all memory allocated on the heap.
See also: Why doesn't C++ have a garbage collector?
Here you have an int array dynamically allocated on the heap with a pointer data pointing to the first element of your array.
Since you explicitly allocated your array on the heap with new, this one will not be deleted from the memory when you go out of scope. When you go out of scope, your pointer will be deleted, and then you will not have access to the array -> Memory leak
Since I can not access this variable outside the function, why doesn't
the compiler call delete by itself every time a call to this function
ends?
Because you should not allocate memory on the heap that you wouldn't use anyway in the first place.
That's how C++ works.
EDIT:
also if compiler would delete the pointer after function returns then there would be way of returning a pointer from function.
The languages in question are C/C++.
My prof said to free memory on the heap when your done using it, because otherwise you can end up with memory that can't be accessed. The problem with this is you might end up with all your memory used up and you are unable to access any of it.
Why doesn't the same concept apply to the stack? I understand that you can always access the memory you used on the stack, but if you keep creating new variables, you will eventually run out of space right? So why can't you free variables on the stack to make room for new variables like you can on the heap?
I get that the compiler frees variables on the stack, but thats at the end of the scope of the variable right. Doesn't it also free a variable on the heap at the end of its scope? If not, why not?
Dynamically allocated objects ("heap objects" in colloquial language) are never variables. Thus, they can never go out of scope. They don't live inside any scope. The only way you can handle them is via a pointer that you get at the time of allocation.
(The pointer is usually assigned to a variable, but that doesn't help.)
To repeat: Variables have scope; objects don't. But many objects are variables.
And to answer the question: You can only free objects, not variables.
The end of the closed "}" braces is where the stack "frees" its memory. So if I have:
{
int a = 1;
int b = 2;
{
int c = 3; // c gets "freed" at this "}" - the stack shrinks
// and c is no longer on the stack.
}
} // a and b are "freed" from the stack at this last "}".
You can think of c as being "higher up" on the stack than "a" and "b", so c is getting popped off before them. Thus, every time you write a "}" symbol, you are effectively shrinking the stack and "freeing" data.
There are already nice answers but I think you might need some more clarification, so I'll try to make this a more detailed answer and also try to make it simple (if I manage to). If something isn't clear (which with me being not a native english speaker and having problems with formulating answers sometimes might be likely) just ask in the comments. Also gonna take the use the Variables vs Objects idea that Kerrek SB uses in his answer.
To make that clearer I consider Variables to be named Objects with an Object being something to store data within your program.
Variables on the stack got automatic storage duration they automatically get destroyed and reclaimed once their scope ends.
{
std::string first_words = "Hello World!";
// do some stuff here...
} // first_words goes out of scope and the memory gets reclaimed.
In this case first_words is a Variable (since it got its own name) which means it is also an Object.
Now what about the heap? Lets describe what you might consider being "something on the heap" as a Variable pointing to some memory location on the heap where an Object is located. Now these things got what's called dynamic storage duration.
{
std::string * dirty = nullptr
{
std::string * ohh = new std::string{"I don't like this"} // ohh is a std::string* and a Variable
// The actual std::string is only an unnamed
// Object on the heap.
// do something here
dirty = ohh; // dirty points to the same memory location as ohh does now.
} // ohh goes out of scope and gets destroyed since it is a Variable.
// The actual std::string Object on the heap doesn't get destroyed
std::cout << *dirty << std::endl; // Will work since the std::string on the heap that dirty points to
// is still there.
delete dirty; // now the object being pointed to gets destroyed and the memory reclaimed
dirty = nullptr; can still access dirty since it's still in its scope.
} // dirty goes out of scope and get destroyed.
As you can see objects don't adhere to scopes and you got to manually manage their memory. That's also a reason why "most" people prefer to use "wrappers" around it. See for example std::string which is a wrapper around a dynamic "String".
Now to clarify some of your questions:
Why can't we destroy objects on the stack?
Easy answer: Why would you want to?
Detailed answer: It would be destroued by you and then destroyed again once it leaves the scope which isn't allowed. Also you should generally only have variables in your scope that you actually need for your computation and how would you destroy it if you actually need that variable to finish your computation? But if you really were to only need a variable for a small time within a computation you could just make a new smaller scope with { } so your variable gets automatically destroyed once it isn't needed anymore.
Note: If you got a lot of variables that you only need for a small part of your computation it might be a hint that that part of the computation should be in its own function/scope.
From your comments: Yeah I get that, but thats at the end of the scope of the variable right. Doesn't it also free a variable on the heap at the end of its scope?
They don't. Objects on the heap got no scope, you can pass their address out of a function and it still persists. The pointer pointing to it can go out of scope and be destroyed but the Object on the heap still exists and you can't access it anymore (memory leak). That's also why it's called manual memory management and most people prefer wrappers around them so that it gets automatically destroyed when it isn't needed anymore. See std::string, std::vector as examples.
From your comments: Also how can you run out of memory on a computer? An int takes up like 4 bytes, most computers have billions of bytes of memory... (excluding embedded systems)?
Well, computer programs don't always just hold a few ints. Let me just answer with a little "fake" quote:
640K [of computer memory] ought to be enough for anybody.
But that isn't enough like we all should know. And how many memory is enough? I don't know but certainly not what we got now. There are many algorithms, problems and other stuff that need tons of memory. Just think about something like computer games. Could we make "bigger" games if we had more memory? Just think about it... You can always make something bigger with more resources so I don't think there's any limit where we can say it's enough.
So why can't you free variables on the stack to make room for new variables like you can on the heap?
All information that "stack allocator" knows is ESP which is pointer to the bottom of stack.
N: used
N-1: used
N-2: used
N-3: used <- **ESP**
N-4: free
N-5: free
N-6: free
...
That makes "stack allocation" very efficient - just decrease ESP by the size of allocation, plus it is locality/cache-friendly.
If you would allow arbitrary deallocations, of different sizes - that will turn your "stack" into "heap", with all associated additional overhead - ESP would not be enough, because you have to remember which space is deallocated and which is not:
N: used
N-1: free
N-2: free
N-3: used
N-4: free
N-5: used
N-6: free
...
Clearly - ESP is not more enough. And you also have to deal with fragmentation problems.
I get that the compiler frees variables on the stack, but thats at the end of the scope of the variable right. Doesn't it also free a variable on the heap at the end of its scope? If not, why not?
One of the reasons is that you don't always want that - sometimes you want to return allocated data to caller of your function, that data should outlive scope where it was created.
That said, if you really need scope-based lifetime management for "heap" allocated data (and most of time it is scope-based, indeed) - it is common practice in C++ to use wrappers around such data. One of examples is std::vector:
{
std::vector<int> x(1024); // internally allocates array of 1024 ints on heap
// use x
// ...
} // at the end of the scope destructor of x is called automatically,
// which does deallocation
Read about function calls - each call pushes data and function address on the stack. Function pops data from stack and eventually pushes its result.
In general, stack is managed by OS, and yes - it can be depleted. Just try doing something like this:
int main(int argc, char **argv)
{
int table[1000000000];
return 0;
}
That should end quickly enough.
Local variables on the stack don't actually get freed. The registers pointing at the current stack are just moved back up and the stack "forgets" about them. And yes, you can occupy so much stack space that it overflows and the program crashes.
Variables on the heap do get freed automatically - by the operating system, when the program exits. If you do
int x;
for(x=0; x<=99999999; x++) {
int* a = malloc(sizeof(int));
}
the value of a keeps getting overwritten and the place in the heap where a was stored is lost. This memory is NOT freed, because the program doesn't exit. This is called a "memory leak". Eventually, you will use up all the memory on the heap, and the program will crash.
The heap is managed by code: Deleting a heap allocation is done by calling the heap manager. The stack is managed by hardware. There is no manager to call.
# include <iostream>
int main()
{
using std::cout;
int *p= new int;
*p = 10;
cout<<*p<<"\t"<<p<<"\n";
delete p;
cout<<*p<<"\t"<<p<<"\n";
return 0;
}
Output:
10 0x237c010
0 0x237c010
Here after deleting p, why the pointer p retains its value? Doesn't delete frees the pointer p?
What exactly is meant by 'freeing the pointer'?
Does 'delete p' simply mean '*p = 0'?(which seems from the output)
Here after deleting p, why the pointer p retains its value?
It's how the language is designed. If you want the pointer you hold to be zeroed, you'll need to assign it zero yourself. The pointer p is another piece of memory, separate from the allocation/object it points to.
Doesn't delete frees the pointer p?
It calls the destructor the object and returns the memory to the system (like free). If it is an array (delete[]), destructors for all elements will be called, then the memory will be returned.
What exactly is meant by 'freeing the pointer'?
When you want a piece of memory from the system, you allocate it (e.g. using new). When you are finished using it, you return it using the corresponding free/delete call. It's a resource, which you must return. If you do not, your program will leak (and nobody wants that).
In order to understand what freeing memory means, you must first understand what allocating memory means. What follows is a simplified explanation.
There exists memory. Memory is a large blob of stuff that you could access. But since it's global, you need some way to portion it out. Some way to govern who can access which pieces of memory. One of the systems that governs the apportionment of memory is called the "heap".
The heap owns some quantity of memory (some is owned by the stack and some is owned by static data, but nevermind that now). At the beginning of your program, the heap says that you have access to no heap-owned memory.
What new int does is two fold. First, it goes to the heap system and says, "I want a piece of memory suitable to store an int into." It gets back a pointer to exactly that: a piece of the heap, into which you can safely store and retrieve exactly one value of type int.
You are now the proud owner of one int's worth of memory. The heap guarantees that as long as its rules are followed, whatever you put there will be preserved until you explicitly change it. This is the covenant between you and the almighty heap.
The other thing new int does is initialize that piece of the heap with an int value. In this case, it is default initialized, because no value was passed (new int(5) would initialize it with the value 5).
From this point forward, you are legally allowed to store exactly one int in this piece of memory. You are allowed to retrieve the int stored there. And you're allowed to do one other thing: tell the heap that you are finished using that memory.
When you call delete p, two things happen. First, p is deinitialized. Again, because it is an int, nothing happens. If this were a class, its destructor would be called.
But after that, delete goes out to the heap and says, "Hey heap: remember this pointer to an int you gave me? I'm done with it now." The heap system can do whatever it wants. Maybe it will clear the memory, as some heaps do in debug-builds. In release builds however, the memory may not be cleared.
Of course, the reason why the heap can do whatever it wants is because, the moment you delete that pointer, you enter into a new agreement with the heap. Previously, you asked for a piece of memory for an int, and the heap obliged. You owned that memory, and the heap guaranteed that it was yours for as long as you wanted. Stuff you put there would remain there.
After you had your fun, you returned it to the heap. And here's where the contract comes in. When you say delete p, for any object p, you are saying the following:
I solemnly swear not to touch this memory address again!
Now, the heap might give that memory address back to you if you call new int again. It might give you a different one. But you only have access to memory allocated by the heap during the time between new and delete.
Given this, what does this mean?
delete p;
cout << *p << "\t" << p << "\n";
In C++ parlance, this is called "undefined behavior". The C++ specification has a lot of things that are stated to be "undefined". When you trigger undefined behavior anything can happen! *p could be 0. *p could be the value it used to be. Doing *p could crash your program.
The C++ specification is a contract between you and your compiler/computer. It says what you can do, and it says how the system responds. "Undefined behavior" is what happens when you break the contract, when you do something the C++ specification says you aren't supposed to. At that point, anything can happen.
When you called delete p, you told the system that you were finished using p. By using it again, you were lying to the system. And therefore, the system no longer has to abide by any rules, like storing the values you want to store. Or continuing to run. Or not spawning demons from your nose. Or whatever.
You broke the rules. And you must suffer the consequences.
So no, delete p is not the equivalent of *p = 0. The latter simply means "set 0 into the memory pointed to by p." The former means "I'm finished using the memory pointed to by p, and I won't use it again until you tell me I can."
Here after deleting p, why the pointer p retains its value? Doesn't delete frees the pointer p?
It frees the memory the pointer points to (after calling any appropriate destructor). The value of the pointer itself is unchanged.
What exactly is meant by 'freeing the pointer'?
As above - it means freeing the memory the pointer points to.
Does 'delete p' simply mean '*p = 0'?(which seems from the output)
No. The system doesn't have to write anything to the memory that's freed, and if it does write something it doesn't have to write 0. However, the system does generally have to manage that memory in some way, and that might actually write to the area of memory that the pointer was pointing to. Also, the just-freed memory can be allocated to something else (and in a multi-threaded application, that could happen before the delete operation even returns). The new owner of that memory block can of course write whatever they want to that memory.
A pointer that is pointing to a freed block of memory is often known as a 'dangling' pointer. It is an error to dereference a dangling pointer (for read or write). You will sometimes see code immediately assign a NULL or 0 to a pointer immediately after deleting the pointer, sometimes using a macro or function template that both deletes and clears the pointer. Note that this won't fix all bugs with dangling pointers, since other pointers may have been set to point to the memory block.
The modern method of dealing with these kinds of problems is to avoid using raw pointers altogether in favor of using smart pointers such as shared_ptr or unique_ptr.
delete p simply frees the memory allocated during a call to the new operator. It does not change the value of the pointer or the content of the deallocated memory.
(Note the following isn't how it actually works so take it with a grain of salt.)
Inside the implementation of new it keeps a list of all available memory when you said "int *p= new int;" it cut a int sized block off of its list of available memory and gave it to you. When you run "delete p;" it gets put back in the list of available memory. If your program called new 30 times without calling delete at all you would get 30 different int sized chunks from new. If you called new then delete 30 times in a row you might (but not necessarily) get the same int sized block. This is because you said you weren't using it any more when you called delete and so new was free to reuse it.
TLDR; Delete notifies new that this section of memory is available again it doesn't touch your variable.
I don't quite get the point of dynamically allocated memory and I am hoping you guys can make things clearer for me.
First of all, every time we allocate memory we simply get a pointer to that memory.
int * dynInt = new int;
So what is the difference between doing what I did above and:
int someInt;
int* dynInt = &someInt;
As I understand, in both cases memory is allocated for an int, and we get a pointer to that memory.
So what's the difference between the two. When is one method preferred to the other.
Further more why do I need to free up memory with
delete dynInt;
in the first case, but not in the second case.
My guesses are:
When dynamically allocating memory for an object, the object doesn't get initialized while if you do something like in the second case, the object get's initialized. If this is the only difference, is there a any motivation behind this apart from the fact that dynamically allocating memory is faster.
The reason we don't need to use delete for the second case is because the fact that the object was initialized creates some kind of an automatic destruction routine.
Those are just guesses would love it if someone corrected me and clarified things for me.
The difference is in storage duration.
Objects with automatic storage duration are your "normal" objects that automatically go out of scope at the end of the block in which they're defined.
Create them like int someInt;
You may have heard of them as "stack objects", though I object to this terminology.
Objects with dynamic storage duration have something of a "manual" lifetime; you have to destroy them yourself with delete, and create them with the keyword new.
You may have heard of them as "heap objects", though I object to this, too.
The use of pointers is actually not strictly relevant to either of them. You can have a pointer to an object of automatic storage duration (your second example), and you can have a pointer to an object of dynamic storage duration (your first example).
But it's rare that you'll want a pointer to an automatic object, because:
you don't have one "by default";
the object isn't going to last very long, so there's not a lot you can do with such a pointer.
By contrast, dynamic objects are often accessed through pointers, simply because the syntax comes close to enforcing it. new returns a pointer for you to use, you have to pass a pointer to delete, and (aside from using references) there's actually no other way to access the object. It lives "out there" in a cloud of dynamicness that's not sitting in the local scope.
Because of this, the usage of pointers is sometimes confused with the usage of dynamic storage, but in fact the former is not causally related to the latter.
An object created like this:
int foo;
has automatic storage duration - the object lives until the variable foo goes out of scope. This means that in your first example, dynInt will be an invalid pointer once someInt goes out of scope (for example, at the end of a function).
An object created like this:
int foo* = new int;
Has dynamic storage duration - the object lives until you explicitly call delete on it.
Initialization of the objects is an orthogonal concept; it is not directly related to which type of storage-duration you use. See here for more information on initialization.
Your program gets an initial chunk of memory at startup. This memory is called the stack. The amount is usually around 2MB these days.
Your program can ask the OS for additional memory. This is called dynamic memory allocation. This allocates memory on the free store (C++ terminology) or the heap (C terminology). You can ask for as much memory as the system is willing to give (multiple gigabytes).
The syntax for allocating a variable on the stack looks like this:
{
int a; // allocate on the stack
} // automatic cleanup on scope exit
The syntax for allocating a variable using memory from the free store looks like this:
int * a = new int; // ask OS memory for storing an int
delete a; // user is responsible for deleting the object
To answer your questions:
When is one method preferred to the other.
Generally stack allocation is preferred.
Dynamic allocation required when you need to store a polymorphic object using its base type.
Always use smart pointer to automate deletion:
C++03: boost::scoped_ptr, boost::shared_ptr or std::auto_ptr.
C++11: std::unique_ptr or std::shared_ptr.
For example:
// stack allocation (safe)
Circle c;
// heap allocation (unsafe)
Shape * shape = new Circle;
delete shape;
// heap allocation with smart pointers (safe)
std::unique_ptr<Shape> shape(new Circle);
Further more why do I need to free up memory in the first case, but not in the second case.
As I mentioned above stack allocated variables are automatically deallocated on scope exit.
Note that you are not allowed to delete stack memory. Doing so would inevitably crash your application.
For a single integer it only makes sense if you need the keep the value after for example, returning from a function. Had you declared someInt as you said, it would have been invalidated as soon as it went out of scope.
However, in general there is a greater use for dynamic allocation. There are many things that your program doesn't know before allocation and depends on input. For example, your program needs to read an image file. How big is that image file? We could say we store it in an array like this:
unsigned char data[1000000];
But that would only work if the image size was less than or equal to 1000000 bytes, and would also be wasteful for smaller images. Instead, we can dynamically allocate the memory:
unsigned char* data = new unsigned char[file_size];
Here, file_size is determined at runtime. You couldn't possibly tell this value at the time of compilation.
Read more about dynamic memory allocation and also garbage collection
You really need to read a good C or C++ programming book.
Explaining in detail would take a lot of time.
The heap is the memory inside which dynamic allocation (with new in C++ or malloc in C) happens. There are system calls involved with growing and shrinking the heap. On Linux, they are mmap & munmap (used to implement malloc and new etc...).
You can call a lot of times the allocation primitive. So you could put int *p = new int; inside a loop, and get a fresh location every time you loop!
Don't forget to release memory (with delete in C++ or free in C). Otherwise, you'll get a memory leak -a naughty kind of bug-. On Linux, valgrind helps to catch them.
Whenever you are using new in C++ memory is allocated through malloc which calls the sbrk system call (or similar) itself. Therefore no one, except the OS, has knowledge about the requested size. So you'll have to use delete (which calls free which goes to sbrk again) for giving memory back to the system. Otherwise you'll get a memory leak.
Now, when it comes to your second case, the compiler has knowledge about the size of the allocated memory. That is, in your case, the size of one int. Setting a pointer to the address of this int does not change anything in the knowledge of the needed memory. Or with other words: The compiler is able to take care about freeing of the memory. In the first case with new this is not possible.
In addition to that: new respectively malloc do not need to allocate exactly the requsted size, which makes things a bit more complicated.
Edit
Two more common phrases: The first case is also known as static memory allocation (done by the compiler), the second case refers to dynamic memory allocation (done by the runtime system).
What happens if your program is supposed to let the user store any number of integers? Then you'll need to decide during run-time, based on the user's input, how many ints to allocate, so this must be done dynamically.
In a nutshell, dynamically allocated object's lifetime is controlled by you and not by the language. This allows you to let it live as long as it is required (as opposed to end of the scope), possibly determined by a condition that can only be calculated at run-rime.
Also, dynamic memory is typically much more "scalable" - i.e. you can allocate more and/or larger objects compared to stack-based allocation.
The allocation essentially "marks" a piece of memory so no other object can be allocated in the same space. De-allocation "unmarks" that piece of memory so it can be reused for later allocations. If you fail to deallocate memory after it is no longer needed, you get a condition known as "memory leak" - your program is occupying a memory it no longer needs, leading to possible failure to allocate new memory (due to the lack of free memory), and just generally putting an unnecessary strain on the system.
Consider the following program:
int main() {
while(...) {
int* foobar = new int;
}
return 0;
}
When does foobar go out of scope?
I know when using new, attributes are allocated on the heap and need to be deleted manually with delete, in the code above, it is causing a memory leak. However, what about scope?
I thought it would go out of scope as soon as the while loop terminates, because you have no direct access to it anymore. For example, you cannot delete it after the loop has terminated.
Be careful here, foobar is local to the while loop, but the allocation on the heap has no scope and will only be destructed if you call delete on it.
The variable and the allocation are not linked in any way as far as the compiler is concerned. Indeed, the allocation happens at run time, so the compiler never even sees it.
foobar is a local variable that goes out of scope at the end of the block.
*foobar is a dynamically allocated object with manual lifetime. Since it doesn't have scoped lifetime, the question makes no sense -- it doesn't have a scope out of which it could go. Its lifetime is managed manually, and the object lives until you delete it.
Your question is dangerously burdened with prejudice and preconceptions. It is best to approach C++ with a clean mind and an open attitude. Only that way will you be able to appreciate the language's wonders to the fullest.
Here's the clean and open approach: Do think about 1) storage classes (automatic, static, dynamic), 2) object lifetime (scoped, permanent, manual), 3) object semantics (value (copies) vs reference (aliases)), 4) RAII and single-responsiblity classes. Purge your mind of a) stack/heap, b) pointers, c) new/delete, d) destructors/copy constructors/assignment operators.
That's a pretty awesome memory leak. You have a variable on the stack pointing to the memory allocated on the heap. You need to delete the memory on the heap before you lose the reference to it when the while loop scope ends. Alternately if you don't want to fuss with memory management always use smart pointers to own the raw memory on the heap and let it clean itself up.
#include <memory>
int main() {
while(...) {
std::unique_ptr<int> foobar = new int;
} // smart pointer foobar deletes the allocated int each iteration
return 0;
}
The pointer (foobar) will go out of scope right as the program gets to the closing brace of the while loop. So if the expression in the ... remains true, memory will be leaked every time the loop executes as you have lost a handle to the allocated object as of that closing brace.
Here foobar is an int pointer occupying memory in the stack. The int instance you are creating dynamically with new goes to heap. When foobar goes out of scope, you lose the reference to it, so you cannot delete the memory allocated in the heap.
The best solution would be:
while(--)
{
int foobar;
}//it goes out of scope here. deleted from stack automatically!!
If you still want to use the dynamic allocation then do this:
while(--)
{
int* foobar=new int;
//do your work here!
delete foobar; //This deletes the heap memory allocated!
foobar=NULL; //avoid dangling pointer! :)
}
foobar goes out of scope after each interation of the loop.
The memory you allocate and assign to foobar is being leaked, in that it is still allocated on the heap but no references to it are available in the program.
Since foobar is declared in the body of the loop, it goes out of scope at the end of every iteration of the loop. It is then redeclared, and new memory is allocated again and again until the loop ends. The actual object that foobar points to never goes out of scope. Scope doesn't apply to dynamically allocated(aka heap) objects, only to automatic(stack) objects.
Foobar the pointer is created on the stack, but the new int is created on the heap. In the case of the while loop, each time the code loops, foobar falls out scope. The newly created int persists on the heap. On each iteration a new int is created, and the pointer is reset, which means the pointer no longer can access any of the previous int(s) on the heap.
What seems to be lacking, in every one of the previous answers, and even in this one, is the heap falling out of scope. Maybe, I am getting the terminology incorrect, but I know that at some point the heap is reset, too. It may occur once the program no longer runs, or when the computer is turned off, but I know it occurs.
Let us look at this question from a different perspective. I have written any number of programs, which leak memory. Over all the years, I have owned my computer, I am positive, I have leaked over 2 gigabytes of memory. My computer only has 1 gig of memory. Therefore, if the heap NEVER falls out of scope, then my computer has some magical memory. Would one of you care to explain when exactly the heap falls out of scope?