I understand that we use this when we need to give a pointer to an array of pointers in the dynamic memory but what I don't understand is that how this works in the stack.
Does this make an array of pointers in the stack too that is pointing to the array of pointers in the heap or does it make a single pointer in the stack that is pointing to the array of pointers in the heap? if yes, then what is the difference between
int **p = new int*[100]
and
int *p = new int[100]
Thanks in advance. I have been trying to understand this for a long time now and have read a lot of documentation online but I still don't understand this.
int **p declares a pointer on the stack which points to pointer(s) on the heap. Each of that pointer(s) point to an integer or array of integers on the heap.
This:
int **p = new int*[100];
means that you declared a pointer on the stack and initialized it so that it points to an array of 100 pointers on heap. For now each of that 100 pointers point nowhere. By "nowhere" I mean that they point neither to a valid chunk of memory, nor they are nullptrs. They are not initialized, thus they contain some garbage value which was in the memory before the pointers were allocated. You should assign something sensible to them in a loop before usage. Note that p[0] - p[99] pointers are not guaranteed to point to adjacent regions of memory if you assign return values of new to them. For example, if you allocate memory for each of them as p[i] = new int[200];, p[0][201] will not reference p[1][2], but will lead to an undefined behavior.
And this:
int *p = new int[100];
is a pointer on the stack which points to an array of 100 integers on the heap.
Don't worry, pointers and arrays in C are always a source of confusion. In general, when you declare an array of, say type int, you create a pointer of type int that points to the first element in a contiguous block of memory that will store ints. For example, if I have a simple array of ints using int *p_to_intarr = new int[3], I get this:
+++++++ <--------- p_to_intarr
| int |
+++++++
| int |
+++++++
| int |
+++++++
In general, if I want an array of type T, I create a pointer to type T like T *ptr_to_Tarr = new T[3].
So what if I want an array of array of ints? Lets just replace the T in with the type of an "array of ints" and this will give us an array of "array of ints". Well we just say in our first example that the type of an array of ints in int *, and so an array of array of ints would be: int* *ptr_to_arrayofintarr = new int*[3]. Note we just replaced the T with int star. This is often written more neatly as int **ptr_to_arrayofintarr = new int*[3].
So int **p could be a pointer to a 2d array. It could also be a reference to a 1d array; depends on the specific case :)
The new expression evaluates to a pointer of some type, pointing to memory that has been allocated in the free store (essentially the heap) but not necessarily in the free store. (It can still be in the free store depending on the context; for instance, consider an initialization-list in a constructor for an object being allocated in the free store.)
The object(s) initialized by new is/are of course in the free store.
In the assignment statements you've shown, you can see the type of the pointer returned by new on the left side of the equals sign, and the type of the free store object(s) to the right of the new. Thus the only object that is locally evaluated (i.e. might be on the stack) is, in the first case, a pointer-to-pointer-to-int, and in the second case, a pointer-to-int. The objects in free space are an array of pointers-to-ints in the first case and a simple array of ints in the second.
Note that just because the array in the first assignment consists of pointers doesn't mean that the pointers themselves actually point to anything yet; new does not magically recursively allocate free space for objects to be targeted by any pointers in an array it creates. (This wouldn't make much sense anyway.)
The ** means that you have a pointer to pointers. In more practical terms, it means you have a two-dimensional array.
Since each element of the array is also a pointer, you need to initialize those pointers as well:
for (int i = 0; i < 100; ++i)
p[i] = new int[200];
This initializes a 100x200 array. You can access the bottom-right corner with p[99][199].
When it's time to delete the pointer you have to reverse the process:
for (int i = 0; i < 100; ++i)
delete [] p[i];
delete [] p;
Related
I am confused about how to create a dynamic defined array:
int *array = new int[n];
I have no idea what this is doing. I can tell it's creating a pointer named array that's pointing to a new object/array int? Would someone care to explain?
new allocates an amount of memory needed to store the object/array that you request. In this case n numbers of int.
The pointer will then store the address to this block of memory.
But be careful, this allocated block of memory will not be freed until you tell it so by writing
delete [] array;
int *array = new int[n];
It declares a pointer to a dynamic array of type int and size n.
A little more detailed answer: new allocates memory of size equal to sizeof(int) * n bytes and return the memory which is stored by the variable array. Also, since the memory is dynamically allocated using new, you should deallocate it manually by writing (when you don't need anymore, of course):
delete []array;
Otherwise, your program will leak memory of at least sizeof(int) * n bytes (possibly more, depending on the allocation strategy used by the implementation).
The statement basically does the following:
Creates a integer array of 'n' elements
Allocates the memory in HEAP memory of the process as you are using new operator to create the pointer
Returns a valid address (if the memory allocation for the required size if available at the point of execution of this statement)
It allocates space on the heap equal to an integer array of size N,
and returns a pointer to it, which is assigned to int* type pointer called "array"
It allocates that much space according to the value of n and pointer will point to the array i.e the 1st element of array
int *array = new int[n];
The new operator is allocating space for a block of n integers and assigning the memory address of that block to the int* variable array.
The general form of new as it applies to one-dimensional arrays appears as follows:
array_var = new Type[desired_size];
In C/C++, pointers and arrays are (almost) equivalent.
int *a; a[0]; will return *a, and a[1]; will return *(a + 1)
But array can't change the pointer it points to while pointer can.
new int[n] will allocate some spaces for the "array"
As of C++11, the memory-safe way to do this (still using a similar construction) is with std::unique_ptr:
std::unique_ptr<int[]> array(new int[n]);
This creates a smart pointer to a memory block large enough for n integers that automatically deletes itself when it goes out of scope. This automatic clean-up is important because it avoids the scenario where your code quits early and never reaches your delete [] array; statement.
Another (probably preferred) option would be to use std::vector if you need an array capable of dynamic resizing. This is good when you need an unknown amount of space, but it has some disadvantages (non-constant time to add/delete an element). You could create an array and add elements to it with something like:
std::vector<int> array;
array.push_back(1); // adds 1 to end of array
array.push_back(2); // adds 2 to end of array
// array now contains elements [1, 2]
I have this code declaring an array of pointers and assigning value of 5 to the location that the first pointer in the array points to:
int *p[10];
*p[0]=5;
However, this is showing an EXC_BAD_ACCESS error. I tried
int **p = new int *[10];
*p[0]=5;
But this is giving the same error. How do I assign a value to the location pointed by an element of my array of pointers? Thank you.
Before dereferencing the first pointer in the array, you need to initialize it to point to a valid memory location. For example:
int *p[10];
p[0] = new int;
*p[0] = 5;
Note that manual memory allocation at such level is almost certainly not what you want to use, except as a learning exercise, or as part of implementation of a larger structure. If you in fact want to an array of 10 integers, and access parts of it through a pointer, you can do it like this:
int array[10]; // allocate room for ten integers
array[0] = 5; // initialize the first one
int *first = &array[0]; // get a pointer to the first one
assert(*first == 5); // work with the pointer
If you do need an array of pointers, look into std::unique_ptr and std::shared_ptr to make sure they do not leak data in case of exceptions, early returns or forgotten delete.
I'm learning C++ and I have a question about pointers.
I have this code:
int* max = new int;
*max = 0;
I think, I have created a pointer on the heap (if I'm not right, please, tell me).
One question:
Is there a way to create an initialize a pointer with one instruction? (now, I'm using two instructions).
Another question:
How can I create a pointer on the stack?
I see code like this one:
int myVar = 20;
int* pVar = &myVar;
I think I haven't created a pointer on the stack but I think it is the only way to not create a pointer on the heap.
I'm very very very new on C++ development.
int* max = new int;
The above line creates a pointer on the stack and initializes it with an integer stored on the heap. Whenever new is involved in an expression, it will return a pointer to its dynamically-created operand:
Per Paragraph 5.3.4/2 of the C++11 Standard:
Entities created by a new-expression have dynamic storage duration (3.7.4). [ -- ] If the entity is a non-array object, the new-expression returns a pointer to the object created. If it is an array, the new-expression returns a pointer to the initial element of the array.
int myVar = 20;
int* pVar = &myVar;
In this example, both the pointer and its value is stored on the stack. A new-expression was not involved in the assignment, so nothing is created on the heap in this situation.
If you want to initialize the value of the pointed-to object in one line, you'd have to value-initialize it like so:
int* max = new int(5);
or in C++11 you can use uniform-initialization:
int* max = new int{5};
It is also important that you remember to delete that which you created with new. Since the memory is in dynamic allocation, its lifetime is not dependent by the scope in which it was created. If you forget to delete your program will get a memory leak.
delete max;
And if max was a pointer set to an array created by a new-expression, you'd use delete[]:
delete[] max;
Note: If a pointer was not initialized by a new-expression, then there is no need to delete.
It's typically recommended that you use containers to mangage the memory for you. Something like std::unique_ptr will do. Once its destructor is called, the memory it holds is deleted:
std::unique_ptr<int> max{new int{5}};
In C++14 we have make_unique:
auto max = std::make_unique<int>(5);
Pointers are normal variables, which content is a memory address. This memory can be heap memory or stack memory. Don't confuse the pointer with the memory space it points to.
Your first code allocates space on the heap which can hold an int. You store a pointer to that memory on the stack.
Your second code allocates space on the stack which can hold an int. You store a pointer to that memory on the stack.
So both pointers are on the stack, but only the second points to the stack.
In both cases, the type you allocate is a primitive type. Primitive types aren't initialized per default, unless you immediately assign a value to it (your second code), or use the constructor syntax, which also works with heap-allocated values:
int *max = new int(0);
The same syntax can be used in your second code, by the way:
int myVar(20);
In case you're interested: You can also define pointers on the heap. int* is the type of pointers to ints, so just go ahead and allocate such a type on the heap:
new int*();
This expression returns an int**, which you can then store somewhere. Again, you typically store this pointer-to-pointer on the stack:
int **pointerToPointer = new int*();
As with ints, you can initialize an int* in the new-expression to some pointer-to-int (here, the max pointer from above):
int **pointerToPointer = new int*(max);
Now you have two pointers with the same address: max (pointer on the stack) and some pointer on the heap which you point to using pointerToPointer, i.e. the following holds (I dereference `pointerToPointer, which leads to the value stored behind this pointer, which is a pointer to an int):
max == *pointerToPointer
Is there a way to create an initialize a pointer with one instruction?
yes:
int * a = new int(20); //initialized with 20;
How can I create a pointer on the stack?
int myVar = 20;
int* pVar = &myVar; //copied your code snipped It's alright
Your first example
int* max = new int;
*max = 0;
is indeed creating a new int on the heap, and your variable max saves a pointer to that int. If you're going to use this you're going to have to use delete max; when you no longer need it to avoid memory leaks.
The second example
int myVar = 20;
int* pVar = &myVar;
is creating a int on the stack, pVar now is a pointer to the address where the int is saved in the memory. However if you're using this you don't have to delete the pVar because its not on the heap (you didn't use the new keyword).
The main difference between the two variables (created on heap and stack) is that the stack variable is going to get deleted automatically when it leaves the scopes. The scope is defined by curly braces {}:
int* somefnc()
{
int e1
{
int* e2 = new int(0);
int e3 = 0;
} // int e3 gets automatically deleted here
return e2;
} // e1 gets automatically deleted here
// e2 still exists and has to be manually deleted.
One of the advantage of pointers is when dealing with arrays. If you were to make a char array of x elements on the stack you have to know the number of elements at compile time. If you want to make a char array of x elements at runtime with a dynamic number of elements you'd have to use char* ar = new char[x]; and then access it by ar[x-1] = '\n';.
In order to initialise a variable when creating it on heap you can use var x = new var(args);.
you can pass the value as argument to initialize the memory in heap.
int *iptr = new int(20);
it is initialized with value 20;
stack: contains local variables. so when you create a pointer and assign it with local objects. its pointing to variables in stack. we are not creating pointer on heap
.
I'm familiar with Java and trying to teach myself C/C++. I'm stealing some curriculum from a class that is hosting their materials here. I unfortunately can't ask the teacher since I'm not in the class. My concern is with the section under "dynamically declared arrays":
If you
want to be able to alter the size of
your array at run time, then declare
dynamic arrays. These are done with
pointers and the new operator. For the
basics on pointers, read the pointers
section.
Allocate memory using new, and then
you access the array in the same way
you would a static array. For example,
int* arrayPtr = new int[10]; for
(int i = 0; i < 10; i++) {
arrayPtr[i] = i; }
The memory picture is identical to the
static array, but you can change the
size if you need to. Don't forget you
must deallocate the memory before
allocating new memory (or you will
have a memory leak).
delete [] arrayPtr; // the []
is needed when deleting array pointers
arrayPtr = new int[50]; . . .
When you're completely done with the
array, you must delete its memory:
delete [] arrayPtr;
Dynamic multi-dimensional arrays are
done in a similar manner to Java. You
will have pointers to pointers. For an
example, see a
My understanding is that an array in C is simply a reference to the memory address of the first element in the array.
So, what is the difference between int *pointerArray = new int[10]; and int array[10]; if any?
I've done some tests that seem to indicate that they do the exact same thing. Is the website wrong or did I read that wrong?
#include <cstdlib>
#include <iostream>
using namespace std;
int main(int argc, char** argv) {
// Initialize the pointer array
int *pointerArray = new int[10];
for (int i = 0; i < 10; i++){
pointerArray[i] = i;
}
// Initialize the regular array
int array[10];
for (int i = 0; i < 10; i++){
array[i]= i;
}
cout << *(pointerArray + 5) << endl;
cout << *(array + 5) << endl;
cout << pointerArray[5] << endl;
cout << array[5] << endl;
cout << pointerArray << endl;
cout << array << endl;
return 0;
}
Output:
5
5
5
5
0x8f94030
0xbfa6a37c
I've tried to "dynamically re-size" my pointer array as described on the site, but my new (bigger) pointer array ends up filled with 0's which is not very useful.
int array[10]; declares the array size statically, that means it is fixed - which is the only major difference. It also might be allocated to be inside the function's stack frame, i.e. on the program's stack. You do not need to worry about using delete [] on that kind of array, in fact, you might crash the program if you delete it.
When you use operator new, you allocate memory dynamically which could be slower and the memory usually comes from the heap rather than the program's stack (though not always). This is better in most cases, as you are more limited in the stack space than the heap space. However, you must watch out for memory leaks and delete[] your stuff when you don't need it anymore.
As to your array being filled with zeros, what your class material does not say is that you have to do this:
int *arr = new int[20]; // old array
//do magic here and decide that we need a bigger array
int *bigger = new int[50]; // allocate a bigger array
for (int i = 0; i < 20; i++) bigger[i] = arr[i]; // copy the elements from the old array into the new array
delete[] arr;
arr = bigger;
That code extends the array arr by 30 more elements. Note that you must copy the old data into the new array, or else it will not be there (in your case, everything becomes 0).
My understanding is that an array in C is simply a reference to the memory address of the first element in the array.
So, what is the difference between int *pointerArray = new int[10]; and int array[10]; if any?
What you mention is the reason for much confusion in any C/C++ beginner.
In C/C++, an array corresponds to a block of memory sufficiently large to hold all of its elements. This is associated to the [] syntax, like in your example:
int array[10];
One feature of C/C++ is that you can refer to an array by using a pointer to its type. For this reason, you are allowed to write:
int* array_pointer = array;
which is the same as:
int* array_pointer = &array[0];
and this allows to access array elements in the usual way: array_pointer[3],
but you cannot treat array as a pointer, like doing pointer arithmetics on it (i.e., array++ miserably fails).
That said, it is also true that you can manage arrays without using the [] syntax at all and just allocate arrays by using malloc and then using them with raw pointers. This makes the "beauty" of C/C++.
Resuming: a distinction must be made between the pointer and the memory that it points to (the actual array):
the [] syntax in declarations (i.e., int array[10];) refers to both aspects at once (it gives you, as to say, a pointer and an array);
when declaring a pointer variable (i.e., int* p;), you just get the pointer;
when evaluating an expression (i.e., int i = p[4];, or array[4];), the [] just means dereferencing a pointer.
Apart from this, the only difference between int *pointerArray = new int[10]; and int array[10]; is that former is allocated dynamically, the latter on the stack.
Dynamically allocated:
int * pointerArray = new int[10];
[BTW, this is a pointer to an array of 10 ints, NOT a pointer array]
Statically allocated (possibly on the stack):
int array[10];
Otherwise they are the same.
The problem with understanding C/C++ arrays when coming from Java is that C/C++ distinguishes between the array variable and the memory used to store the array contents. Both concepts are important and distinct. In Java, you really just have a reference to an object that is an array.
You also need to understand that C/C++ has two ways of allocating memory. Memory can be allocated on the help or the stack. Java doesn't have this distinction.
In C and C++, an array variable is a pointer to the first element of the array. An array variable can exist on the heap or the stack, and so can the memory that contains its contents. And they can be difference. Your examples are int arrays, so you can consider the array variable to be an int*.
There are two differences between int *pointerArray = new int[10]; and int array[10];:
The first difference is that the memory that contains the contents of the first array is allocated on the heap. The second array is more tricky. If array is a local variable in a function then its contents are allocated on the stack, but if it is a member variable of a class then its contents are allocated wherever the containing object is allocated (heap or stack).
The second difference is that, as you've realised, the first array is dynamic: its size can be determined at run-time. The second array is fixed: the compiler must be able to determine its size at compile time.
First, I'd look for some other place to learn C++. The page you cite is
very confusing, and has little to do with the way one actually programs
in C++. In C++, most of the time, you'd use std::vector for an array,
not the complex solutions proposed on the page you cite. In practice,
you never use operator new[] (an array new).
In fact, std::vector is in some ways more like ArrayList than simple
arrays in Java; unlike an array in Java, you can simply grow the vector
by inserting elements into it, preferrably at the end. And it supports
iterators, although C++ iterators are considerably different than Java
iterators. On the other hand, you can access it using the []
operator, like a normal array.
The arrays described on the page you cite are usually called C style
arrays. In C++, their use is mostly limited to objects with static
lifetime, although they do occasionally appear in classes. In any case, they are never allocated dynamically.
The main difference is that some operations that are allowed on pointers are not allowed on arrays.
On the one hand:
int ar[10];
is using memory allocated on the stack. You can think of it also locally available and while it is possible to pass a pointer / reference to otehr functions, the memory will be freed as soon as it goes out of scope (in your example at the end of the main method but that's usually not the case).
On the other hand:
int ar* = new int[10];
allocates the memory for the array on the heap. It is available until your whole program exits or it is deleted using
delete[] ar;
note, that for delete you need the "[]" if and only if the corresponding new has had them, too.
There is a difference but not in the area that you point to. *pointerArray will point to the beginning of a block of memory of size 10 bytes. So will array. The only difference will be where it is stored in memory. pointerArray is dynamically assigned memory (at run-time) and hence will go on the heap, while array[10] will be allocated at compile-time and will go to the stack.
It is true that you can get most of array functionality by using a pointer to its first element. But compiler knows that a static array is composed of several elements and the most notable difference is the result of the sizeof operator.
sizeof(pointerArray) = sizeof int*
sizeof(array) = 10 * sizeof int
In many tutorials, the first code samples about dynamic memory start along the lines of:
int * pointer;
pointer = new int; // version 1
//OR
pointer = new int [20]; // version 2
They always proceed to explain how the second version works, but totally avoid talking about the first version.
What I want to know is, what does pointer = new int create? What can I do with it? What does it mean? Every tutorial without fail will avoid talking about the first version entirely. All I've found out (through messing about) is this:
#include <iostream>
using namespace std;
int main()
{
int * pointer;
pointer = new int;
pointer[2] = 1932; // pointer [2] exists? and i can assign to it?!
cout << pointer[2] << endl; // ... and access it successfully?!
};
The fact that I can subscript pointer tells me so far that pointer = new int implicitly creates an array. But if so, then what size is it?
If someone could help clear this all up for me, I'd be grateful...
My teacher explained it like this.
Think of cinema. The actual seats are memory allocations and the ticket you get are the pointers.
int * pointer = new int;
This would be a cinema with one seat, and pointer would be the ticket to that seat
pointer = new int [20]
This would be a cinema with 20 seats and pointer would be the ticket to the first seat. pointer[1] would be the ticket to the second seat and pointer[19] would be the ticket to the last seat.
When you do int* pointer = new int; and then access pointer[2] you're letting someone sit in the aisle, meaning undefined behaviour
This is a typical error in C and C++ for beginners. The first sentence, creates a space for holding just an int. The second one creates a space for holding 20 of those ints. In both cases, however, it assigns the address of the beginning of the dynamically-reserved area to the pointer variable.
To add to the confusion, you can access pointers with indices (as you put pointer[2]) even when the memory they're pointing is not valid. In the case of:
int* pointer = new int;
you can access pointer[2], but you'd have an undefined behavior. Note that you have to check that these accesses don't actually occur, and the compiler can do usually little in preventing this type of errors.
This creates only one integer.
pointer = new int; // version 1
This creates 20 integers.
pointer = new int [20] // version 2
The below is invalid, since pointer[2] translates as *(pointer + 2) ; which is not been created/allocated.
int main()
{
int * pointer;
pointer = new int;
pointer[2] = 1932; // pointer [2] exists? and i can assign to it?!
cout << pointer[2] << endl; // ... and access it succesfuly?!
};
Cheers!
new int[20] allocates memory for an integer array of size 20, and returns a pointer to it.
new int simply allocates memory for one integer, and returns a pointer to it. Implicitly, that is the same as new int[1].
You can dereference (i.e. use *p) on both pointers, but you should only use p[i] on the pointer returned by the new int[20].
p[0] will still work on both, but you might mess up and put a wrong index by accident.
Update: Another difference is that you must use delete[] for the array, and delete for the integer.
pointer = new int allocates enough memory on the heap to store one int.
pointer = new int [20] allocates memory to store 20 ints.
Both calls return a pointer to the newly allocated memory.
Note: Do not rely on the allocated memory being initialized, it may contain random values.
pointer = new int; allocates an integer and stores it's address in pointer. pointer[2] is a synonym for pointer + 2. To understand it, read about pointer arithmetic. This line is actually undefined behavior, because you are accessing memory that you did not previously allocate, and it works because you got lucky.
int* p = new int allocates memory for one integer. It does not implictly create an array. The way you are accessing the pointer using p[2] will cause the undefined behavior as you are writing to an invalid memory location. You can create an array only if you use new[] syntax. In such a case you need to release the memory using delete[]. If you have allocated memory using new then it means you are creating a single object and you need to release the memory using delete.
*"The fact that i can subscript pointer tells me so far that I pointer = new int implicitly creates an array. but if so, then what size is it?"
*
This was the part of the question which I liked the most and which you emphasize upon.
As we all know dynamic memory allocation makes use of the space on the Stack which is specific to the given program.
When we take a closer look onto the definition of new operator :-
void* operator new[] (std::size_t size) throw (std::bad_alloc);
This actually represents an array of objects of that particular size and if this is successful, then it automatically Constructs each of the Objects in the array. Thus we are free to use the objects within the bound of the size because it has already been initialized/constructed.
int * pointer = new int;
On the other hand for the above example there's every possibility of an undefined behaviour when any of
*(pointer + k) or *(k + pointer)
are used. Though the particular memory location can be accessed with the use of pointers, there's no guarantee because the particular Object for the same was not created nor constructed.This can be thought of as a space which was not allocated on the Stack for the particular program.
Hope this helps.
It does not create array. It creates a single integer and returns the pointer to that integer. When you write pointer[2] you refer to a memory which you have not allocated. You need to be carefull and not to do this. That memory can be edited from the external program which you, I belive, don't want.
int * pointer; pointer = new int; // version 1
//OR
pointer = new int [20] // version 2
what I want to know is, what does pointer = new int create? what can I do with it? what does it mean? Every tutorial without fail will avoid talking about the first version entirely
The reason the tutorial doesn't rell you what to do with it is that it really is totally useless! It allocates a single int and gives you a pointer to that.
The problem is that if you want an int, why don't you just declare one?
int i;