I come across a code as follow:
int* list = new int[5];
I never saw such declaration. What does it do? I suppose list is a pointer to an int? Using VS debug environment, I can see list having one int value. So how are the other element accessed? Thanks.
The statement is called badly-mannered dynamic (memory) allocation, involving two operators, new (new[]) and delete (delete[]). The new operator allocates memory and the delete operator deallocates memory.
Since it's a chunk of memory assigned to a pointer, it behaves mostly the same as an array (int list[5]), so you should access it as usually as you do with a pointer. Like
int *list = new int[5];
list[0] = 10; list[3] = 25; // Whatever
for (int i = 0; i < 5; i++)
list[i] = 1+1;
You can see a list of topics about dynamic allocation in C++ on Stack Overflow here.
This is not a linked list, but a construct known as dynamic memory. If you want to create an array normally, you would write something like
int myArray[3] = {
1, 2, 3
};
Notice that when I write the size of myArray (3), I use a constant expression; If I were to write
int x = 3;
int myArray[x] = {
1, 2, 3
};
It wouldn’t work, because x is a variable, not a constant expression. Note that even if x is a const, this still wouldn’t work. This is because before the program is run, memory needs to be allocated for the arrays.
When you say int * myData = new int[3], you can create memory on the fly, without constant expressions. In this case, it would be perfectly fine to do this
int x = 3;
int * myData = new int[x];
However, myData is a pointer, not an array, so the common sizeof(x) / sizeof(*x) trick doesn’t work here.
Make sure that after your dynamic memory is no longer of use, you write delete[] myData. This de-allocates the memory block and allows other memory to be stored in its place.
This memory is a block of memory just like an array. A linked list, however, is totally different, and would take a while to explain. Check out the Wikipedia on it.
Related
I have a simple swapping function to take an integer array, and return a new array with swapped values.
int* Node::dataSwap(int *data, int n_index, int swap_index){
printDatt(data);
int *path = data;
int swapped = data[n_index];
int to_swap = data[swap_index];
path[n_index] = to_swap;
path[swap_index] = swapped;
printDatt(data);
return path;
}
However, the reference to the original data is being altered by this function. The output looks something like this (printing the should be the same data to console).
0, 1, 2
3, 4, 5
6, 7, 8
0, 1, 2
3, 4, 8
6, 7, 5
Why is "data" being changed when I am not changing it? Is "path" a reference to the actual mem addr of "data"?
The type of the argument data and the local variable path is int *. You can read this as "pointer to int".
A pointer is a variable holding a memory address. Nothing more, nothing less. Since you set path = data, those two pointers are equal.
In your mind, data is an array. But that's not what the function dataSwap is seeing. To the function dataSwap, its argument data is just a pointer to an int. This int is the first element of your array. You accessed elements of the array using data[n_index]; but that's just a synonym for *(data + n_index).
How to remedy to your problem?
The C way: malloc and memcpy
Since you want to return a new array, you should return a new array. To do this, you should allocate a new region of memory with malloc, and then copy the values of the original array to the new region of memory, using memcpy.
Note that it is impossible to do this using only the current arguments of the function, since none of those arguments indicate the size of the array:
data is a pointer to the first element of the array;
n_index is the index of one of the elements in the array;
swap_index is the index of another element in the array.*
So you should add a fourth element to the function, int size, to specify how many elements are in the array. You can use size as argument to malloc and memcpy, or to write a for loop iterating over the elements of the array.
New problem arising: if you call malloc to allocate new memory, then the user will have to call free to free the memory at some point.
C++ has the cool keyword new whose syntax is somewhat lighter than the syntax of malloc. But this doesn't solve the main problem; if you allocate new memory with the keyword new, then the user will have to free the memory with the keyword delete at some point.
Urgh, so much burden!
But this was the C way. A good rule of thumb in C++ is: never handle arrays manually. The standard library has std::vector for that. There are situations where using new might be the best solution; but in most simple cases, it isn't.
The C++ way: std::vector
Using the class std::vector from the standard library, your code becomes:
#include <vector>
std::vector<int> Node::dataSwap(std::vector<int> data, int n_index, int swap_index)
{
std::vector<int> new_data = data;
int swapped = data[n_index];
int to_swap = data[swap_index];
new_data[n_index] = to_swap;
new_data[swap_index] = swapped;
return (new_data);
}
No malloc, no new, no free and no delete. The class std::vector handles all that internally. You don't need to manually copy the data either; the initialisation new_data = data calls the copy constructor of class std::vector and does that for you.
Avoid using new as much as you can; use a class that handles all the memory internally, like you would expect it in a higher-level language.
Or, even simpler:
The C++ way: std::vector and std::swap
#include <vector>
#include <algorithm>
std::vector<int> Node::dataSwap(std::vector<int> data, int n_index, int swap_index)
{
std::vector<int> new_data = data;
std::swap(new_data[n_index], new_data[swap_index]);
return (new_data);
}
Is "path" a reference to the actual mem addr of "data"?
Yes! In order to create a new array that is a copy of the passed data (only with one pair of values swapped over), then your function would need to create the new array (that is, allocate data for it), copy the passed data into it, then perform the swap. The function would then return the address of that new data, which should be freed later on, when it is no longer needed.
However, in order to do this, you would need to also pass the size of the data array to the function.
One way to do this, using 'old-style' C++, is with the new operator. With the added 'size' parameter, your function would look something like this:
int* Node::dataSwap(int *data, int n_index, int swap_index, int data_size)
{
printDatt(data);
int *path = new int[data_size]; // Create new array...
for (int i = 0; i < data_size; ++i) path[i] = data[i]; // ... and copy data
int swapped = data[n_index];
int to_swap = data[swap_index];
path[n_index] = to_swap;
path[swap_index] = swapped;
printDatt(data);
return path; // At some point later on, your CALLING code would "delete[] path"
}
You are changing the memory at which the pointer path point and that is data. I think try to understand better how the pointers works will help you. :)
Then you can use the swap function from the std library:
std::swap(data[n_index], data[swap_index]);
It will make your code nicer.
I am working on an assignment for my c++ class where I need to dynamically assign a new int, array, and pointer from another pointer to practice dynamic memory allocation.
At first, I was struggling with creating a new int to provide an int for my new array, but I got it to compile and was wondering if my declarations were correct.
int *dmaArray = new int;
*dmaArray = 4;
I then took that and put it into a dynamically created array, but I don't know how to declare values of the array as it errors out saying "cannot convert to int". I did some thinking and I believe it's because it was declared and needs to be initialized at the declaration; I can't because the declaration is already a declaration (new) in itself.
int * nodeValues = new int[*dmaArray];
nodeValues[*dmaArray] = {6, 2, 28, 1};
A loop wouldn't work to assign values after since the values aren't consecutive or in any pattern. (well, regardless, I would need to use an array because the assignment said so.
This is not how to to declare a dynamic array and initialize it:
int * nodeValues = new int[*dmaArray];
nodeValues[*dmaArray] = {6, 2, 28, 1};
So you declare it this way:
int* nodeValues = new int[dmaArray];
And to assign values to it use loops or manually:
nodeValues[0] = 6;
nodeValues[1] = 2;
nodeValues[2] = 28,
nodeValues[3] = 1;
Remember arrays use indexes to read/write its elements as the fact being some sort of data of the same type contiguous to each other in memory.
So if you want to print the array:
for(auto i(0); i != dmaArray; ++i)
std::cout << nodeValues[i] << ", ";
Finally you should clean memory dynamically allocated after finishing with it because the compiler doesn't do it for you:
delete[] nodeValues;
I figured out how:
int * nodeValues = new int[*dmaArray]{6,5,28,1};
I am a c++ newbie. While learning I came across this.
if I have a pointer like this
int (*a)[2][3]
cdecl.org describe this as declare a as pointer to array 2 of array 3 of int:
When I try
int x[2][3];
a = &x;
this works.
My question is how I can initialize a when using with new() say something like
a = new int [] [];
I tried some combinations but doesn't get it quite right.
Any help will be appreciated.
You will have to do it in two steps - first allocate an array of pointers to pointers(dynamically allocated arrays) and then, allocate each of them in turn. Overall I believe a better option is simply to use std::vector - that is the preferred C++ way of doing this kind of things.
Still here is an example on how to achieve what you want:
int a**;
a = new int*[2];
for (int i =0; i< 2;++i){
a[i] = new int[3]
}
... use them ...
// Don't forget to free the memory!
for (int i = 0; i< 2; ++i) {
delete [] a[i];
}
delete [] a;
EDIT: and as requested by Default - the vector version:
std::vector<std::vector<int> > a(2, std::vector<int>(3,0));
// Use a and C++ will take care to free the memory.
It's probably not the answer you're looking for, but what you
need is a new expression whose return type is (*)[2][3] This
is fairly simple to do; that's the return type of new int
[n][2][3], for example. Do this, and a will point to the
first element of an array of [2] of array of [3] int. A three
dimensional array, in sum.
The problem is that new doesn't return a pointer to the top
level array type; it returns a pointer to the first element of
the array. So if you do new int[2][3], the expression
allocates an array of 2 array of 3 int, but it returns
a pointer to an array of 3 int (int (*a)[3]), because in C++,
arrays are broken (for reasons of C compatibility). And there's
no way of forcing it to do otherwise. So if you want it to
return a pointer to a two dimensional array, you have to
allocate a three dimensional array. (The first dimension can be
1, so new [1][2][3] would do the trick, and effectively only
allocate a single [2][3].)
A better solution might be to wrap the array in a struct:
struct Array
{
int data[2][3];
};
You can then use new Array, and everything works as expected.
Except that the syntax needed to access the array will be
different.
I would like to know if there is a way to delete a pointer array without touching the pointed objects in memory.
I'm writing a restriction routine for a HashSet I implemented a couple of days ago, so when the hash table is full it gets replaced by another double sized table. I'm representing the hash table using an array of pointers to an object (User), and the array itself is declared dynamically in my HashSet class, so it can be deleted after copying all its content to the new table using a hash function.
So basically I need to:
Declare another table with a size that equals the double of the original array size.
Copy every pointer to User objects from my original array to the new one applying my hash function (it gets the User object from memory and it calculates the index using a string that represents the user's name).
After inserting all the pointers from the original array to the new one, I will have to free the allocated memory for the original array and replace the pointer in my HashSet class (member private userContainer) with the location of the new one (array).
The problem is that if I use delete[] userContainer to free the allocated memory for it, it will also delete every object in memory so the newly created replacement array will point to freed positions in memory!
What you describe does not sound right.
Let's say you have a class A and you create an array of As with:
A** array1 = new A*[32];
Then fill it:
for(int i = 0; i < 32; ++i)
array1[i] = new A();
Doing a delete[] array1 does not free the elements of array1.
So this is safe:
A** array1 = new A*[32];
for(int i = 0; i < 32; ++i)
array1[i] = new A();
A** arary2 = new A*[64];
for(i = 0; i < 32; ++i)
array2[i] = array1[i];
delete [] array1;
for(i = 0; i < 32; ++i)
// do something with array2[i]
In general, when you delete an array of pointers, whatever objects the pointers pointed to remain in existence. In fact, this is a potential source of large memory leaks.
But in some sort of reference-counted environment (eg, Objective-C or Qt), when you delete an array OBJECT (vs a simple [] array) then the reference counts are decremented and the objects will be deleted if the count goes to zero.
But if you're restructuring a hash table you'd better have somehow saved the pointer values before you delete the array, or else all the addressed objects will be lost. As you save them you can increment their reference counts (if you do it right).
(It would help to know what language you're dealing with, and what you mean by "array".)
I don't think your problem exists. Here's a baby example to show that there's nothing to worry about:
Foo * brr[10];
{
Foo * arr[10];
// This is not touching the objects!
for (Foo * it = arr; it != arr + 10; ++it) *it = new Foo;
std::copy(arr, arr + 10, brr);
} // no more arr
for (Foo * it = brr; it != brr + 10; ++it) delete *it; // fine
You can copy the pointers around freely as much as you like. Just remember to delete the object to which the pointers point when they're no longer needed.
A perhaps trivial reminder: Pointers don't have destructors; in particular, when a pointer goes out of scope, nothing happens.
Do you know the difference between malloc/free, new/delete and new[]/delete[]?
I figure that you might want to not use new[]/delete[] in your situation, as you don't want destructors to be called I guess?
I have 2 doubts regarding basics of pointers usage.
With the following code
int (*p_b)[10];
p_b = new int[3][10];
// ..do my stuff
delete [] p_b
p_b is pointing to an array of 3 elements, each having fixed-size length of 10 int.
Q1:
How to declare p_b if I want that each element be a pointer to a fixed array size?
Basically I want the following
p_b[0] = pointer to a fixed-array size of 10
p_b[1] = pointer to a fixed-array size of 10
// ... and so on
I was thinking to int (** p_b)[10] but then I don't know how to use new to allocate it? I would like to avoid falling back to more general int** p_b
Q2:
Is per my original code sample above, how to call new so that p_b points to a unique fixed-size array of 10 int other than calling p_b = new int[1][10] ? To free memory I have to call delete[] while I cannot find an expression where I can call only simply delete.
p_b is pointing to an array of 3 elements, each having fixed-size length of 10 int.
How to declare p_b if I want that each element be a pointer to a fixed array size?
Does your first sentence not completely cover that question?
Is per my original code sample above, how to call new so that p_b points to a unique fixed-size array of 10 int other than calling p_b = new int[1][10]? To free memory I have to call delete[] while I cannot find an expression where I can call only simply delete.
I completely do not understand why this is a problem, but you could do it by wrapping your array inside another type... say std::array, boost::array or std::vector.
First of all, if your new expression has square brackets (new somtype[somesize]), your delete has to have square brackets as well (delete [] your_pointer).
Second, right now you've defined p_b to be a single pointer to some data. If what you really want is an array of pointers, then you need to define it as an array. Since you apparently want three independent arrays, you'll have to allocate each of them separately. It's probably easiest if you start with a typedef:
typedef int *p_int;
p_int p_b[3];
Then you'll allocate your three arrays:
for (int i=0; i<3; i++)
p_b[i] = new int[10];
To delete those, you'll need to delete each one separately:
for (int i=0; i<3; i++)
delete [] p_b[i];
I definitely agree with #Tomalak that you should almost never mess with things like this yourself though. It's not clear what you really want to accomplish, but it's still pretty easy to guess that chances are quite good that a standard container is likely to be a simpler, cleaner way to do it anyway.
Here's an example of how to implement Q1:
int main()
{
typedef int foo[10];
foo* f = new foo[3];
f[0][5] = 5;
f[2][7] = 10;
delete [] f;
}
As for Q2, the only way to delete memory allocated with new[] is with delete[]. If you personally don't want to write delete [], you can use a vector or another STL container. Really, unless this is some hardcore uber-optimisation, you should be using vectors anyway. Never manage memory manually unless you are absolutely forced to.
To use a raw pointer to manage a 2-d array you must first create a pointer to a pointer to array element type that will point to each row of the array. Next, each row pointer must be assigned to the actual array elements for that row.
int main()
{
int **p;
// declare an array of 3 pointers
p = new int*[3];
// declare an array of 10 ints pointed to by each pointer
for( int i = 0; i < 3; ++i ) {
p[i] = new int[10];
}
// use array as p[i][j]
// delete each array of ints
for( int i = 0; i < 3; ++i ) {
delete[] p[i];
}
// delete array of pointers
delete[] p;
}
A far easier solution is to use std::array. If your compiler does not provide that class you can use std::vector also.
std::array<std::array<int,10>,3> myArr;
myArr[0][0] = 1;
For Q1, I think you want
int (*p[3])[10];
Try cdecl when you're unsure.
Your other question seems to be well answered by other answers.
regards,
Yati Sagade
Actually, nobody posted an answer to your exact question, yet.
Instead of
int (*p_arr)[10] = new int[3][10];
// use, then don't forget to delete[]
delete[] p_arr;
I suggest using
std::vector<std::array<int, 10>> vec_of_arr(3);
or if you don't need to move it around and don't need runtime length:
std::array<std::array<int, 10>, 3> arr_of_arr;
Q1
How to declare p_b if I want that each element be a pointer to a fixed array size?
int(**pp_arr)[10] = new std::add_pointer_t<int[10]>[3];
for (int i = 0; i < 3; ++i)
pp_arr[i] = new int[1][10];
// use, then don't forget to delete[]
for (int i = 0; i < 3; ++i)
delete[] pp_arr[i];
delete[] pp_arr;
The modern variant of that code is
std::vector<std::unique_ptr<std::array<int, 10>>> vec_of_p_arr(3);
for (auto& p_arr : vec_of_p_arr)
p_arr = std::make_unique<std::array<int, 10>>();
or if you don't need to move it around and don't need runtime length:
std::array<std::unique_ptr<std::array<int, 10>>, 3> arr_of_p_arr;
for (auto& p_arr : arr_of_p_arr)
p_arr = std::make_unique<std::array<int, 10>>();
Q2
Is per my original code sample above, how to call new so that p_b points to a unique fixed-size array of 10 int other than calling p_b = new int[1][10]?
Not without wrapping the array into another type.
std::array<int, 10>* p_arr = new std::array<int, 10>;
// use, then don't forget to delete
delete p_arr;
You can replace std::array<int, 10> with your favourite array-wrapping type, but you cannot replace it with a fixed-size array alias. The modern variant of that code is:
auto p_arr = std::make_unique<std::array<int, 10>>();