In C++, I want my class to have a char** field that will be sized with user input. Basically, I want to do something like this -
char** map;
map = new char[10][10];
with the 10's being any integer number. I get an error saying cannot convert char*[10] to char**. Why can it not do this when I could do -
char* astring;
astring = new char[10];
?
Because an array is not a pointer. Arrays decay into pointers to their first elements, but that happens only at the first level: a 2D array decays into a pointer to a 1D array, but that's it—it does not decay into a pointer to a pointer.
operator new[] allows to allocate a dynamic array of a size only known at runtime, but it only lets you allocate 1D arrays. If you want to allocate a dynamic 2D array, you need to do it in two steps: first allocate an array of pointers, then for each pointer, allocate another 1D array. For example:
char **map = new char*[10]; // allocate dynamic array of 10 char*'s
for(int i = 0; i < 10; i++)
map[i] = new char[10]; // allocate dynamic array of 10 char's
Then to free the array, you have to deallocate everything in reverse:
for(int i = 0; i < 10; i++)
delete [] map[i];
delete [] map;
On other hand, what the
map = new char[x][y];
really do is that, you new a array contain x items whose type is an array containing y items. And y must be a const integer.
So, if you really want your code pass without any problem, the correct way is
char (* map)[10] ;
map = new char[some_value_you_want][10];
Or more clearly,
typedef char array10[10];
array10 * map = new array10[some_value_you_want];
Multi-dimentional arrays are not supported in that manner. You need to allocate the two dimentions seperately.
char** map = new (char*)[10];
for( int i = 0; i < 10; ++i ) {
map[i] = new char[10];
}
Related
Consider the following case:
int **my_array = new int*[10];
What do we assign to my_array here?
my_array is a pointer that points to what?
Is there any way to iterate through my_array (the pointer) and set up a two-dimensional array of integers (and not int*)?
We assign to my_array the address of an array. The array contains pointers which can point to other arrays, but don't yet.
Yes, we can do this:
int **my_array = new int*[10];
for(int i=0; i<10; ++i)
my_array[i] = new int[13];
my_array[2][11] = 500;
What do we assign to my_array here?
You can assign an int* to the elements of my_array. E.g.
my_array[0] = new int[20];
or
int i;
my_array[0] = &i;
my_array is a pointer that points to what?
It points to an an array of 10 int* objects.
Is there any way to iterate through my_array (the pointer) and set up a two-dimensional array of integers (and not int*)?
Not sure what you are expecting to see here. An element of my_array can only be an int*.
If you want my_array to be a pointer to a 2D array, you may use:
int (*my_array)[20] = new int[10][20];
Now you can use my_array[0][0] through my_array[9][19].
PS If this is your attempt to understand pointers and arrays, it's all good. If you are trying to deploy this code in a working program, don't use raw arrays any more. Use std::vector or std::array.
For a 1D array, use:
// A 1D array with 10 elements.
std::vector<int> arr1(10);
For a 2D array, use:
// A 2D array with 10x20 elements.
std::vector<std::vector<int>> arr2(10, std::vector<int>(20));
how to allocate run time memory to an array of size[4][3]?
i.e int a[4][3]
If need is to allocate memory to an array at run time than how to allocate memory to 2D array or 3D array.
Editing the answer based on comments. Allocate separately for each dimension. For a 2D array a 2 level allocation is required.
*a = (int**)malloc(numberOfRows*sizeof(int*));
for(int i=0; i<numberOfRows; i++)
{
(*arr)[i] = (int*)malloc(numberOfColumns*sizeof(int));
}
The simplest way to allocate dynamically an array of type int[4][3] is the following
int ( *a )[3] = new int[4][3];
// some stuff using the array
delete []a;
Another way is to allocate several arrays. For example
int **a = new int * [4];
for ( size_t i = 0; i < 4; i++ ) a[i] = new int[3];
// some stuff using the array
for ( size_t i = 0; i < 4; i++ ) delete []a[i];
delete []a;
What have you tried. new int[4][3] is a perfectly valid
expression, and the results can be assigned to a variable with the
appropriate type:
int (*array2D)[3] = new int[4][3];
Having said that: I can't really think of a case where this
would be appropriate. Practically speaking, anytime you need
a 2 dimensional array, you should define a class which
implements it (using std::vector<int> for the actual memory).
A pure C approach is the following:
int (*size)[4][3];
size = malloc(sizeof *size);
/* Verify size is not NULL */
/* Example of access */
(*size)[1][2] = 89;
/* Do something useful */
/* Deallocate */
free(size);
The benefit is that you consume less memory by not allocating intermediate pointers, you deal with a single block of memory and deallocation is simpler. This is especially important if you start to have more than 2 dimensions.
The drawback is that the access syntax is more complicated, as you need to dereference a pointer before being able to index.
Use calloc, i guess this will do.
int **p;
p=(int**)calloc(4,sizeof(int));
In C you can use pointer to pointer
AS #Lundin mentioned this is not 2D array. It is a lookup table using pointers to fragmented memory areas allocated all over the heap.
You need to allocate how many pointers you need and then allocate each pointer. you can allocate fixed size or varaible size depending on your requirement
//step-1: pointer to row
int **a = malloc(sizeof(int *) * MAX_NUMBER_OF_POINTERS);
//step-2: for each rows
for(i = 0; i < MAX_NUMBER_OF_POINTERS; i++){
//if you want to allocate variable sizes read them here
a[i] = malloc(sizeof(int) * MAX_SIZE_FOR_EACH_POINTER); // where as if you use character pointer always allocate one byte extra for null character
}
Where as if you want to allocate char pointers avoid using sizeof(char) inside for loop. because sizeof(char) == 1 and do not cast malloc result.
see How to declare a 2d array in C++ using new
You could use std::vector<> since it is a templated container (meaning array elements can be whatever type you need). std::vector<> allows for dynamic memory usage (you can change the size of the vector<> whenever you need to..the memory is allocated and free'd automatically).
For example:
#include <iostream>
#include <vector>
using namespace std; // saves you from having to write std:: in front of everthing
int main()
{
vector<int> vA;
vA.resize(4*3); // allocate memory for 12 elements
// Or, if you prefer working with arrays of arrays (vectors of vectors)
vector<vector<int> > vB;
vB.resize(4);
for (int i = 0; i < vB.size(); ++i)
vB[i].resize(3);
// Now you can access the elements the same as you would for an array
cout << "The last element is " << vB[3][2] << endl;
}
You can use malloc() in c or new in c++ for dynamic memory allocation.
This question already has an answer here:
How to properly work with dynamically-allocated multi-dimensional arrays in C++ [duplicate]
(1 answer)
Closed 7 years ago.
I've been having bad luck with with dynamic pointers when I range them to 2 dimensions and higher. For example I want a pointer to a 2D array. I know that:
int A[3][4];
int (*P)[4] = A;
Is completely legit (even if I don't completely understand why). Taking into consideration that:
int *P = new int[4];
works, I imagined that:
int **P = new int[5][7];
Would also work, but it's not. This code states the error:
Error: A value of type "(*)[7]" cannot be used to initialize an entity of
type "int **"
By seeing this the new part becomes a pointer to an array of 7 integers I made:
int (*P)[4] = new int[7][4];
And this does work but it's not what I want to accomplish. By doing it like that I'm limited to at least using a constant value for any subsequent dimension, but I want it to be fully defined at run time and therefore "dynamic".
How could I go and make this multidimensional pointer work??
Let's start with some basic examples.
When you say int *P = new int[4];
new int[4]; calls operator new function()
allocates a memory for 4 integers.
returns a reference to this memory.
to bind this reference, you need to have same type of pointer as that of return reference so you do
int *P = new int[4]; // As you created an array of integer
// you should assign it to a pointer-to-integer
For a multi-idimensional array, you need to allocate an array of pointers, then fill that array with pointers to arrays, like this:
int **p;
p = new int*[5]; // dynamic `array (size 5) of pointers to int`
for (int i = 0; i < 5; ++i) {
p[i] = new int[10];
// each i-th pointer is now pointing to dynamic array (size 10)
// of actual int values
}
Here is what it looks like:
To free the memory
For one dimensional array,
// need to use the delete[] operator because we used the new[] operator
delete[] p; //free memory pointed by p;`
For 2d Array,
// need to use the delete[] operator because we used the new[] operator
for(int i = 0; i < 5; ++i){
delete[] p[i];//deletes an inner array of integer;
}
delete[] p; //delete pointer holding array of pointers;
Avoid memory leakage and dangling pointers!
You want something like:
int **P = new int*[7];
p[0] = new int[5];
p[1] = new int[5];
...
Another approach would be to use a 1D array as an 2D array. This way you only have to allocate the memory once (one continous block);
int *array;
size_t row=5,col=5;
array = (int*)malloc(row*col*sizeof(int)) //or new int[row*col]
This would result in the same as "int array[5][5]".
to access the fields you just do:
array[1 //the row you want
* col //the number of columns
+2//the column you want
] = 4;
This is equal to:
array[1][2];
This performs bounds checking on some debug compilers, uses dynamic size and deletes itself automatically. The only gotcha is x and y are the opposite way round.
std::vector<std::vector<int>> array2d(y_size, std::vector<int>(x_size));
for (int y = 0; y < y_size; y++)
{
for (int x = 0; x < x_size; y++)
{
array2d[y][x] = 0;
}
}
I'm trying to create a multi-dimensional array, the size of which the user will supply.
So far I have this:
int definedgroups; // for number of groups needed
cout << "Enter the Number of Groups you require: " << endl;
cin >> definedgroups;
const int definedgroups = definedgroups;
int User_Groups [definedgroups] [4];
I believe the array needs constant values, so i tried assigning my variable as a constant but still no luck.
In C++, static arrays, that is, those defined like this:
foo arrayStatic[bar];
require bar to be a constant integer. In other words, the programmer needs to know its value beforehand.
Whenever bar is unknown, a dynamic array could be used instead. They're defined like this:
foo* arrayDynamic;
arrayDynamic = new foo[bar];
Here, bar could be an integer variable.
Don't forget that dynamic memory must be deallocated eventually. So, in this case, we can deallocate arrayDynamic like this:
delete [] arrayDynamic;
A two-dimensional dynamic array is defined analogously:
foo** arrayDynamic2D;
arrayDynamic2D = new foo*[bar];
for (int i = 0; i < bar; i++)
arrayDynamic2D[i] = new foo[baz];
and deallocated in a similar fashion:
for (int i = 0; i < bar; i++)
delete [] arrayDynamic2D[i];
delete [] arrayDynamic2D;
Static memory is allocated in the stack whereas dynamic memory is allocated in the heap.
It's not possible to do it in C++ using static arrays. Use std::vector in a hierarchical way (i.e. vectors of vectors) to implement a multi-dimensional array easily (though not necessarily very efficiently).
E.g.
std::vector<std::vector<double> > array(nrows, std::vector<double>(ncols));
creates a nrows x ncols matrix.
You need dynamic memory allocation using new:
int **User_Groups = new int*[definedgroups];
//Allocate memory for 2nd dimension
for (int i = 0; i < 4; ++i)
User_Groups[i] = new int[4];
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>>();