I have a dynamically populated array of strings in C++:
string** A;
it is populated like this:
A = new string*[size1];
and then:
for (unsigned int i = 0; i < size1; i++)
{
A[i] = new string[size2];
for (unsigned int j = 0; j < size2; j++)
{
A[i][j] = whatever[j];
}
}
elsewhere, I want to find out the dimensions (size1 and size2).
I tries using this:
sizeof(A[i]) / sizeof(A[i][0])
but it doesn't work.
Any ideas ?
Thanks
When you allocate memory via new T[N], the value N is not stored anywhere . If you need to know it later, you will need to keep track of it in your code.
There are pre-existing classes for allocating memory that also remember the length that was allocated. In your code:
vector<vector<string>> A(size1, vector<string>(size2));
// (code to populate...)
then you can access A.size() to get size1, and A[0].size() to get size2.
If the dimensions are known at compile-time you may use array instead of vector.
It is very simple to find the size of a two dimensional (more exactly of one-dimensional dynamically allocated arrays) array. Just declare it like
std::vector<std::vector<std::string>> A;
and use
std::cout << A.size() << std::endl;
As for your approach then you have to store the sizes in some variables when the array is allocated.
If you are learning C++, I would recommend that you learn Classes. With a class you can encapsulate int variables along with your 2D array that you can use to store the dimensions of your array. For example:
class 2Darray{
string **array;
int rows;
int cols;
}
You can then get the dimensions of your 2Darray object anytime by reading these member variables.
vectors will do this for you behind the scenes but its good for you to learn how to do this.
You can't create an array just using pointer operator. Every array is basically a pointer with allocated memory. That's why compiler wants constant before creating array.
Basically; sizeof(A[i]) won't give you the size of array. Because sizeof() function will return the a pointers size which is points to A[i] location. sizeof(A[i]) / sizeof(A[i][1]) will probably give you 1 because you are basically doing sizeof(int)/sizeof(int*)
So you need to store the boundary yourself or use vectors. I would prefer vectors.
Can't get array dimensions through pointer(s)
Related
First of all, happy new year!
So, I'd like to ask if I could use some input values as the size of a bidimensional array, for example:
I'd like to know, if instead of doing this:
const int N = 10;
const int M = 10;
typedef int IntMatrix[N][M];
Let's say that would be the max size of the array I could create, but then the user inputs that the size must have a size of 5x5. I know I could then use 5x5 as a limit when doing stuff, but could I do like the same, but using the input values as the dimension of the Matrix?
Something like:
cin >> N >> M;
And then use that as the MAX size of each dimension.
Thanks for your help!
No. The size of an array must be known at compile time and can not be determined at runtime as described in this tutorial for example. Therefore, the size of the array cannot depend on user input.
What you can do, is allocate an array dynamically and store it's address in a pointer. The size of a dynamic array can be determined at runtime. However, there is a problem. Only the outermost dimension of a dynamically allocated 2D array may be determined at runtime. You have 2 options: Either you allocate a flat array of size NxM where the rows are stored continuously one after the other and you calculate the index using maths. Or, you use an array of pointers and assign each pointer to a dynamically allocated array column. The first option is more efficient.
There is another problem. Dynamic memory management is hard, and it's never a good idea to do it manually even if you know what you're doing. Much less if you don't. There is a container class in the standard library which takes care of memory management of dynamic arrays. It's std::vector. Always use it when you need a dynamic array. Your options stay similar. Either use a flat, NxM size vector, or a vector of vectors.
The array should be dynamically allocatedn because array size should be known at compile-time. You can do this way:
int N,M; // Dimensions
int** intMatrix; // Array of array
std::cin << N << M;
intMatrix = new int*[N]; // Allocate N the row
for(int i=0; i<N; i++){
intMatrix[i] = new int[M]; // For each row, allocate the col
}
// aaaand don't forget to free memory like this:
for(int i=0; i<N; i++){
delete [] intMatrix[i];
}
delete [] intMatrix;
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.
In c++, I can create a 2D array with fixed number of columns, say 5, as follows:
char (*c)[5];
then I can allocate memory for rows as follows
c = new char[n][5];
where n can be any variable which can be assigned value even at run time. I would like to know whether and how can I dynamically allocate variable amount of memory to each row with this method. i.e. I want to use first statement as such but can modify the second statement.
Instead of a pointer to an array, you'd make a pointer to a pointer, to be filled with an array of pointers, each element of which is in turn to be filled with an array of chars:
char ** c = new char*[n]; // array of pointers, c points to first element
for (unsigned int i = 0; i != n; ++i)
c[i] = new char[get_size_of_array(i)]; // array of chars, c[i] points to 1st element
A somewhat more C++ data structure would be a std::vector<std::string>.
As you noticed in the comment, dynamic arrays allocated with new[] cannot be resized, since there is no analogue of realloc in C++ (it doesn't make sense with the object model, if you think about it). Therefore, you should always prefer a proper container over any manual attempt at dynamic lifetime management.
In summary: Don't use new. Ever. Use appropriate dynamic containers.
You need to declare c as follows: char** c; then, allocate the major array as follows: c = new char*[n]; and then, allocate each minor array as follows: c[i] = new char[m]
#include <iostream>
using namespace std;
main()
{
int row,col,i,j;
cout<<"Enter row and col\n";
cin>>row>>col;
int *a,(*p)[col]=new (int[row][col]);
for(i=0;i<row;i++)
for(j=0;j<col;j++)
p[i][j]=i+j;
for(i=0;i<row;i++)
for(j=0;j<col;j++)
cout<<i<<" "<<j<<" "<<p[i][j]<<endl;
//printf("%d %d %d\n",i,j,p[i][j]);
}
const int ROWS = 3;
const int COLUMNS = 4;
void fillArray(double a[ROWS][COLUMNS], double value);
void deleteArray(double a[ROWS][COLUMNS]);
int main () {
double a[ROWS][COLUMNS];
fillArray(a, 0);
deleteArray(a);
}
In C++, how do you delete (or fill with specific values) a static n-dimension array?
In C++ we generally do not use arrays. We use std::vector.
You can use memset or std::fill to fill the array with specific values.
BTW you can use delete on dynamically allocated arrays not on static ones.
memset( a, 0 ,ROWS * COLUMNS * sizeof( double ));
or
std::fill(&a[0][0], &a[0][0]+sizeof(a)/sizeof(double), 0);
You can delete only an object created by new (and that object will be allocated in the heap). What do you mean by "deleting a static POD variable"? It has no sense:
1) It doesn't have any destructor to perform additional tasks before freeing the memory,
2) The stack memory will be "freed" as you exit the current block.
And to set it: either loop, either simple memset(a, 0, sizeof(a)); .
Also, the array in your example is not static.
std::vector is what is generally used for C++ arrays (especially when you're new at it). One of vector's constructors will fill it for you to:
std::vector<type> myVector(initialSize, defaultValue);
If you want multidimensional, you could do a vector of vectors, or boost::multi_array:
boost::multi_array<type, numberOfDimensions> myArray(boost::extents[firstSize][secondSize][thirdSize]);
In that case, you'll need to use the multiple-for-loops approach, because it doesn't seem to have a constructor that does that.
EDIT: Actually you can use std::vector to make a multidimensional array with default values:
std::vector<std::vector<double> > a(3, std::vector<double>(4, 0));
Where 3 is the number of rows, 4 is the number of columns and 0 is the default value.
What it's doing is create a vector of vectors with 3 rows, where the default value for each row is a vector with 4 zeroes.
Filling arrays in C++ is the same as filling them using C, namely nested for loops
int i, j;
for (i = 0; i < ROWS; i++)
for (j = 0; j < COLS; j++)
a[i][j] = 0
Arrays aren't "deleted" but they can use free if they've been allocated on the heap (if they've been allocated on the stack within a function, this is unnecessary).
int i;
for (i = 0; i < ROWS; i++)
free(a[i]);
free(a);
Firstly, the code you posted seems confused. What is it that you think "deleteArray" is supposed to do? 'a' is an auto variable and therefore cannot be deleted or freed.
Secondly, wrap your array in a class. There is a nice one in the FAQ that you can start with, but it can be improved. The first improvement is to use a vector rather than newing a block of memory. Then std::fill can be used to fill the array.
Use std::fill
#include <algorithm>
And then your implementation is simply:
std::fill(&a[0][0], &a[0][0]+sizeof(a)/sizeof(a[0][0], value);
You don't delete the array since it is stack allocated.
int *x = new int[5]();
With the above mentality, how should the code be written for a 2-dimensional array - int[][]?
int **x = new int[5][5] () //cannot convert from 'int (*)[5]' to 'int **'
In the first statement I can use:
x[0]= 1;
But the second is more complex and I could not figure it out.
Should I use something like:
x[0][1] = 1;
Or, calculate the real position then get the value
for the fourth row and column 1
x[4*5+1] = 1;
I prefer doing it this way:
int *i = new int[5*5];
and then I just index the array by 5 * row + col.
You can do the initializations separately:
int **x = new int*[5];
for(unsigned int i = 0; i < 5; i++)
x[i] = new int[5];
There is no new[][] operator in C++. You will first have to allocate an array of pointers to int:
int **x = new int*[5];
Then iterate over that array. For each element, allocate an array of ints:
for (std::size_t i = 0; i < 5; ++i)
x[i] = new int[5];
Of course, this means you will have to do the inverse when deallocating: delete[] each element, then delete[] the larger array as a whole.
This is how you do it:
int (*x)[5] = new int[7][5] ;
I made the two dimensions different so that you can see which one you have to use on the lhs.
Ff the array has predefined size you can write simply:
int x[5][5];
It compiles
If not why not to use a vector?
There are several ways to accomplish this:
Using gcc's support for flat multidimensional arrays (TonyK's answer, the most relevant to the question IMO). Note that you must preserve the bounds in the array's type everywhere you use it (e.g. all the array sizes, except possibly the first one), and that includes functions that you call, because the produced code will assume a single array. The allocation of $ new int [7][5] $ causes a single array to be allocated in memory. indexed by the compiler (you can easily write a little program and print the addresses of the slots to convince yourself).
Using arrays of pointers to arrays. The problem with that approach is having to allocate all the inner arrays manually (in loops).
Some people will suggest using std::vector's of std::vectors, but this is inefficient, due to the memory allocation and copying that has to occur when the vectors resize.
Boost has a more efficient version of vectors of vectors in its multi_array lib.
In any case, this question is better answered here:
How do I use arrays in C++?