I was wondering how i can access multidimensional rows in a 3D via pointer like this:
int ccc[8][7][2] = ....;
for(int i=0;i<8;i++)
{
int** cc_i = ccc[i];
for(int j=0;j<7;j++)
{
int* c_j = cc_i[j];
int th0 = c_j[0];
int th1 = c_j[0];
}
}
You can't, because a pointer to a pointer is not the same as an array of arrays. The layout in memory is radically different.
You can however declare e.g. cc_i as a pointer to an array, like
int (*cc_i)[2] = ccc[i];
Like this
int ccc[8][7][2] = ....;
for(int i=0;i<8;i++)
{
int (*cc_i)[2] = ccc[i];
for(int j=0;j<7;j++)
{
int *c_j = cc_i[j];
int th0 = c_j[0];
int th1 = c_j[0];
}
}
Related
I want to use a function made for dynamic arrys for an pre-initialised array too.
int MinInRow(int** fieldArray, int value, int currentColumn, int maxNumColumns)
{
int minVal = 0;
for (int inkrCol = 0; inkrCol < maxNumColumns; inkrCol++)
{
if (feldArray[currentColumn][inkrSpalte] < value)
{
minVal = feldArray[currentColumn][inkrCol];
}
}
return minVal;
}
So if I try
int testArray[3][4] =
{ {4,5,6,7},
{0,1,2,3},
{9,8,10,11}, };
int (*bufferArray)[4] = testArray;
or...
int** bufferArray = testArray;
or...
...
int main()
{
/*
read in other needed parameters
....
*/
std::cout << MinInRow((*bufferArray)[4], int value, int currentColumn, int maxNumColumns);
retrun 0;
}
I cant compile. But shouldnt int** bufferArray = testarray; just work fine? I mean there is always an double pointer to the first address of an 2D-Array.
how can I hand it over to the MinInRow-function? Thanks!
There is a little problem. "*" That one tell you that you want to make a pointer, but a pointer doesn't want a variable, he want the address. And then you've written
int** bufferArray = testarray.
You should write
int* bufferArray = new int [size of array] It's for 1D array
and ....
int** bufferArray = new int* [rows];
for (size_t = 0 ; i != rows ; ++i){
bufferArray[i] = new int [colons];
}
for 2D array
"I cant compile. But shouldnt int** bufferArray = testarray; just work fine? I mean there is alway an double pointer to the first address of an 2D-Array."
And Yes, it will be work nice. If testarray already 2D array
I made a function the returns an array like so
void array_function(int i){
int* a = NULL;
a = new int[3];
a = {i-1, i, i+1};
return a;
}
Now I want to call this function in the a new function
int main(){
int n = 3
for(int i = 0; i < n; i++){
//call the function
}
}
I am not sure how I can call the function to give me the array, any help will be appreciated
Use std::array instead. It has more friendly value semantics:
#include <array>
std::array<int, 3> array_function(int const i) {
return {{ i - 1, i, i + 1 }};
}
int main() {
for(int i = 0; i < 3; i++){
auto arr = array_function(i);
// Use array
}
}
First your function is void, which translates as no-return-function. Make it return int* like
int* array_function(int i)
Now, to call the function you need to assign it to a temporary variable, which you can do work and then you should delete it. Full code:
int* array_function(int i){
int* a = new int[3];
a[0] = i-1, a[1] = i, a[2] = i+1;
return a;
}
int main(){
int n = 3;
for(int i = 0; i < n; i++){
int* a = array_function(i); // if you are going to do something with this array, which you will
// some work with a
delete[] a; // delete it to release memory from heap, everytime you do new, you should use delete at the end of your program
}
}
You want something like this to create the array, since the method in your question does not have a return type assigned (void means that it returns nothing), you have to define a return type for the method to work, in this case a pointer to an array:
int* array_function(int i){
int* a = NULL;
a = new int[3];
a = {i-1, i, i+1};
return a;
}
And then just store the result of the method in a local variable like this:
int* myArray = array_function(n);
My weekend assignment was to make a function that gets an array of integers and the size of the array, and creates an array of pointers so that the pointers will be sorted using bubble sort (without changing the original array).
While debugging I found out that it works just fine, but when the function goes back to main() the pointers array gets initialized and everything's gone.
#include <iostream>
using namespace std;
void pointerSort(int arr[], int size, int* pointers[]);
void swap(int a, int b);
void main()
{
int arr[5]={7,2,5,9,4};
int size = 5;
int* pointers[5];
pointerSort(arr, size, pointers);
for (int i = 0; i < 5 ; i++)
cout << *pointers[i] << endl;
}
void pointerSort(int arr[], int size, int* pointers[])
{
int j, i;
bool change = true;
pointers = new int*[size];
for (i = 0; i < size; i++)
pointers[i] = &arr[i];
i = 0;
j = 1;
while (i <= size-1 && change == true)
{
change = false;
for (i = 0; i < size-j; i++)
{
if (*pointers[i] > *pointers[i+1])
{
swap(pointers[i], pointers[i+1]);
change = true;
}
}
j++;
}
}
void swap(int&a, int&b)
{
int temp;
temp = a;
a = b;
b = temp;
}
pointers = new int*[size];
At this point pointers is already an array of pointers, no allocation is needed.
After this line pointers IS NO LONGER THE ARRAY IN YOUR MAIN FUNCTION.
This is why your function is failing, because you are reassigning the array to which pointers is pointing to. The original array ISNT getting reinitialized, its just ignored throughout the entire code.
It is also a memory leak as ATaylor mentions, since you do not delete the allocated space, and cannot delete the space after the function finishes.
To fix everything: just remove the above line.
I'm trying to make a pointer point to a 2D array of pointers. What is the syntax and how would I access elements?
By the letter of the law, here's how to do it:
// Create 2D array of pointers:
int*** array2d = new (int**)[rows];
for (int i = 0; i < rows; ++i) {
array2d[i] = new (int*)[cols];
}
// Null out the pointers contained in the array:
for (int i = 0; i < rows; ++i) {
for (int j = 0; j < cols; ++j) {
array2d[i][j] = NULL;
}
}
Be careful to delete the contained pointers, the row arrays, and the column array all separately and in the correct order.
However, more frequently in C++ you'd create a class that internally managed a 1D array of pointers and overload the function call operator to provide 2D indexing. That way you're really have a contiguous array of pointers, rather than an array of arrays of pointers.
It depends. It can be as simple as:
int main()
{
int* data[10][20]; // Fixed size known at compile time
data[2][3] = new int(4);
}
If you want dynamic sizes at runtime you need to do some work.
But Boost has you covered:
int main()
{
int x;
int y;
getWidthAndHeight(x,y);
// declare a 2D array of int*
boost::multi_array<int*,2> data(boost::extents[x][y]);
data[2][3] = new int(6);
}
If you are fine with jagged arrays that can grow dynamically:
int main()
{
std::vector<std::vector<int*> > data;
data.push_back(std::vector<int*>(10,NULL));
data[0][3] = new int(7);
}
Note: In all the above. I assume that the array does not own the pointer. Thus it has not been doing any management on the pointers it contains (though for brevity I have been using new int() in the examples). To do memory management correctly you need to do some more work.
int *pointerArray[X][Y];
int **ptrToPointerArray = pointerArray;
That's how you make a true (contiguous in memory) multidimensional array.
But realize that once you cast a multidimensional array to a pointer like that, you lose the ability to index it automatically. You would have to do the multidimensional part of the indexing manually:
int *pointerArray[8][6]; // declare array of pointers
int **ptrToPointerArray = &pointerArray[0][0]; // make a pointer to the array
int *foo = pointerArray[3][1]; // access one element in the array
int *bar = *(ptrToPointerArray + 3*8 + 1); // manually perform row-major indexing for 2d array
foo == bar; // true
int *baz = ptrToPointerArray[3][1]; // syntax error
double** array = new double*[rowCnt];
for (int row = 0; row < rowCnt; ++row)
array[row] = new double[colCnt];
for (int row = 0; row < rowCnt; ++row)
for (int col = 0; col < colCnt; ++col)
array[row][col] = 0;
You could try Boost::MultiArray.
Check out this page for details.
:)
I had these once in a piece of code I wrote.
I was the laughing stock of the team when the first bugs leaked out. On top of that we use Hungarian notation, leading to a name like papChannel - a pointer to an array of pointers...
It's not nice. It's nicer to use typedefs to define a 'row of columns' or vice versa. Makes indexing more clear, too.
typedef int Cell;
typedef Cell Row[30];
typedef Row Table[20];
Table * pTable = new Table;
for( Row* pRow = *pTable; pRow != *pTable+_countof(*pTable); ++pRow ) {
for( Cell* pCell = *pRow; pCell != *pRow + _countof(*pRow); ++pCell ) {
... do something with cells.
}
}
You can define a vector of vectors:
typedef my_type *my_pointer;
typedef vector<vector<my_pointer> > my_pointer2D;
Than create a class derived from my_pointer2D, like:
class PointersField: public my_pointer2D
{
PointsField(int n, int m)
{
// Resize vectors....
}
}
PointsField pf(10,10); // Will create a 10x10 matrix of my_pointer
I prefer to use the () operator. There are lots of reasons for this (C++ FAQs 13.10). Change the internal representation to a std::vector if you like:
template <class T, int WIDTH, int HIEGHT>
class Array2d
{
public:
const T& operator ()(size_t col, size_t row) const
{
// Assert col < WIDTH and row < HIEGHT
return m_data [( row * WIDTH + col)];
}
T& operator ()(size_t col, size_t row)
{
// Assert col < WIDTH and row < HIEGHT
return m_data [( row * WIDTH + col)];
}
private:
T m_data[WIDTH * HIEGHT];
};
You can use it like this:
Array2d< Object*, 10, 10 > myObjectArray;
myObjectArray(5,6) = new Object();
See my code. It works on my FC9 x86_64 system:
#include <stdio.h>
template<typename t>
struct array_2d {
struct array_1d {
t *array;
array_1d(void) { array = 0; }
~array_1d()
{
if (array) {
delete[] array;
array = 0;
}
}
t &operator[](size_t index) { return array[index]; }
} *array;
array_2d(void) { array = 0; }
array_2d(array_2d<t> *a) { array = a->array; a->array = 0; }
void init(size_t a, size_t b)
{
array = new array_1d[a];
for (size_t i = 0; i < a; i++) {
array[i].array = new t[b];
}
}
~array_2d()
{
if (array) {
delete[] array;
array = 0;
}
}
array_1d &operator[](size_t index) { return array[index]; }
};
int main(int argc, char **argv)
{
array_2d<int> arr = new array_2d<int>;
arr.init(16, 8);
arr[8][2] = 18;
printf("%d\n",
arr[8][2]
);
return 0;
}
Effo UPD: a response to "Isn't that an array of pointers to arrays?", adding the example of array of pointers, very simple:
int main(int argc, char **argv)
{
array_2d<int*> parr = new array_2d<int*>;
int i = 10;
parr.init(16, 8);
parr[10][5] = &i;
printf("%p %d\n",
parr[10][5],
parr[10][5][0]
);
return 0;
}
Did I still misunderstand your question?
And you could even
typedef array_2d<int*> cell_type;
typedef array_2d<cell_type*> array_type;
int main(int argc, char **argv)
{
array_type parr = new array_type;
parr.init(16, 8);
parr[10][5] = new cell_type;
cell_type *cell = parr[10][5];
cell->init(8, 16);
int i = 10;
(*cell)[2][2] = &i;
printf("%p %d\n",
(*cell)[2][2],
(*cell)[2][2][0]
);
delete cell;
return 0;
}
It also works on my FC9 x86_64 system.
Is there some way to delay defining the size of an array until a class method or constructor?
What I'm thinking of might look something like this, which (of course) doesn't work:
class Test
{
private:
int _array[][];
public:
Test::Test(int width, int height);
};
Test::Test(int width, int height)
{
_array[width][height];
}
What Daniel is talking about is that you will need to allocate memory for your array dynamically when your Test (width, height) method is called.
You would declare your two dimensional like this (assuming array of integers):
int ** _array;
And then in your Test method you would need to first allocate the array of pointers, and then for each pointer allocate an array of integers:
_array = new *int [height];
for (int i = 0; i < height; i++)
{
_array [i] = new int[width];
}
And then when the object is released you will need to explicit delete the memory you allocated.
for (int i = 0; i < height; i++)
{
delete [] _array[i];
_array [i] = NULL;
}
delete [] _array;
_array = NULL;
vector is your best friend
class Test
{
private:
vector<vector<int> > _array;
public:
Test(int width, int height) :
_array(width,vector<int>(height,0))
{
}
};
I think it is time for you to look up the new/delete operators.
Seeing as this is a multidimensional array, you're going to have to loop through calling 'new' as you go (and again not to forget: delete).
Although I am sure many will suggest to use a one-dimensional array with width*height elements.
(Months later) one can use templates, like this:
// array2.c
// http://www.boost.org/doc/libs/1_39_0/libs/multi_array/doc/user.html
// is professional, this just shows the principle
#include <assert.h>
template<int M, int N>
class Array2 {
public:
int a[M][N]; // vla, var-len array, on the stack -- works in gcc, C99, but not all
int* operator[] ( int j )
{
assert( 0 <= j && j < M );
return a[j];
}
};
int main( int argc, char* argv[] )
{
Array2<10, 20> a;
for( int j = 0; j < 10; j ++ )
for( int k = 0; k < 20; k ++ )
a[j][k] = 0;
int* failassert = a[10];
}