This question already has answers here:
Passing two-dimensional array via pointer
(9 answers)
Closed 9 years ago.
I've got array: int i[20][30]. What is the fastest way to make int** pointer to use on it?
My friend just wrote function which uses same size arrays and my function is on pointers and I need to pass his data.
int a[20][30]; // data as 2D array
int *ap[20]; // array of pointers
for (j = 0; j < 20; ++j)
ap[j] = a[j]; // convert array to pointers
// can now pass `ap` to function expecting `int **`...
you cant change int i[20][30] to int ** except of manually go over the data and build it.
but, something that may help you, every xD array in c++ is just a 1D array with a somehow complexed way to access a member.
Example:
#define MaxX 20
#define MaxY 30
int i[MaxX][MaxY];
...
//When you write:
i[x][y] = something;
//It's actually been translate to:
i[x * MaxY + y] = something;
You can't have an int ** pointer to that array directly, it wouldn't make any sense. If you want to access the array as a block of memory you can just use:
int *p = (int *)i; // you can go to up (p + 20*30)
Otherwise you need a pointer to an array of int[30], i.e.
int (*p)[30] = &i[0]; // points to first row
With the first one, because of the cast, you can then define:
int **p2 = &p;
And you can access the array with *(*p2 + n). You've cast away the array dimensions, but as long as you use them in the pointer calculations, it's all valid, i.e. *(*p2 + 35) refers to row 2, column 6 (one row of 30, plus element 5 of the next). This is very quick, but if the function needs to operate on the array using the original dimensions, this wouldn't work (since it isn't an array of pointers).
Related
I'm mostly just documenting this question as someone may stumble upon it, and may find it useful. And also, I'm very curios with, how does std::swap works on a 2D array like: Arr[10][10].
My question arised because as to my understanding an array like this is just a 1D array with some reindexing.
For reference:
How are 2-Dimensional Arrays stored in memory?
int main()
{
const int x = 10;
const int y = 10;
int Arr[y][x];
// fill the array with some elements...
for (int i = 0; i < x*y; i++)
{
Arr[i / y][i % x] = i;
}
// swap 'row 5 & 2'
// ??? how does swap know how many elements to swap?
// if it is in fact stored in a 1D array, just the
// compiler will reindex it for us
std::swap(Arr[5], Arr[2]);
return 0;
}
I could understand swapping two 'rows' if our data type is, say a pointer to a pointer like int** Arr2D then swap with std::swap(Arr2D[2], Arr2D[5]) as we do not need to know the length here, we just need to swap the two pointers, pointing to '1D arrays'.
But how does std::swap work with Arr[y][x]?
Is it using a loop maybe, to swap all elements within x length?
std::swap has an overload for arrays that effectively swaps each two elements, again, using std::swap.
As for the size information, it is embedded within the array type (Arr[i] is int[x]), so the compiler knows to deduce T2 as int and N as 10.
OT: Why aren't variable-length arrays part of the C++ standard? (but this particular case is OK)
This question already has answers here:
Range-based for loop on a dynamic array?
(6 answers)
Closed 6 years ago.
if I create a new int array object using new like below:
int *array = new int[20];
and fill the array with some integers, then attempting to scan through that array with a for-each loop will throw me an error:
for (int x : array) // error
why does this happen and how can I fix it? I tried playing around with the referencer and address notations (* and &) but every combination I tried fails.
ex.
for (int &x : *array) // does not work either.
array is not an array, but a pointer, which can't be used for Range-based for loop, because there's no information about the size of the "array", the begin and end of the "array" can't be deduced.
range_expression - any expression that represents a suitable
sequence (either an array or an object for which begin and end member
functions or free functions are defined, see below) or a
braced-init-list.
You might make it real array, (or use std::array instead of the raw array)
int array[20];
for (int x : array) { processing with x... }
Or write the loop manually:
for (int i = 0; i < 20; i++) { processing with array[i]... }
in for (int x : array), it requires that begin(array) and end(array) are valid. which is not for int*.
it would work for int array[20], std::vector, std::array<int, 20>.
If I declare a 2D array
int A[sz][sz];
How can I create a pointer to this object?
I ask because I want to return an array via pointer to a pointer, int**, from a function but I want to build the array without knowing the size beforehand. The size will be passed as an argument. I want to know if there is a way to do this without using dynamic allocation.
The problem is if I do something like int** A inside the function this gives A no information about the size I want.
How can I create the array and then assign a pointer to this array, if it's a 2D array.
I should be more clear. I want return a pointer to a pointer so it wouldn't be a pointer to the 2D array but a something like int**.
Your problem is, that a 2D array in the form int** requires an array of int* for the two step dereferencing, which simply does not exist when you declare an array with int A[sz][sz];.
You can build it yourself like this:
int* pointers[sz];
for(size_t i = sz; i--; ) pointers[i] = A[i];
This might seem absurd, but is rooted in the way C handles arrays: A[i] is of type int ()[sz], which is the subarray of row i. But when you use that array in the assignment, it decays to a pointer to the first element in that subarray, which is of type int*. After the loop, A and pointers are two very different things (the type of A is int ()[sz][sz])
Sidenote: You say that you want to return this from a function. If your array is allocated on the stack, you must not return a pointer to its data, it will disappear the moment your function returns. You can only return pointers/references to objects that have either static storage or are part of another existing object. If you fail to comply with this, you are likely to get stack corruption.
Edit:
A little known fact about C is, that you can actually pass around pointers to real C arrays, not just the pointer types that an array decays to. Here is a small program to demonstrate this:
#include <stddef.h>
#include <stdio.h>
int (*foo(int size, int (*bar)[size][size], int y))[] {
return &(*bar)[y];
}
int main() {
int mySize = 30;
int baz[mySize][mySize];
int (*result)[mySize];
result = foo(mySize, &baz, 15);
printf("%ld\n", (size_t)result - (size_t)baz);
}
The expected output of this example program is 1800. The important thing is that the actual size of the array must be known, either by being a compile time constant, or by being passed along with the array pointer (and if it's passed along with the array pointer, the size argument must appear before the array pointer does).
Let me flesh out your question a little bit. You mention:
I ask because I want to return an array [...] from a function but I
want to build the array without knowing the size beforehand. The size
will be passed as an argument. I want to know if there is a way to do
this without using dynamic allocation.
For the I want to return an array from a function [...] size passed as an argument, it seems reasonable to me that you can use std::vector everywhere, and call its .data() method when you need access to the underlying array (which is guaranteed to be contiguous). For example:
std:vector<double> myfun(size_t N) {
std::vector<double> r(N);
// fill r[0], r[1], ..., r[N-1]
return r;
}
// later on:
r.data(); // gives you a pointer to the underlying double[N]
And for the I want to to do this without dynamic allocation, that is not possible unless you know the size at compile time. If that is the case, then do exactly as before but use std::array, which can implement optimizations based on known compile-time size:
std::array<double, N> myfun() {
std::array<double, N> r;
// fill r[0], r[1], ..., r[N-1]
return r;
}
// later on:
r.data(); // gives you a pointer to the underlying double[N]
And to be generic, I would actually use a template function capable of working with arbitrary containers:
template<typename T>
void myfun(T& data) {
for(int k=0; k<data.size(); k++) {
// do stuff to data[k]
}
}
// call as, for example:
std::vector<double> data(10);
myfun(data);
// or equally valid:
std::array<double, 10> data;
myfun(data);
Finally, if you are working with two-dimensional data, please remember that when you store the Matrix in row-major order that is:
Matrix [1, 2; 3 4] is stored as [1 2 3 4]
then you can refer to element (i, j) of the matrix by calling data[i * ncols + j]. For example: consider a three by four matrix:
a b c d
e f g h
i j k l
The element (2, 2) (that is: third row, third column because we assume zero-based C-type indexing) is calculated as: M[2][2] = M[2 * 4 + 2] = M[10] = k. This is the case because it was stored as:
[a b c d e f g h i j k l]
[0 1 2 3 4 5 6 7 8 9 10 11]
and k is the element with index 10.
the responses to your question are weird. Just do this:
int A[2][2];
int**p =NULL;
*p = A[0]; // **p==A[0][0] , *(*p+1)==A[0][1]
Why can't you declare a 2D array argument in a function as you do with a normal array?
void F(int bar[]){} //Ok
void Fo(int bar[][]) //Not ok
void Foo(int bar[][SIZE]) //Ok
Why is it needed to declare the size for the column?
Static Arrays:
You seem not to have got the point completely. I thought to try to explain it somewhat. As some of the above answers describe, a 2D Array in C++ is stored in memory as a 1D Array.
int arr[3][4] ; //consider numbers starting from zero are stored in it
Looks somewhat like this in Memory.
1000 //ignore this for some moments 1011
^ ^
^ ^
0 1 2 3 4 5 6 7 8 9 10 11
|------------| |-----------| |-------------|
First Array Second Array Third Array
|----------------------------------------------|
Larger 2D Array
Consider that here, the Bigger 2D Array is stored as contiguous memory units. It consists of total 12 elements, from 0 to 11. Rows are 3 and columns are 4. If you want to access the third array, you need to skip the whole first and second arrays. That is, you need to skip elements equal to the number of your cols multiplied by how many arrays you want skip. It comes out to be cols * 2.
Now when you specify the dimensions to access any single index of the array, you need to tell the compiler beforehand exactly how much elements to skip. So you give it the exact number of cols to perform the rest of the calculation.
So how does it perform the calculation? Let us say it works on the column major order, that is, it needs to know the number of columns to skip. When you specify one element of this array as...
arr[i][j] ;
Compiler performs this calculation automatically.
Base Address + (i * cols + j) ;
Let us try the formula for one index to test its veracity. We want to access the 3rd element of the 2nd Array. We would do it like this...
arr[1][2] ; //access third element of second array
We put it in the formula...
1000 + ( 1 * 4 + 2 )
= 1000 + ( 6 )
= 1006 //destination address
And we reach at the address 1006 where 6 is located.
In a nutshell, we need to tell the compiler the number of cols for this calculation. So we send it as a parameter in a function.
If we are working on a 3D Array, like this...
int arr[ROWS][COLS][HEIGHT] ;
We would have to send it the last two dimensions of the array in a function.
void myFunction (int arr[][COLS][HEIGHT]) ;
The formula now would become this..
Base Address + ( (i * cols * height) + (j * height) + k ) ;
To access it like this...
arr[i][j][k] ;
COLS tell the compiler to skip the number of 2D Array, and HEIGHT tells it to skip the number of 1D Arrays.
And so on and so forth for any dimension.
Dynamic Arrays:
As you ask about different behavior in case of dynamic arrays which are declared thus..
int ** arr ;
Compiler treats them differently, because each index of a Dynamic 2D Array consists of an address to another 1D Array. They may or may not be present on contiguous locations on heap. Their elements are accessed by their respective pointers. The dynamic counterpart of our static array above would look somewhat like this.
1000 //2D Pointer
^
^
2000 2001 2002
^ ^ ^
^ ^ ^
0 4 8
1 5 9
2 6 10
3 7 11
1st ptr 2nd ptr 3rd ptr
Suppose this is the situation. Here the 2D Pointer or Array on the location 1000. It hold the address to 2000 which itself holds address of a memory location. Here pointer arithmetic is done by the compiler by virtue of which it judges the correct location of an element.
To allocate memory to 2D Pointer, we do it..
arr = new int *[3] ;
And to allocate memory to each of its index pointer, this way..
for (auto i = 0 ; i < 3 ; ++i)
arr[i] = new int [4] ;
At the end, each ptr of the 2D Array is itself an array. To access an element you do...
arr[i][j] ;
Compiler does this...
*( *(arr + i) + j ) ;
|---------|
1st step
|------------------|
2nd step
In the first step, the 2D Array gets dereferenced to its appropriate 1D Array and in the second step, the 1D Array gets dereferenced to reach at the appropriate index.
That is the reason why Dynamic 2D Arrays are sent to the function without any mention of their row or column.
Note:
Many details have been ignored and many things supposed in the description, especially the memory mapping just to give you an idea.
You can't write void Foo(int bar[][]), because bar decays to a pointer. Imagine following code:
void Foo(int bar[][]) // pseudocode
{
bar++; // compiler can't know by how much increase the pointer
// as it doesn't know size of *bar
}
So, compiler must know size of *bar, therefore size of rightmost array must be provided.
Because when you pass an array, it decays to a pointer, so excluding the outer-most dimension is ok and that's the only dimension you can exclude.
void Foo(int bar[][SIZE])
is equivalent to:
void Foo(int (*bar)[SIZE])
The compiler needs to know how long the second dimension is to calculate the offsets. A 2D array is in fact stored as a 1D array.
If you want to send an array with no known dimensions, consider using pointer to pointers and some sort of way to know the dimension yourself.
This is different from e.g. java, because in java the datatype also contains the dimension.
Since static 2D arrays are like 1D arrays with some sugar to better access data, you have to think about the arithmetic of pointers.
When the compiler tries to access element array[x][y], it has to calculate the address memory of the element, that is array+x*NUM_COLS+y. So it needs to know the length of a row (how many elements it contains).
If you need more information I suggest this link.
there are basically three ways to allocate a 2d array in C/C++
allocate on heap as a 2d array
you can allocate a 2d array on the heap using malloc such as:
const int row = 5;
const int col = 10;
int **bar = (int**)malloc(row * sizeof(int*));
for (size_t i = 0; i < row; ++i)
{
bar[i] = (int*)malloc(col * sizeof(int));
}
this is actually stored as an array of arrays therefore isn't necessarily
contiguous in memory. note that this also means there will be a pointer for
each array costing yout extra memory usage (5 pointers in this example, 10
pointers if you allocate it the other way around). you can pass this array to
a function with the signature:
void foo(int **baz)
allocate on heap as 1d array
for various reasons (cache optimizations, memory usage etc.) it may be
desirable to store the 2d array as a 1d array:
const int row = 5;
const int col = 10;
int *bar = (int*)malloc(row * col * sizeof(int));
knowing second dimension you can access the elements using:
bar[1 + 2 * col] // corresponds semantically to bar[2][1]
some people use preprocessor magic (or method overloading of () in C++) to
handle this automatically such as:
#define BAR(i,j) bar[(j) + (i) * col]
..
BAR(2,1) // is actually bar[1 + 2 * col]
you need to have the function signature:
void foo(int *baz)
in order to pass this array to a function.
allocate on stack
you can allocate a 2d array on stack using something like:
int bar[5][10];
this is allocated as a 1d array on the stack therefore compiler needs to know
the second dimension to reach the element you need just like we did in the
second example, therefore the following is also true:
bar[2][1] == (*bar)[1 + 2 * 10]
function signature for this array should be:
void foo(int baz[][10])
you need to provide the second dimension so that compiler would know where to reach in memory. you don't have to give the first dimension since C/C++ is not a safe language in this respect.
let me know if I mixed up rows and columns somewhere..
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++?