I'm trying to figure out how to loop through a 2 dimensional array as a single dimensional array. Since a 2 dimensional array will occupy continuous memory, is there a way to address the two dimensional array as a single dimensional array by changing the index by 4 bytes. I'm assuming an integer array. Could some one provide an example? I tried the following but it doesn't work:
for (int i = 0; i < 2; i++){
for (int j = 0;j < 2; j++){
z[i][j] = count;
count++;
}
}
for (int i = 0; i < 4; i++)
printf("%d\n", z[i]);
2D arrays can be iterated through in a single loop like this:
#include <stdio.h>
int main()
{
int a[2][2], *p;
a[0][0] = 100;
a[0][1] = 200;
a[1][0] = 300;
a[1][1] = 400;
p = &a[0][0];
while(p!=&a[0][4])
printf("%d\n", *p++);
return 0;
}
Remember that an array index is just an offset from the first element of the array, so there is no real difference between a[0][3] and a[1][1] - they both refer to the same memory location.
Access 2D arrays like this
int *array; // note one level of pointer indirection
array = malloc(width * height * sizeof(int)); / allocate buffer somehow
for(y=0;y<height;y++)
for(x=0;x<width;x++)
array[y*width+x] = 0; // address the element by calculation
In three dimensions
int *array; // note one level of pointer indirection
array = malloc(width * height * depth * sizeof(int)); / allocate buffer somehow
for(z=0;z<depth;z++)
for(y=0;y<height;y++)
for(x=0;x<width;x++)
array[z*width*height + y*width+x] = 0; // address the element by calculation
Generally it's easier to use flat buffers than to mess around with C and C++ convoluted rules for multi-dimensional arrays. You can also of course
iterate through the entire array with a single index. If you want to set
up a 2D array, then cast the array to a single pointer, it behaves the
same way.
#define HEIGHT 50
#define WIDTH 90
int array2D[HEIGHT][WIDTH}:
int * array = reinterpret_cast<int *>(array2D):
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
int arr[4][3]={{1,2,3},{4,5,6},{7,8,9},{10,11,12}};
void my_fun(int(*a)[],int m,int n)
{
for(int i=0;i<m*n;i++)
{
printf("%d\n",(*a)[i]);
}
}
int main()
{
my_fun(arr,3,4);
return 0;
}
Related
How do I edit a value in an array with pointer in C?
int *pointer;
int array[3][1];
I tried this:
int *Pointer
int array[2][2];
Pointer[1][1]= 6;
but when compiling, I get a segmentation fault error. What to do?
Given some array int Array[Rows][Columns], to make a pointer to a specific element Array[r][c] in it, define int *Pointer = &Array[r][c];.
Then you may access that element using *Pointer in an expression, including assigning to *Pointer to assign values to that element. You may also refer to the element as Pointer[0], and you may refer to other elements in the same row as Pointer[y], where y is such that 0 ≤ y+c < Columns, i.e., Pointer[y] remains in the same row of the array.
You may also use Pointer[y] to refer to elements of the array in other rows as long as none of the language lawyers see you doing it. (In other words, this behavior is technically not defined by the C standard, but many compilers allow it.) E.g., after Pointer = &Array[r][c];, Pointer[2*Columns+3] will refer to the element Array[r+2][c+3].
To make a pointer you can use to access elements of the array using two dimensions, define int (*Pointer)[Columns] = &Array[r];.
Then Pointer[x][y] will refer to element Array[r+x][y]. In particularly, after int (*Pointer)[Columns] = &Array[0]; or int (*Pointer)[Columns] = Array;, Pointer[x][y] and Array[x][y] will refer to the same element.
You can access any given element with this syntax: array[x][y].
By the same token, you can assign your pointer to any element with this syntax: p = &array[x][y].
In C, you can often treat arrays and pointers as "equivalent". Here is a good explanation:
https://eli.thegreenplace.net/2009/10/21/are-pointers-and-arrays-equivalent-in-c
However, you cannot treat a simple pointer as a 2-d array. Here's a code example:
/*
* Sample output:
*
* array=0x7ffc463d0860
* 1 2 3
* 4 5 6
* 7 8 9
* p=0x7ffc463d0860
* 0x7ffc463d0864:1 0x7ffc463d0868:2 0x7ffc463d086c:3
* 0x7ffc463d0870:4 0x7ffc463d0874:5 0x7ffc463d0878:6
* 0x7ffc463d087c:7 0x7ffc463d0880:8 0x7ffc463d0884:9
*/
#include <stdio.h>
int main()
{
int i, j, *p;
int array[3][3] = {
{1,2,3},
{4,5,6},
{7,8,9}
};
// Dereference 2-D array using indexes
printf("array=%p\n", array);
for (i=0; i < 3; i++) {
for (j=0; j < 3; j++)
printf ("%d ", array[i][j]);
printf ("\n");
}
// Dereference 2-D array using pointer
p = &array[0][0];
printf("p=%p\n", p);
for (i=0; i < 3; i++) {
for (j=0; j < 3; j++)
printf ("%p:%d ", p, *p++);
printf ("\n");
}
/* Compile error: subscripted value p[0][0] is neither array nor pointer nor vector
p = &array[0][0];
printf("p=%p, p[0]=%p, p[0][0]=%p\n", p, &p[0], &p[0][0]);
*/
return 0;
}
Cast the 2D-array into 1D-array to pass it to a pointer,
And then, You are ready to access array with pointer. You can use this method to pass 2D-array to a function too.
#include <stdio.h>
int main()
{
int arr[2][2];
for (int i = 0; i < 2; i++)
{
for (int j = 0; j < 2; j++)
{
arr[i][j] = (2 * i) + j;
}
}
int *Pointer = (int *)arr; // Type conversion
/*
&arr[0][0] = Pointer + 0
&arr[0][1] = Pointer + 1
&arr[1][2] = Pointer + 2
&arr[2][2] = Pointer + 3
Dereference Pointer to access variable behind the address
*(Pointer + 0) = arr[0][0]
*(Pointer + 1) = arr[0][1]
*(Pointer + 2) = arr[1][2]
*(Pointer + 3) = arr[2][2]
*/
for (int i = 0; i < 2; i++)
{
for (int j = 0; j < 2; j++)
{
printf("%d ", *(Pointer + (2 * i) + j)); // Accessing array with pointer
}
printf("\n");
}
return 0;
}
Using the function wv_matalloc from https://www.ratrabbit.nl/ratrabbit/content/sw/matalloc/introduction , you can write the following code:
#include <stdio.h>
#include "wv_matalloc.h"
int main()
{
double **matrix;
int m = 3;
int n = 4;
// allocate m*n matrix:
matrix = wv_matalloc(sizeof(double),0,2,m,n);
// example of usage:
int i,j;
for (i=0; i<m; i++)
for (j=0; j<n; j++)
matrix[i][j] = i*j;
printf("2 3: %f\n",matrix[2][3]);
}
Compile with:
cc -o main main.c wv_matalloc.c
1.
You never assigned a value to Pointer in your example. Thus, attempting to access array by Pointer invokes undefined behavior.
You need to assign Pointer by the address of the first element of array if the pointer shall be a reference:
Pointer = *array;
2.
You can't use 2D notation (p[1][1]) for a pointer to int. This is a C syntax violation.
3.
Since rows of static 2D arrays are allocated subsequent in memory, you also can count the number of array elements until the specific element of desire. You need to subtract the count by 1 since indexing start at 0, not 1.
How does it work?
Each row of array contains 2 elements. a[1][1] (the first element of the second row) is directly stored after the first two.
Note: This is not the best approach. But worth a note beside all other answers as possible solution.
#include <stdio.h>
int main (void)
{
int *Pointer;
static int array[2][2];
Pointer = *array;
Pointer[2] = 6;
printf("array[1][1] (by pointer) = %d\n", Pointer[3]);
printf("array[1][1] (by array istelf) = %d\n", array[1][1]);
}
Output:
array[2][2] (by pointer) = 6
array[2][2] (by array istelf) = 6
Side Notes:
To address the first element of the second row by array[1][2] invokes undefined behavior. You should not use this way.
"but when compiling, I get a segmentation fault error."
Segmentation fault error do not occur at compile time. They occur at run time. It just gives you the impression because high probably your implementation immediately executes the program after compilation.
I am exploring pointer "mechanics" in C/C++. I try to understand if and how is possible to implement a 2D matrix using two pointers (one for "rows" and one for "columns") instead of a single double pointer. I am aware that a matrix with rows*columns number of values could be stored in memory sequentially, but i am looking to comprehend deeper the mechanics of pointers and eventually to implement a function similar to
int value=getValue(vectorNr,vectorValue)
that is able to "simulate" the construct
value=Matrix[vectorNr][vectorValue]
vectorPointer vectorValue
| AddressV1 |------|valAddr11 valAddr12 valAddr13 |
| AddressV2 |------|valAddr21 valAddr22 valAddr23 |
| AddressV3 |------|valAddr31 valAddr32 valAddr33 |
I tried to begin writing a code like this but I quickly get stuck on pointer arithmetic and address offsetting. I also might chose a very dirty approach so any comment is welcome.
CODE TO IMPLEMENT A 2D ARRAY WITH POINTERS (BUT NOT USING DOUBLE POINTERS). To avoid confusion between rows and columns I refer to "Vectors as rows" and "Columns as vector values"
int vectorsNumber = 3; //Number of Vectors
int valuesNumber = 3; //Number of values stored in one Vector
//Addresses of Vectors. Since Vectors holds a reference to set of values, vectorPointer will hold an address for every set.
void* vectorPointer = malloc(vectorsNumber *sizeof(void*));
//Populating the vectorPointer with the address generated by allocating memory for every set of values
for (int i = 0; i < vectorsNumber; i++)
{
vectorPointer = (int*)malloc(valuesNumber * sizeof(int)); //Values shall be of int types
vectorPointer++; //ILLEGAL since cannot perform arithmetic on pointers of type void. What do do??
}
//Restore the initial address. In any case...ILLEGAL arithmetic. What do do??
for (int i = 0; i < vectorsNumber; i++)
{
vectorPointer--; //Restore the initial address. In any case...ILLEGAL arithmetic.
}
//Declaring the pointer to hold the address of one value. Memory was already allocated before
int* valueAddress;
for (int j = 0; j < vectorsNumber; j++)
{
//Getting the address of the first value of the first Vector
valueAddress = (int*)vectorPointer; //Is this casting valid in C language?
//Populating the value with whatever operation
for (int k = 0; k < valuesNumber; k++)
{
*valueAddress = (k + 1)*(j + 1); //populate the Vector with int values
}
vectorPointer++; //Switch to next Vector.ILLEGAL arithmetic
}
Actually, you only need one pointer. One way of doing it is by allocating enough memory to hold all the values, and then have functions that map the x/y values in the array to the respective memory location. Assume we want those to be the dimensions and our array variable:
int dimX = 10, dimY = 5;
int *array;
You can set a value this way:
void arraySet(int value, int x, int y) {
array[x + dimX * y] = value;
}
And get a value this way:
int arrayGet(int x, int y) {
return array[x + dimX * y];
}
Allocate the memory beforehand such as in the main function:
array = malloc(sizeof(int)*dimX*dimY);
Use it like this:
arraySet(123, 9, 3); // sets the value of [9, 3] to 123
printf("Memory at 9, 3 is %d\n", arrayGet(9, 3));
This "two pointers" idea doesn't make any sense and the code you posted cannot be salvaged. What you should do instead is to use a pointer to a 2D array:
int (*ptr)[x][y] = malloc(sizeof *ptr);
...
free(ptr);
That's it. However, a pointer to a 2D array is cumbersome, since we have to de-reference it before accessing the actual array. That is, we'd end up writing (*ptr)[i][j] = ...; which is ugly.
To dodge this, we can instead still allocate a 2D array, but instead of pointing at "the whole array", we point at the first element, which is a 1D array:
int (*ptr)[y] = malloc( sizeof(int[x][y]) );
...
ptr[i][j] = ... ; // more convenient syntax for access
...
free(ptr);
More info: Correctly allocating multi-dimensional arrays
You can simulate int a[2][3]; with
one dimensional array and index computing:
int* matrix = (int*) malloc(6 * sizeof(int));
int get_matrix_2_3(int* matrix, int i, int j) { return matrix[3 * i + j]; }
2-dimensional array:
int** matrix = (int**) malloc(2 * sizeof(int*));
for (int i = 0; i != 2; ++i) {
matrix[i] = (int*) malloc(3 * sizeof(int));
}
matrix[1][2] = 42;
I'm trying to create a shared memory for IPC. I want to put a structure with dynamic 2D array in it into the shared memory. Here is the structure.
/* struct with 2 2D arrays. */
struct _test {
uint16_t **A;
uint16_t **B;
} test;
I know that a double pointer is not actually a 2D array and I should use a pointer to array like int (*ptr)[3], but the problem is that I can only get the size of the 2D array during runtime. So I have to declare the 2D array this way(at least what I know of).
Then I calculate the size of those two arrays in run time, say they both are 2x2 arrays, which takes 16 bytes(uint16_t is 2 byte). So I did this:
#include <sys/ipc.h>
#include <sys/shm.h>
#include <memory.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
int main()
{
size_t size = 16; //size of two uint16_t 2D arrays.
key_t key;
key = ftok("dev/null", 1);
int s_shm_id = shmget(key, size, IPC_CREAT|0600); //get 16 bytes of shared memory.
test *shm = (test*)shmat(s_shm_id,(const void*)0,0); //attach it.
//I want pointers in this struct to point at shared memory.
test *ptr = malloc(sizeof(test));
//Array A starts at the beginning of shared memory.
ptr -> A = (uint16_t **)shm; //My main confusion is here. Is this right?
int i;
for(i=0; i<2; i++)
ptr->A[i] =(uint16_t *)((uint16_t *)ptr->A + i*2);
//Array B starts right after A.
ptr -> B = ptr -> A[i-1] + 2;
for(i=0; i<2; i++)
ptr -> B[i] = (uint16_t *)((uint16_t *)ptr->B + i*2);
}
I understand this is basically wrong, I got segfault, but how? What a pointer needs is an address to point at, since I have already created a space(using shmget), why can't I make a pointer to point at it? Thanks for any feed back in advance!
What you are after are "jagged" or "scattered" arrays, not "linear" arrays. A scattered 2D-array in fact is not one array, but 1+N arrays, with N being the dimension of the 2D-matrix you are after.
The code you show misses to allocated this 1 array inside "1+N".
Assuming you successfully allocated enough memory to hold two 2D arrays with dimension N of uint16_t, that is 2 * N*N * sizeof (uint16_t) bytes, then the code to prepare the access to this memory might look like this (error checking left out for readability):
void * p = ... /* Allocate memory here; does not necessarily needs to be SHM. */
uint16_t ** A = malloc(N * sizeof *A);
for (size_t i = 0; i < N; ++i)
{
A[i] = ((uint16_t*) p) + i*N;
}
uint16_t ** B = malloc(N * sizeof *B);
for (size_t i = N; i < 2*N; ++i)
{
B[i - N] = ((uint16_t*) p) + i*N;
}
/* Access A[0 .. N-1][0 .. N-1] and B[0 .. N-1][0 .. N-1] here ... */
Placing A and B inside a struct which in turn is allocated dynamically is left to you as an exercise.
Accessing the arrays' elements is done straight forward: A[0][0] accesses the 1st array's 1st row's 1st element.
For clarity the same code for NxM arrays
uint16_t ** A = malloc(N * sizeof *A);
for (size_t i = 0; i < N; ++i)
{
A[i] = ((uint16_t*) p) + i*M;
}
uint16_t ** B = malloc(N * sizeof *B);
for (size_t i = N; i < 2*N; ++i)
{
B[i - N] = ((uint16_t*) p) + i*M;
}
Does someone know how I can use dynamically allocated multi-dimensional arrays using C? Is that possible?
Since C99, C has 2D arrays with dynamical bounds. If you want to avoid that such beast are allocated on the stack (which you should), you can allocate them easily in one go as the following
double (*A)[n] = malloc(sizeof(double[n][n]));
and that's it. You can then easily use it as you are used for 2D arrays with something like A[i][j]. And don't forget that one at the end
free(A);
Randy Meyers wrote series of articles explaining variable length arrays (VLAs).
With dynamic allocation, using malloc:
int** x;
x = malloc(dimension1_max * sizeof(*x));
for (int i = 0; i < dimension1_max; i++) {
x[i] = malloc(dimension2_max * sizeof(x[0]));
}
//Writing values
x[0..(dimension1_max-1)][0..(dimension2_max-1)] = Value;
[...]
for (int i = 0; i < dimension1_max; i++) {
free(x[i]);
}
free(x);
This allocates an 2D array of size dimension1_max * dimension2_max. So, for example, if you want a 640*480 array (f.e. pixels of an image), use dimension1_max = 640, dimension2_max = 480. You can then access the array using x[d1][d2] where d1 = 0..639, d2 = 0..479.
But a search on SO or Google also reveals other possibilities, for example in this SO question
Note that your array won't allocate a contiguous region of memory (640*480 bytes) in that case which could give problems with functions that assume this. So to get the array satisfy the condition, replace the malloc block above with this:
int** x;
int* temp;
x = malloc(dimension1_max * sizeof(*x));
temp = malloc(dimension1_max * dimension2_max * sizeof(x[0]));
for (int i = 0; i < dimension1_max; i++) {
x[i] = temp + (i * dimension2_max);
}
[...]
free(temp);
free(x);
Basics
Arrays in c are declared and accessed using the [] operator. So that
int ary1[5];
declares an array of 5 integers. Elements are numbered from zero so ary1[0] is the first element, and ary1[4] is the last element. Note1: There is no default initialization, so the memory occupied by the array may initially contain anything. Note2: ary1[5] accesses memory in an undefined state (which may not even be accessible to you), so don't do it!
Multi-dimensional arrays are implemented as an array of arrays (of arrays (of ... ) ). So
float ary2[3][5];
declares an array of 3 one-dimensional arrays of 5 floating point numbers each. Now ary2[0][0] is the first element of the first array, ary2[0][4] is the last element of the first array, and ary2[2][4] is the last element of the last array. The '89 standard requires this data to be contiguous (sec. A8.6.2 on page 216 of my K&R 2nd. ed.) but seems to be agnostic on padding.
Trying to go dynamic in more than one dimension
If you don't know the size of the array at compile time, you'll want to dynamically allocate the array. It is tempting to try
double *buf3;
buf3 = malloc(3*5*sizeof(double));
/* error checking goes here */
which should work if the compiler does not pad the allocation (stick extra space between the one-dimensional arrays). It might be safer to go with:
double *buf4;
buf4 = malloc(sizeof(double[3][5]));
/* error checking */
but either way the trick comes at dereferencing time. You can't write buf[i][j] because buf has the wrong type. Nor can you use
double **hdl4 = (double**)buf;
hdl4[2][3] = 0; /* Wrong! */
because the compiler expects hdl4 to be the address of an address of a double. Nor can you use double incomplete_ary4[][]; because this is an error;
So what can you do?
Do the row and column arithmetic yourself
Allocate and do the work in a function
Use an array of pointers (the mechanism qrdl is talking about)
Do the math yourself
Simply compute memory offset to each element like this:
for (i=0; i<3; ++i){
for(j=0; j<3; ++j){
buf3[i * 5 + j] = someValue(i,j); /* Don't need to worry about
padding in this case */
}
}
Allocate and do the work in a function
Define a function that takes the needed size as an argument and proceed as normal
void dary(int x, int y){
double ary4[x][y];
ary4[2][3] = 5;
}
Of course, in this case ary4 is a local variable and you can not return it: all the work with the array must be done in the function you call of in functions that it calls.
An array of pointers
Consider this:
double **hdl5 = malloc(3*sizeof(double*));
/* Error checking */
for (i=0; i<3; ++i){
hdl5[i] = malloc(5*sizeof(double))
/* Error checking */
}
Now hdl5 points to an array of pointers each of which points to an array of doubles. The cool bit is that you can use the two-dimensional array notation to access this structure---hdl5[0][2] gets the middle element of the first row---but this is none-the-less a different kind of object than a two-dimensional array declared by double ary[3][5];.
This structure is more flexible then a two dimensional array (because the rows need not be the same length), but accessing it will generally be slower and it requires more memory (you need a place to hold the intermediate pointers).
Note that since I haven't setup any guards you'll have to keep track of the size of all the arrays yourself.
Arithmetic
c provides no support for vector, matrix or tensor math, you'll have to implement it yourself, or bring in a library.
Multiplication by a scaler and addition and subtraction of arrays of the same rank are easy: just loop over the elements and perform the operation as you go. Inner products are similarly straight forward.
Outer products mean more loops.
If you know the number of columns at compile time, it's pretty simple:
#define COLS ...
...
size_t rows;
// get number of rows
T (*ap)[COLS] = malloc(sizeof *ap * rows); // ap is a *pointer to an array* of T
You can treat ap like any 2D array:
ap[i][j] = x;
When you're done you deallocate it as
free(ap);
If you don't know the number of columns at compile time, but you're working with a C99 compiler or a C2011 compiler that supports variable-length arrays, it's still pretty simple:
size_t rows;
size_t cols;
// get rows and cols
T (*ap)[cols] = malloc(sizeof *ap * rows);
...
ap[i][j] = x;
...
free(ap);
If you don't know the number of columns at compile time and you're working with a version of C that doesn't support variable-length arrays, then you'll need to do something different. If you need all of the elements to be allocated in a contiguous chunk (like a regular array), then you can allocate the memory as a 1D array, and compute a 1D offset:
size_t rows, cols;
// get rows and columns
T *ap = malloc(sizeof *ap * rows * cols);
...
ap[i * rows + j] = x;
...
free(ap);
If you don't need the memory to be contiguous, you can follow a two-step allocation method:
size_t rows, cols;
// get rows and cols
T **ap = malloc(sizeof *ap * rows);
if (ap)
{
size_t i = 0;
for (i = 0; i < cols; i++)
{
ap[i] = malloc(sizeof *ap[i] * cols);
}
}
ap[i][j] = x;
Since allocation was a two-step process, deallocation also needs to be a two-step process:
for (i = 0; i < cols; i++)
free(ap[i]);
free(ap);
malloc will do.
int rows = 20;
int cols = 20;
int *array;
array = malloc(rows * cols * sizeof(int));
Refer the below article for help:-
http://courses.cs.vt.edu/~cs2704/spring00/mcquain/Notes/4up/Managing2DArrays.pdf
Here is working code that defines a subroutine make_3d_array to allocate a multidimensional 3D array with N1, N2 and N3 elements in each dimension, and then populates it with random numbers. You can use the notation A[i][j][k] to access its elements.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
// Method to allocate a 2D array of floats
float*** make_3d_array(int nx, int ny, int nz) {
float*** arr;
int i,j;
arr = (float ***) malloc(nx*sizeof(float**));
for (i = 0; i < nx; i++) {
arr[i] = (float **) malloc(ny*sizeof(float*));
for(j = 0; j < ny; j++) {
arr[i][j] = (float *) malloc(nz * sizeof(float));
}
}
return arr;
}
int main(int argc, char *argv[])
{
int i, j, k;
size_t N1=10,N2=20,N3=5;
// allocates 3D array
float ***ran = make_3d_array(N1, N2, N3);
// initialize pseudo-random number generator
srand(time(NULL));
// populates the array with random numbers
for (i = 0; i < N1; i++){
for (j=0; j<N2; j++) {
for (k=0; k<N3; k++) {
ran[i][j][k] = ((float)rand()/(float)(RAND_MAX));
}
}
}
// prints values
for (i=0; i<N1; i++) {
for (j=0; j<N2; j++) {
for (k=0; k<N3; k++) {
printf("A[%d][%d][%d] = %f \n", i,j,k,ran[i][j][k]);
}
}
}
free(ran);
}
There's no way to allocate the whole thing in one go. Instead, create an array of pointers, then, for each pointer, create the memory for it. For example:
int** array;
array = (int**)malloc(sizeof(int*) * 50);
for(int i = 0; i < 50; i++)
array[i] = (int*)malloc(sizeof(int) * 50);
Of course, you can also declare the array as int* array[50] and skip the first malloc, but the second set is needed in order to dynamically allocate the required storage.
It is possible to hack a way to allocate it in a single step, but it would require a custom lookup function, but writing that in such a way that it will always work can be annoying. An example could be L(arr,x,y,max_x) arr[(y)*(max_x) + (x)], then malloc a block of 50*50 ints or whatever and access using that L macro, e.g.
#define L(arr,x,y,max_x) arr[(y)*(max_x) + (x)]
int dim_x = 50;
int dim_y = 50;
int* array = malloc(dim_x*dim_y*sizeof(int));
int foo = L(array, 4, 6, dim_x);
But that's much nastier unless you know the effects of what you're doing with the preprocessor macro.
int rows, columns;
/* initialize rows and columns to the desired value */
arr = (int**)malloc(rows*sizeof(int*));
for(i=0;i<rows;i++)
{
arr[i] = (int*)malloc(cols*sizeof(int));
}
// use new instead of malloc as using malloc leads to memory leaks
`enter code here
int **adj_list = new int*[rowsize];
for(int i = 0; i < rowsize; ++i)
{
adj_list[i] = new int[colsize];
}
Does someone know how I can use dynamically allocated multi-dimensional arrays using C? Is that possible?
Since C99, C has 2D arrays with dynamical bounds. If you want to avoid that such beast are allocated on the stack (which you should), you can allocate them easily in one go as the following
double (*A)[n] = malloc(sizeof(double[n][n]));
and that's it. You can then easily use it as you are used for 2D arrays with something like A[i][j]. And don't forget that one at the end
free(A);
Randy Meyers wrote series of articles explaining variable length arrays (VLAs).
With dynamic allocation, using malloc:
int** x;
x = malloc(dimension1_max * sizeof(*x));
for (int i = 0; i < dimension1_max; i++) {
x[i] = malloc(dimension2_max * sizeof(x[0]));
}
//Writing values
x[0..(dimension1_max-1)][0..(dimension2_max-1)] = Value;
[...]
for (int i = 0; i < dimension1_max; i++) {
free(x[i]);
}
free(x);
This allocates an 2D array of size dimension1_max * dimension2_max. So, for example, if you want a 640*480 array (f.e. pixels of an image), use dimension1_max = 640, dimension2_max = 480. You can then access the array using x[d1][d2] where d1 = 0..639, d2 = 0..479.
But a search on SO or Google also reveals other possibilities, for example in this SO question
Note that your array won't allocate a contiguous region of memory (640*480 bytes) in that case which could give problems with functions that assume this. So to get the array satisfy the condition, replace the malloc block above with this:
int** x;
int* temp;
x = malloc(dimension1_max * sizeof(*x));
temp = malloc(dimension1_max * dimension2_max * sizeof(x[0]));
for (int i = 0; i < dimension1_max; i++) {
x[i] = temp + (i * dimension2_max);
}
[...]
free(temp);
free(x);
Basics
Arrays in c are declared and accessed using the [] operator. So that
int ary1[5];
declares an array of 5 integers. Elements are numbered from zero so ary1[0] is the first element, and ary1[4] is the last element. Note1: There is no default initialization, so the memory occupied by the array may initially contain anything. Note2: ary1[5] accesses memory in an undefined state (which may not even be accessible to you), so don't do it!
Multi-dimensional arrays are implemented as an array of arrays (of arrays (of ... ) ). So
float ary2[3][5];
declares an array of 3 one-dimensional arrays of 5 floating point numbers each. Now ary2[0][0] is the first element of the first array, ary2[0][4] is the last element of the first array, and ary2[2][4] is the last element of the last array. The '89 standard requires this data to be contiguous (sec. A8.6.2 on page 216 of my K&R 2nd. ed.) but seems to be agnostic on padding.
Trying to go dynamic in more than one dimension
If you don't know the size of the array at compile time, you'll want to dynamically allocate the array. It is tempting to try
double *buf3;
buf3 = malloc(3*5*sizeof(double));
/* error checking goes here */
which should work if the compiler does not pad the allocation (stick extra space between the one-dimensional arrays). It might be safer to go with:
double *buf4;
buf4 = malloc(sizeof(double[3][5]));
/* error checking */
but either way the trick comes at dereferencing time. You can't write buf[i][j] because buf has the wrong type. Nor can you use
double **hdl4 = (double**)buf;
hdl4[2][3] = 0; /* Wrong! */
because the compiler expects hdl4 to be the address of an address of a double. Nor can you use double incomplete_ary4[][]; because this is an error;
So what can you do?
Do the row and column arithmetic yourself
Allocate and do the work in a function
Use an array of pointers (the mechanism qrdl is talking about)
Do the math yourself
Simply compute memory offset to each element like this:
for (i=0; i<3; ++i){
for(j=0; j<3; ++j){
buf3[i * 5 + j] = someValue(i,j); /* Don't need to worry about
padding in this case */
}
}
Allocate and do the work in a function
Define a function that takes the needed size as an argument and proceed as normal
void dary(int x, int y){
double ary4[x][y];
ary4[2][3] = 5;
}
Of course, in this case ary4 is a local variable and you can not return it: all the work with the array must be done in the function you call of in functions that it calls.
An array of pointers
Consider this:
double **hdl5 = malloc(3*sizeof(double*));
/* Error checking */
for (i=0; i<3; ++i){
hdl5[i] = malloc(5*sizeof(double))
/* Error checking */
}
Now hdl5 points to an array of pointers each of which points to an array of doubles. The cool bit is that you can use the two-dimensional array notation to access this structure---hdl5[0][2] gets the middle element of the first row---but this is none-the-less a different kind of object than a two-dimensional array declared by double ary[3][5];.
This structure is more flexible then a two dimensional array (because the rows need not be the same length), but accessing it will generally be slower and it requires more memory (you need a place to hold the intermediate pointers).
Note that since I haven't setup any guards you'll have to keep track of the size of all the arrays yourself.
Arithmetic
c provides no support for vector, matrix or tensor math, you'll have to implement it yourself, or bring in a library.
Multiplication by a scaler and addition and subtraction of arrays of the same rank are easy: just loop over the elements and perform the operation as you go. Inner products are similarly straight forward.
Outer products mean more loops.
If you know the number of columns at compile time, it's pretty simple:
#define COLS ...
...
size_t rows;
// get number of rows
T (*ap)[COLS] = malloc(sizeof *ap * rows); // ap is a *pointer to an array* of T
You can treat ap like any 2D array:
ap[i][j] = x;
When you're done you deallocate it as
free(ap);
If you don't know the number of columns at compile time, but you're working with a C99 compiler or a C2011 compiler that supports variable-length arrays, it's still pretty simple:
size_t rows;
size_t cols;
// get rows and cols
T (*ap)[cols] = malloc(sizeof *ap * rows);
...
ap[i][j] = x;
...
free(ap);
If you don't know the number of columns at compile time and you're working with a version of C that doesn't support variable-length arrays, then you'll need to do something different. If you need all of the elements to be allocated in a contiguous chunk (like a regular array), then you can allocate the memory as a 1D array, and compute a 1D offset:
size_t rows, cols;
// get rows and columns
T *ap = malloc(sizeof *ap * rows * cols);
...
ap[i * rows + j] = x;
...
free(ap);
If you don't need the memory to be contiguous, you can follow a two-step allocation method:
size_t rows, cols;
// get rows and cols
T **ap = malloc(sizeof *ap * rows);
if (ap)
{
size_t i = 0;
for (i = 0; i < cols; i++)
{
ap[i] = malloc(sizeof *ap[i] * cols);
}
}
ap[i][j] = x;
Since allocation was a two-step process, deallocation also needs to be a two-step process:
for (i = 0; i < cols; i++)
free(ap[i]);
free(ap);
malloc will do.
int rows = 20;
int cols = 20;
int *array;
array = malloc(rows * cols * sizeof(int));
Refer the below article for help:-
http://courses.cs.vt.edu/~cs2704/spring00/mcquain/Notes/4up/Managing2DArrays.pdf
Here is working code that defines a subroutine make_3d_array to allocate a multidimensional 3D array with N1, N2 and N3 elements in each dimension, and then populates it with random numbers. You can use the notation A[i][j][k] to access its elements.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
// Method to allocate a 2D array of floats
float*** make_3d_array(int nx, int ny, int nz) {
float*** arr;
int i,j;
arr = (float ***) malloc(nx*sizeof(float**));
for (i = 0; i < nx; i++) {
arr[i] = (float **) malloc(ny*sizeof(float*));
for(j = 0; j < ny; j++) {
arr[i][j] = (float *) malloc(nz * sizeof(float));
}
}
return arr;
}
int main(int argc, char *argv[])
{
int i, j, k;
size_t N1=10,N2=20,N3=5;
// allocates 3D array
float ***ran = make_3d_array(N1, N2, N3);
// initialize pseudo-random number generator
srand(time(NULL));
// populates the array with random numbers
for (i = 0; i < N1; i++){
for (j=0; j<N2; j++) {
for (k=0; k<N3; k++) {
ran[i][j][k] = ((float)rand()/(float)(RAND_MAX));
}
}
}
// prints values
for (i=0; i<N1; i++) {
for (j=0; j<N2; j++) {
for (k=0; k<N3; k++) {
printf("A[%d][%d][%d] = %f \n", i,j,k,ran[i][j][k]);
}
}
}
free(ran);
}
There's no way to allocate the whole thing in one go. Instead, create an array of pointers, then, for each pointer, create the memory for it. For example:
int** array;
array = (int**)malloc(sizeof(int*) * 50);
for(int i = 0; i < 50; i++)
array[i] = (int*)malloc(sizeof(int) * 50);
Of course, you can also declare the array as int* array[50] and skip the first malloc, but the second set is needed in order to dynamically allocate the required storage.
It is possible to hack a way to allocate it in a single step, but it would require a custom lookup function, but writing that in such a way that it will always work can be annoying. An example could be L(arr,x,y,max_x) arr[(y)*(max_x) + (x)], then malloc a block of 50*50 ints or whatever and access using that L macro, e.g.
#define L(arr,x,y,max_x) arr[(y)*(max_x) + (x)]
int dim_x = 50;
int dim_y = 50;
int* array = malloc(dim_x*dim_y*sizeof(int));
int foo = L(array, 4, 6, dim_x);
But that's much nastier unless you know the effects of what you're doing with the preprocessor macro.
int rows, columns;
/* initialize rows and columns to the desired value */
arr = (int**)malloc(rows*sizeof(int*));
for(i=0;i<rows;i++)
{
arr[i] = (int*)malloc(cols*sizeof(int));
}
// use new instead of malloc as using malloc leads to memory leaks
`enter code here
int **adj_list = new int*[rowsize];
for(int i = 0; i < rowsize; ++i)
{
adj_list[i] = new int[colsize];
}