I have an integer variable x that I need to use to make two 2D arrays but I get an error of "cannot allocate an array of constant size 0". After doing some research I apparently need to use malloc but I have no idea how to apply it to my currently situation.
My two arrays I need:
int firMat[x][5];
int secMat[5][x];
#include <stdio.h>
#include <stdlib.h>
int main(void) {
int x = 2;
int **firMat;//int firMat[x][5];
int **secMat;//secMat[5][x];
int i;
firMat = malloc(x * sizeof(int*));
for(i = 0; i< x; ++i)
firMat[i] = malloc(5 * sizeof(int));
secMat = malloc(5 * sizeof(int*));
for(i = 0; i< 5; ++i)
secMat[i] = malloc(x * sizeof(int));
//do stuff E.g. fir[2][1] = 21;
//release E.g.
//for(i = 0; i< x; ++i)
// free(firMat[i]);
//free(firMat);
return 0;
}
If you're using C99, this will work. It will create a "variable-length array", sadly VLAs have been reduced to "optional" in C11.
To use malloc for this, typically I'd abandon the double-array notation, and treat the memory as a flat one-dimensional array, then array[i][j] becomes ptr[ i*cols + j ].
Try to initialize x like in the example below
#define x 2 //outside the function
and then use x like this
int firMat[x][5];
int secMat[5][x];
Related
I was wondering how to properly use scanf to fill out a multidimensional array.
Here's my code:
#include <stdio.h>
#include <stdlib.h>
int main(int argc, const char * argv[]) {
int n; //number of rounds
int* sArray; //multidimensional array that holds the scores of both players
int i;
scanf("%d", &n);
sArray = (int*) calloc (n * 2, sizeof(int));
for(i=0; i<n; i++) {
scanf("%d %d", &sArray[i][1], &sArray[i][2]);
}
return 0;
}
It gives me an error, "Subscripted value is not an array, pointer, or vector." Any help would be much appreciated!
A two dimentional array is defined as follows: int sArray[N][M], but since you wanted to work with the dynamic memory I offer you to take a look at a pointer to pointer at int:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int n;
scanf("%d", &n);
int **sArray;
sArray = (int **)malloc(n * sizeof(int *));
int i;
for(i = 0; i < n; i++)
{
sArray[i] = (int *)malloc(2 * sizeof(int));
scanf("%d %d", &sArray[i][1], &sArray[i][2]);
}
return 0;
}
Don't forget to clean-up after you are done with the array.
As mentioned in the commentaries, You don't need to cast the result of malloc if you work with pure c. I did this because my c++ compiler refused to compile it without this cast.
You might need to check errors during a dynamic allocation of the array. Read more here
There are already a lot of good answers here on how to define your dynamic 2D array. But this variant was not yet mentionned, so I put it for the records.
As the last dimension of your array is fixed, you could define your 2D array as follows:
int (*sArray)[2]; //multidimensional array that holds the scores of both players
...
sArray = (int(*)[2]) calloc (n, sizeof(int)*2); // self explaining
In this way, all the elements will be stored contiguously (each n element of the allocated array, is 2 contiguous integers), without the need for an array to arrays.
The rest of your code remains identical. Except that you shoud address sArray[i][0] and ..[1] instead of [1] and [2] and free memory at the end. In C array indexing starts always from 0 and goes to size-1.
Of course, this approach is strictly limited to 2D arrays where the last dimension is fixed.
Live demo with addressing
Usually to fill a bidimensional array you will use two nested for loops.
For example :
int array[2][3] = {0};
for (i = 0; i < 2; i++)
for (k = 0; k < 3; k++)
scanf("%d", &array [i][k]);
You could do this too:
#include <stdio.h>
#include <stdlib.h>
int main(int argc, const char * argv[]) {
int n; //number of rounds
int** sArray = malloc(2 * sizeof(int*)); //multidimensional array that holds the scores of both players
scanf("%d", &n);
sArray[0] = (int*)calloc(n , sizeof(int));
sArray[1] = (int*)calloc(n , sizeof(int));
int i;
for (i = 0; i < n; i++) {
scanf("%d %d", &sArray[0][i], &sArray[1][i]);
}
free(sArray[0]);
free(sArray[1]);
free(sArray);
return 0;
}
I am trying to dynamically allocate a 2D array, put some values, and print output. However it seems that I am making mistake in getting input to program in atoi() function.
Basically when we assign a static 2D array, we declare it as say int a [3][3]. So 3*3 units if int, that much memory gets allocated. Is same thing holds for allocating dynamic array as well?
Here is my code:
#include<stdio.h>
#include<stdlib.h>
int main(int arg,char* argv)
{
int rows = atoi(argv[1]);
int col = atoi(argv[2]);
int rows =3;
int col=3;
int i,j;
int (*arr)[col] = malloc(sizeof (*arr)*rows);
int *ptr = &(arr[0][0]);
int ct=1;
for (i=0;i<rows;i++)
{
for(j=0;j<col;j++)
{
arr[i][j]=ct;
ct++;
}
}
printf("printing array \n");
for (i=0;i<rows;i++)
{
for(j=0;j<col;j++)
{
printf("%d \t",arr[i][j]);
}
printf("\n");
}
free(arr);
return (0);
}
Program crashes in runtime. Can someone comment?
Try to change the third line to:
int main(int arg,char **argv)
The common method to use dynamic matrices is to use a pointer to pointer to something, and then allocate both "dimensions" dynamically:
int **arr = malloc(sizeof(*arr) * rows);
for (int i = 0; i < rows; ++i)
arr[i] = malloc(sizeof(**arr) * col);
Remember that to free the matrix, you have to free all "rows" in a loop first.
int rows = atoi(argv[1]);
int col = atoi(argv[2]);
int rows =3;
int col=3;
int i,j;
You are defining rows and col twice.... that would never work!
With traditional C, you can only have the array[][] structure for multiple dimension arrays work with compile time constant values. Otherwise, the pointer arithmetic is not correct.
For dynamically sized multi dimensional arrays (those where rows and cols are determined at runtime), you need to do additional pointer arithmetic of this type:
int *a;
int rows=3;
int cols=4;
a = malloc(rows * cols * sizeof(int));
for (int i = 0; i < rows; ++i)
for (int j = 0; j < cols; ++j)
a[i*rows + j] = 1;
free(a);
Alternatively, you can use double indirection and have an array of pointers each pointing to a one dimensional array.
If you are using GCC or any C99 compiler, dynamic calculation of multiple dimension arrays is simplified by using variable length arrays:
// This is your code -- simplified
#include <stdio.h>
int main(int argc, const char * argv[])
{
int rows = atoi(argv[1]);
int col = atoi(argv[2]);
// you can have a rough test of sanity by comparing rows * col * sizeof(int) < SIZE_MAX
int arr[rows][col]; // note the dynamic sizing of arr here
int ct=1;
for (int i=0;i<rows;i++)
for(int j=0;j<col;j++)
arr[i][j]=ct++;
printf("printing array \n");
for (int i=0;i<rows;i++)
{
for(int j=0;j<col;j++)
{
printf("%d \t",arr[i][j]);
}
printf("\n");
}
return 0;
} // arr automatically freed off the stack
With a variable length array ("VLA"), dynamic multiple dimension arrays in C become far easier.
Compare:
void f1(int m, int n)
{
// dynamically declare an array of floats n by m size and fill with 1.0
float *a;
a = malloc(m * n * sizeof(float));
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j)
a[i*n + j] = 1.0;
free(a);
}
With VLA you can write to do the same:
void f2(int m, int n)
{
// Use VLA to dynamically declare an array of floats n by m size and fill with 1.0
float a[m][n];
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j)
a[i][j] = 1.0;
}
Be aware that unlike malloc / free VLA's handling of requesting a size larger than what is available on the stack is not as easily detected as using malloc and testing for a NULL pointer. VLA's are essentially automatic variables and have similar ease and restrictions.
VLA's are better used for smaller data structures that would be on the stack anyway. Use the more robust malloc / free with appropriate detection of failure for larger data structures.
If you are not using a fairly recent vintage C compiler that supports C99 -- time to get one.
I want to dynamically allocate 1 dimension of a 2D array (the other dimension is given). Does this work:
int NCOLS = 20;
// nrows = user input...
double *arr[NCOLS];
arr = (double *)malloc(sizeof(double)*nrows);
and to free it:
free(arr)
Not quite -- what you've declared is an array of pointers. You want a pointer to an array, which would be declared like this:
double (*arr)[NCOLS];
Then, you'd allocate it like so:
arr = malloc(nrows * sizeof(double[NCOLS]));
It can then be treated as a normal nrows by NCOLS 2D array. To free it, just pass it to free like any other pointer.
In C, there's no need to cast the return value of malloc, since there's an implicit cast from void* to any pointer type (this is not true in C++). In fact, doing so can mask errors, such as failing to #include <stdlib.h>, due to the existence of implicit declarations, so it's discouraged.
The data type double[20] is "array 20 of double, and the type double (*)[20] is "pointer to array 20 of double". The cdecl(1) program is very helpful in being able to decipher complex C declarations (example).
An example:
#include <stdio.h>
#include <stdlib.h>
#define COLS 2
void func(int (**arr)[COLS], int rows)
{
int i, j;
*arr = malloc(sizeof(int[COLS]) * rows);
printf("Insert number: \n");
for(i = 0; i < rows; i++)
for(j = 0; j < COLS; j++)
scanf("%d", &(*arr)[i][j]);
for(i = 0; i < rows; i++)
for(j = 0; j < COLS; j++)
printf("%d\n", (*arr)[i][j]);
}
int main(void)
{
int (*arr)[COLS];
func(&arr, 2);
free(arr);
return 0;
}
You have to allocate a new array for each element (each element is a pointer to an array) on the first dimension. You can use a loop for that:
for(i = 0; i < NCOLS; i++)
arr[i] = (double *)malloc(sizeof(double)*nrows);
Do the same to free.
I want to have a large 2 dimensional array such like
int myArray[10000][2];
I was told that the array built in such way is not appropriate, and should use malloc to build in heap. Could someone show me how to accomplish this? thanks!
#include <stdlib.h>
//alloc
int **vv = malloc(2 * sizeof(int *));
for(int i = 0; i < 2; i++)
vv[i] = malloc(10000 * sizeof(int));
//free
for(int i = 0; i < 2; i++)
free(vv[i]);
free(vv);
I have been asked in an interview how do i allocate a 2-D array and below was my solution to it.
#include <stdlib.h>
int **array;
array = malloc(nrows * sizeof(int *));
for(i = 0; i < nrows; i++)
{
array[i] = malloc(ncolumns * sizeof(int));
if(array[i] == NULL)
{
fprintf(stderr, "out of memory\n");
exit or return
}
}
I thought I had done a good job but then he asked me to do it using one malloc() statement not two. I don't have any idea how to achieve it.
Can anyone suggest me some idea to do it in single malloc()?
Just compute the total amount of memory needed for both nrows row-pointers, and the actual data, add it all up, and do a single call:
int **array = malloc(nrows * sizeof *array + (nrows * (ncolumns * sizeof **array));
If you think this looks too complex, you can split it up and make it a bit self-documenting by naming the different terms of the size expression:
int **array; /* Declare this first so we can use it with sizeof. */
const size_t row_pointers_bytes = nrows * sizeof *array;
const size_t row_elements_bytes = ncolumns * sizeof **array;
array = malloc(row_pointers_bytes + nrows * row_elements_bytes);
You then need to go through and initialize the row pointers so that each row's pointer points at the first element for that particular row:
size_t i;
int * const data = array + nrows;
for(i = 0; i < nrows; i++)
array[i] = data + i * ncolumns;
Note that the resulting structure is subtly different from what you get if you do e.g. int array[nrows][ncolumns], because we have explicit row pointers, meaning that for an array allocated like this, there's no real requirement that all rows have the same number of columns.
It also means that an access like array[2][3] does something distinct from a similar-looking access into an actual 2d array. In this case, the innermost access happens first, and array[2] reads out a pointer from the 3rd element in array. That pointer is then treatet as the base of a (column) array, into which we index to get the fourth element.
In contrast, for something like
int array2[4][3];
which is a "packed" proper 2d array taking up just 12 integers' worth of space, an access like array[3][2] simply breaks down to adding an offset to the base address to get at the element.
int **array = malloc (nrows * sizeof(int *) + (nrows * (ncolumns * sizeof(int)));
This works because in C, arrays are just all the elements one after another as a bunch of bytes. There is no metadata or anything. malloc() does not know whether it is allocating for use as chars, ints or lines in an array.
Then, you have to initialize:
int *offs = &array[nrows]; /* same as int *offs = array + nrows; */
for (i = 0; i < nrows; i++, offs += ncolumns) {
array[i] = offs;
}
Here's another approach.
If you know the number of columns at compile time, you can do something like this:
#define COLS ... // integer value > 0
...
size_t rows;
int (*arr)[COLS];
... // get number of rows
arr = malloc(sizeof *arr * rows);
if (arr)
{
size_t i, j;
for (i = 0; i < rows; i++)
for (j = 0; j < COLS; j++)
arr[i][j] = ...;
}
If you're working in C99, you can use a pointer to a VLA:
size_t rows, cols;
... // get rows and cols
int (*arr)[cols] = malloc(sizeof *arr * rows);
if (arr)
{
size_t i, j;
for (i = 0; i < rows; i++)
for (j = 0; j < cols; j++)
arr[i][j] = ...;
}
How do we allocate a 2-D array using One malloc statement (?)
No answers, so far, allocate memory for a true 2D array.
int **array is a pointer to pointer to int. array is not a pointer to a 2D array.
int a[2][3] is an example of a true 2D array or array 2 of array 3 of int
To allocate memory for a true 2D array, with C99, use malloc() and save to a pointer to a variable-length array (VLA)
// Simply allocate and initialize in one line of code
int (*c)[nrows][ncolumns] = malloc(sizeof *c);
if (c == NULL) {
fprintf(stderr, "out of memory\n");
return;
}
// Use c
(*c)[1][2] = rand();
...
free(c);
Without VLA support, if the dimensions are constants, code can use
#define NROW 4
#define NCOL 5
int (*d)[NROW][NCOL] = malloc(sizeof *d);
You should be able to do this with (bit ugly with all the casting though):
int** array;
size_t pitch, ptrs, i;
char* base;
pitch = rows * sizeof(int);
ptrs = sizeof(int*) * rows;
array = (int**)malloc((columns * pitch) + ptrs);
base = (char*)array + ptrs;
for(i = 0; i < rows; i++)
{
array[i] = (int*)(base + (pitch * i));
}
I'm not a fan of this "array of pointers to array" to solve the multi dimension array paradigm. Always favored a single dimension array, at access the element with array[ row * cols + col]? No problems encapsulating everything in a class, and implementing a 'at' method.
If you insist on accessing the members of the array with this notation: Matrix[i][j], you can do a little C++ magic. #John solution tries to do it this way, but he requires the number of column to be known at compile time. With some C++ and overriding the operator[], you can get this completely:
class Row
{
private:
int* _p;
public:
Row( int* p ) { _p = p; }
int& operator[](int col) { return _p[col]; }
};
class Matrix
{
private:
int* _p;
int _cols;
public:
Matrix( int rows, int cols ) { _cols=cols; _p = (int*)malloc(rows*cols ); }
Row operator[](int row) { return _p + row*_cols; }
};
So now, you can use the Matrix object, for example to create a multiplication table:
Matrix mtrx(rows, cols);
for( i=0; i<rows; ++i ) {
for( j=0; j<rows; ++j ) {
mtrx[i][j] = i*j;
}
}
You should now that the optimizer is doing the right thing and there is no call function or any other kind of overhead. No constructor is called. As long as you don't move the Matrix between function, even the _cols variable isn't created. The statement mtrx[i][j] basically does mtrx[i*cols+j].
It can be done as follows:
#define NUM_ROWS 10
#define NUM_COLS 10
int main(int argc, char **argv)
{
char (*p)[NUM_COLS] = NULL;
p = malloc(NUM_ROWS * NUM_COLS);
memset(p, 81, NUM_ROWS * NUM_COLS);
p[2][3] = 'a';
for (int i = 0; i < NUM_ROWS; i++) {
for (int j = 0; j < NUM_COLS; j++) {
printf("%c\t", p[i][j]);
}
printf("\n");
}
} // end of main
You can allocate (row*column) * sizeof(int) bytes of memory using malloc.
Here is a code snippet to demonstrate.
int row = 3, col = 4;
int *arr = (int *)malloc(row * col * sizeof(int));
int i, j, count = 0;
for (i = 0; i < r; i++)
for (j = 0; j < c; j++)
*(arr + i*col + j) = ++count; //row major memory layout
for (i = 0; i < r; i++)
for (j = 0; j < c; j++)
printf("%d ", *(arr + i*col + j));