#define max 40
...
void transpose(int matrix[][max], int* row, int* col)
{
int data[*row][max]; //expression must have a constant value at *row.
for (int i = 0; i < *row; i++)
{
for (int j = 0; j < *col; j++)
{
data[i][j] = matrix[i][j];
}
}
int _col = *row; //this *row works fine.
*row = *col; //also this *row works fine.
*col = _col;
for (int i = 0; i < *row; i++) //this *row is fine too.
{
for (int j = 0; j < *col; j++)
{
matrix[i][j] = data[j][i];
}
}
}
int main()
{
...
if (...)
{
int row = 0, col = 0;
int matrix[30][max];
if (FunctionReadFile(Parameters[0], matrix, &row, &col))
{
...
transpose(matrix, &row, &col);
...
}
...
}
...
return 0;
}
I tried put 'const' before int but it still show this error at [*row], why does this error occured and how to fix it? Does declaring an dynamic array is the only to fix this problem, any possible solution easier?
Your compiler does not support VLA:s, so you should use dynamic allocation:
void transpose(int matrix[][max], int* row, int* col)
{
// Parenthesis matters. This is a pointer to array of size max. Without
// the parenthesis, it would be an array of pointers to int.
int (*data)[max] = malloc(max * sizeof (*data));
for (int i = 0; i < *row; i++)
{
for (int j = 0; j < *col; j++)
{
data[i][j] = matrix[i][j];
}
}
free(data);
}
Do note that I did not check if the allocation failed. You can do that with a simple check. If the pointer is NULL, then the allocation failed. Also, remember to free the memory when you're done with it as shown.
You can use _alloca() or _malloca() in MSVC for stack allocation.
Return Value
The _alloca routine returns a void pointer to the allocated space,
which is guaranteed to be suitably aligned for storage of any type of
object. If size is 0, _alloca allocates a zero-length item and
returns a valid pointer to that item.
A stack overflow exception is generated if the space cannot be
allocated. The stack overflow exception is not a C++ exception; it is
a structured exception. Instead of using C++ exception handling, you
must use Structured Exception Handling (SEH).
Remarks
_alloca allocates size bytes from the program stack. The allocated
space is automatically freed when the calling function exits (not when
the allocation merely passes out of scope). Therefore, do not pass the
pointer value returned by _alloca as an argument to free.
There are restrictions to explicitly calling _alloca in an exception
handler (EH). EH routines that run on x86-class processors operate in
their own memory frame: They perform their tasks in memory space that
is not based on the current location of the stack pointer of the
enclosing function. ...
For example:
void transpose(int matrix[][max], int* row, int* col)
{
int ( *data )[ max ] = _alloca( max * sizeof( *data ) );
for (int i = 0; i < *row; i++)
{
for (int j = 0; j < *col; j++)
{
data[i][j] = matrix[i][j];
}
}
...
Or, with _malloca(), which requires you to call _freea():
void transpose(int matrix[][max], int* row, int* col)
{
int ( *data )[ max ] = _malloca( max * sizeof( *data ) );
for (int i = 0; i < *row; i++)
{
for (int j = 0; j < *col; j++)
{
data[i][j] = matrix[i][j];
}
}
...
_freea( data );
...
Related
Related to dynamic allocation inside a function, most questions & answers are based on double pointers.
But I was recommended to avoid using double pointer unless I have to, so I want to allocate a 'array pointer' (not 'array of pointer') and hide it inside a function.
int (*arr1d) = calloc(dim1, sizeof(*arr1d));
int (*arr2d)[dim2] = calloc(dim1, sizeof(*arr2d));
Since the above lines are the typical dynamic-allocation of pointer of array, I tried the following.
#include <stdio.h>
#include <stdlib.h>
int allocateArray1D(int n, int **arr) {
*arr = calloc(n, sizeof(*arr));
for (int i = 0; i < n; i++) {
(*arr)[i] = i;
}
return 0;
}
int allocateArray2D(int nx, int ny, int *(*arr)[ny]) {
*arr[ny] = calloc(nx, sizeof(*arr));
for (int i = 0; i < nx; i++) {
for (int j = 0; j < ny; j++) {
(*arr)[i][j] = 10 * i + j;
}
}
return 0;
}
int main() {
int nx = 3;
int ny = 2;
int *arr1d = NULL; // (1)
allocateArray1D(nx, &arr1d);
int(*arr2d)[ny] = NULL; // (2)
allocateArray2D(nx, ny, &arr2d);
for (int i = 0; i < nx; i++) {
printf("arr1d[%d] = %d \n", i, arr1d[i]);
}
printf("\n");
printf("arr2d \n");
for (int i = 0; i < nx; i++) {
for (int j = 0; j < ny; j++) {
printf(" %d ", arr2d[i][j]);
}
printf("\n");
}
return 0;
}
And the error message already comes during the compilation.
03.c(32): warning #167: argument of type "int (**)[ny]" is incompatible with parameter of type "int *(*)[*]"
allocateArray2D(nx, ny, &arr2d);
^
It is evident from the error message that it has been messed up with the argument types (that I wrote as int *(*arr)[ny]) but what should I have to put there? I tried some variants like int *((*arr)[ny]), but didn't work).
And if I remove the 2D parts, then the code well compiles, and run as expected. But I wonder if this is the right practice, at least for 1D case since there are many examples where the code behaves as expected, but in fact there were wrong or un-standard lines.
Also, the above code is not satisfactory in the first place. I want to even remove the lines in main() that I marked as (1) and (2).
So in the end I want a code something like this, but all with the 'array pointers'.
int **arr2d;
allocateArray2D(nx, ny, arr2d);
How could this be done?
You need to pass the array pointer by reference (not pass an array pointer to an array of int*):
int *(*arr)[ny] -> int (**arr)[ny]
The function becomes:
int allocateArray2D(int nx, int ny, int (**arr)[ny]) {
*arr = calloc(nx, sizeof(int[ny])); // or sizeof(**arr)
for (int i = 0; i < nx; i++) {
for (int j = 0; j < ny; j++) {
(*arr)[i][j] = 10 * i + j;
}
}
return 0;
}
For details, check out Correctly allocating multi-dimensional arrays
Best practices with malloc family is to always check if allocation succeeded and always free() at the end of the program.
As a micro-optimization, I'd rather recommend to use *arr = malloc( sizeof(int[nx][ny]) );, since calloc just creates pointless overhead bloat in the form of zero initialization. There's no use of it here since every item is assigned explicitly anyway.
Wrong parameter type
Strange allocation
Wrong size type
I would return the array as void * too (at least to check if allocation did not fail).
void *allocateArray2D(size_t nx, size_t ny, int (**arr)[ny]) {
//*arr = calloc(nx, sizeof(**arr)); calloc is not needed here as you assign values to the array
*arr = malloc(nx * sizeof(**arr));
for (size_t i = 0; i < nx; i++) {
for (size_t j = 0; j < ny; j++) {
(*arr)[i][j] = 10 * i + j;
}
}
return *arr;
}
I've written a piece of code but I'm not sure about how it works.
I want to create an array of pointers and pass it as argument to a function, like the following:
int main()
{
int *array[10] = {0};
for(int i = 0; i < 3; i++)
{
array[i] = (int *)malloc(3*sizeof(int *));
}
testFunction(array);
for(int i = 0; i < 3; i++)
{
free(array[i]);
}
return 0;
}
void testFunction(int *array[3])
{
//do something
return;
}
What I don't understand is the following. I declare array as an array of pointers, allocate memory to it by using malloc and then proceed to call testFunction. I want to pass the array by reference, and I understand that when I call the function by using testFunction(array), the array decays to a pointer to its first element (which will be a pointer also). But why in the parameters list I have to write (int *array[3]) with * and not just (int array[3])?
A parameter of type * can accept an argument of type [], but not anything in type.
If you write void testFunction(int arg[3]) it's fine, but you won't be able to access array[1] and array[2] and so on, only the first 3 elements of where array[0] points to. Also a comversion is required (call with testFunction((int*)array);.
As a good practice, it's necessary to make the function parametera consistent with what's passed as arguments. So int *array[10] can be passed to f(int **arg) or f(int *arg[]), but neither f(int *arg) nor f(int arg[]).
void testFunction(int **array, int int_arr_size, int size_of_array_of_pointers)
{
for(int j = 0; j < size_of_array_of_pointers; j++)
{
int *arrptr = array[j]; // this pointer only to understand the idea.
for(int i = 0; i < int_arr_size; i++)
{
arrptr[i] = i + 1;
}
}
}
and
int main()
{
int *array[10];
for(int i = 0; i < sizeof(array) / sizeof(int *); i++)
{
array[i] = malloc(3*sizeof(int));
}
testFunction(array, 3, sizeof(array) / sizeof(int *));
for(int i = 0; i < sizeof(array) / sizeof(int *); i++)
{
free(array[i]);
}
return 0;
}
Evering depends on what // do something means in your case.
Let's start from simple : perhaps, you need just array of integers
If your function change only values in array but does not change size, you can pass it as int *array or int array[3].
int *array[3] allows to work only with arrays of size 3, but if you can works with any arrays of int option int *array require additional argument int size:
void testFunction(int *array, int arr_size)
{
int i;
for(i = 0; i < arr_size; i++)
{
array[i] = i + 1;
}
return;
}
Next : if array of pointers are needed
Argument should be int *array[3] or better int **array (pointer to pointer).
Looking at the initialization loop (I changed sizeof(int *) to sizeof(int))
for(int i = 0; i < 3; i++)
{
array[i] = (int *)malloc(3*sizeof(int));
}
I suppose you need 2-dimension array, so you can pass int **array but with sizes of two dimensions or one size for case of square matrix (height equal to width):
void testFunction(int **array, int wSize, int hSize)
{
int row, col;
for(row = 0; row < hSize; row++)
{
for(col = 0; col < wSize; col++)
{
array[row][col] = row * col;
}
}
}
And finally : memory allocation for 2D-array
Consider the following variant of your main:
int main()
{
int **array;
// allocate memory for 3 pointers int*
array = (int *)malloc(3*sizeof(int *));
if(array == NULL)
return 1; // stop the program
// then init these 3 pointers with addreses for 3 int
for(int i = 0; i < 3; i++)
{
array[i] = (int *)malloc(3*sizeof(int));
if(array[i] == NULL) return 1;
}
testFunction(array, 3, 3);
// First, free memory allocated for int
for(int i = 0; i < 3; i++)
{
free(array[i]);
}
// then free memory allocated for pointers
free(array);
return 0;
}
Pay attention, that value returned by malloc should be checked before usage (NULL means memory was not allocated).
For the same reasons check can be added inside function:
void testFunction(int **array, int wSize, int hSize)
{
int row, col;
if(array == NULL) // check here
return;
for(row = 0; row < hSize; row++)
{
if(array[row] == NULL) // and here
return;
for(col = 0; col < wSize; col++)
{
array[row][col] = row * col;
}
}
}
I'm trying to generate a matrix of some arbitrary dimensions. I can do it just fine by calling scanf in main and then assigning matrix elements on a row by row basis, but trying to do it in a single function, outside of main, (and only if scanf() is called outside of main) gives me a segfault error:
int **genmat(int nrow, int ncol){
int i,j;
int **mat = (int**) malloc(sizeof(int)*ncol*nrow);
char rowbuff[16];
for(i=0; i < nrow; i++){
INPUT: scanf("%[^\n]%*c",rowbuff);
if(strlen(rowbuff) != ncol){
printf("Error: Input must be string of length %d\n", ncol);
goto INPUT;
}
else{
for(j=0; j < ncol; j++){
if(rowbuff[j] == '1'){
mat[i][j] = 1;
}
else{
mat[i][j] = 0;
}
}
}
}
return(mat);
}
The following works just fine:
int *genrow(int ncol, char *rowbuff){
int i;
int *row = malloc(sizeof(int)*ncol);
for(i=0;i<ncol;i++){
row[i] = rowbuff[i]%2;
}
return(row);
}
with the following in my main function to call genrow() for each row of the matrix:
for(i=0; i < row; i++){
INPUT: scanf("%[^\n]%*c",rowbuff);
if(strlen(rowbuff) != col){
printf("Error: Input must be string of length %d\n", col);
goto INPUT;
}
else{
int *newrow = genrow(col, rowbuff);
for(j=0; j < col; j++){
matrix[i][j] = newrow[j];
}
free(newrow);
newrow = NULL;
}
}
Why is the behavior different in these two contexts?
Dynamically allocated 2D arrays are unfortunately burdensome and ugly in C. To properly allocate one, it is very important that you do so with a single call to malloc, just as you tried to do. Otherwise it won't be a 2D array, but instead some segmented, slow look-up table.
However, the result of that malloc call will be a pointer to a 2D array, not a pointer-to-pointer. In fact, pointer-pointers have nothing to do with 2D arrays whatsoever - this is a widespread but incorrect belief.
What you should have done is this:
int (*mat)[nrow][ncol] = malloc( sizeof(int[nrow][ncol] );
This is an array pointer to a 2D array. This syntax is already a bit burdensome, but to make things worse, it is not easy to pass this array pointer back to main, because it is a local pointer variable. So you would need to use a pointer to an array pointer... and there's no pretty way to do that. It goes like this:
void genmat (size_t nrow, size_t ncol, int (**mat)[nrow][ncol] )
{
*mat = malloc( sizeof(int[nrow][ncol]) );
To ease usage a bit, you can create a temporary pointer to rows, which doesn't require multiple levels of indirection and is therefore much easier to work with:
int (*matrix)[ncol] = *mat[0]; // in the pointed-at 2D array, point at first row
for(size_t r=0; r<nrow; r++) // whatever you want to do with this matrix:
{
for(size_t c=0; c<ncol; c++)
{
matrix[r][c] = 1; // much more convenient syntax than (**mat)[r][c]
}
}
From main, you'll have to call the code like this:
size_t row = 3;
size_t col = 4;
int (*mat)[row][col];
genmat(row, col, &mat);
Example:
#include <stdio.h>
#include <stdlib.h>
void genmat (size_t nrow, size_t ncol, int (**mat)[nrow][ncol] )
{
*mat = malloc( sizeof(int[nrow][ncol]) );
int (*matrix)[ncol] = *mat[0];
for(size_t r=0; r<nrow; r++)
{
for(size_t c=0; c<ncol; c++)
{
matrix[r][c] = 1;
}
}
}
void printmat (size_t nrow, size_t ncol, int mat[nrow][ncol])
{
for(size_t r=0; r<nrow; r++)
{
for(size_t c=0; c<ncol; c++)
{
printf("%d ", mat[r][c]);
}
printf("\n");
}
}
int main (void)
{
size_t row = 3;
size_t col = 4;
int (*mat)[row][col];
genmat(row, col, &mat);
printmat(row, col, *mat);
free(mat);
return 0;
}
Please note that real code needs to address the case where malloc returns NULL.
I assume the problems is withint **mat = (int**) malloc(sizeof(int)*ncol*nrow);
You are trying to allocate the a 2D array right? But this isn't the correct method to allocate the memory. You can't allocate the whole chunk of memory one short.
What you should be doing here is, allocate the memory for all the rows(basically pointer to store the column address) and then for columns
int **mat= (int **)malloc(nrow * sizeof(int *));
for (i=0; i<nrow; i++)
mat[i] = (int *)malloc(ncol * sizeof(int));
Refer this link for more info http://www.geeksforgeeks.org/dynamically-allocate-2d-array-c/
I would like a function to return an array, or a pointer to an array, in either c or c++, that does something like the following:
double[][] copy(double b[][], int mx, int my, int nx, int ny)
{
double[nx][ny] a;
int i,j;
for(i = mx; i<= nx; ++i)
for(j = my; j<= ny; ++j)
a[i][j] = b[i][j];
return a;
}
void main(void){
double A[2][3];
double B[2][3] = {{1,2}{3,4}{5,6}};
A = copy(B,0,0,1,2);
}
This is the proper method for returning an array from a function is as follows:
#define NUM_ROWS (5)
#define NUM_COLS (3)
char **createCharArray(void)
{
char **charArray = malloc(NUM_ROWS * sizeof(*charArray));
for(int row = 0; row < NUM_ROWS; row++)
charArray[row] = malloc(NUM_COLS * sizeof(**charArray));
return charArray;
}
In this example, the above function can be called like this:
char **newCharArray = createCharArray();
newCharArray can now be used:
char ch = 'a';
for(int row = 0; row < NUM_ROWS; row++)
for(int col = 0; col < NUM_COLS; col++)
newCharArray[row][col] = ch++;
An array can be passed as an argument to function similarly:
void freeCharArray(char **charArr)
{
for(int row = 0; row < NUM_ROWS; row++)
free(charArr[row]);
free(charArr);
}
You can return the double ** from your copy function like this.
double ** copy(double *src, int row, int col)
{
// first allocate the array with required size
double **copiedArr = (double **) malloc(sizeof(double *)*row);
for(int i=0;i<row;i++)
{
// create space for the inner array
*(copiedArr+i) = (double *) malloc(sizeof(double)*col);
for(int j=0; j<col; j++)
{
// copy the values from source to destination.
*(*(copiedArr+i)+j) = (*(src+i+j));
}
}
// return the newly allocated array.
return copiedArr;
}
call to this function is done like this.
double src[3][3] = {{1,2,3},{4,5,6},{7,8,9}};
double **dest = copy(&src[0][0],3,3); //here is your new array
Here you have to assign returned value of copy() to double** not to double[][].
If you try to assign the returned value to array then it will generate "Incompatible types" error (detail).
As memory allocated to copiedArray on the heap so you have to take responsibility to clear the memory.
void freemem(double **mem, int row)
{
for(int i=0;i<row; i++)
{
free(*(mem+i));
}
free(mem);
}
I also want to point out some correction in your sample code:
return type of main should be int.
one should put the return statement at the end of main.
you can't return the stack allocated value, it is cause of crash
in most of cases.
#include <stdio.h>
#include <stdlib.h>
#define MAX_ROWS 5
#define MAX_COLS 5
int globalvariable = 100;
void CreateMatrix(int ***Matrix)
{
int **ptr;
char *cp;
int i = 0;
*Matrix = (int**)malloc((sizeof(int*) * MAX_ROWS) + ((MAX_ROWS * MAX_COLS)*sizeof(int)));
ptr = *Matrix;
cp = (char*)((char*)*Matrix + (sizeof(int*) * MAX_ROWS));
for(i =0; i < MAX_ROWS; i++)
{
cp = (char*)(cp + ((sizeof(int) * MAX_COLS) * i));
*ptr = (int*)cp;
ptr++;
}
}
void FillMatrix(int **Matrix)
{
int i = 0, j = 0;
for(i = 0; i < MAX_ROWS; i++)
{
for(j = 0; j < MAX_COLS; j++)
{
globalvariable++;
Matrix[i][j] = globalvariable;
}
}
}
void DisplayMatrix(int **Matrix)
{
int i = 0, j = 0;
for(i = 0; i < MAX_ROWS; i++)
{
printf("\n");
for(j = 0; j < MAX_COLS; j++)
{
printf("%d\t", Matrix[i][j]);
}
}
}
void FreeMatrix(int **Matrix)
{
free(Matrix);
}
int main()
{
int **Matrix1, **Matrix2;
CreateMatrix(&Matrix1);
FillMatrix(Matrix1);
DisplayMatrix(Matrix1);
FreeMatrix(Matrix1);
getchar();
return 0;
}
If the code is executed, I get the following error messages in a dialogbox.
Windows has triggered a breakpoint in sam.exe.
This may be due to a corruption of the heap, which indicates a bug in sam.exe or any of the DLLs it has loaded.
This may also be due to the user pressing F12 while sam.exe has focus.
The output window may have more diagnostic information.
I tried to debug in Visual Studio, when printf("\n"); statement of DisplayMatrix() is executed, same error message is reproduced.
If I press continue, it prints 101 to 125 as expected. In Release Mode, there is no issue !!!.
please share your ideas.
In C it is often simpler and more efficient to allocate a numerical matrix with calloc and use explicit index calculation ... so
int width = somewidth /* put some useful width computation */;
int height = someheight /* put some useful height computation */
int *mat = calloc(width*height, sizeof(int));
if (!mat) { perror ("calloc"); exit (EXIT_FAILURE); };
Then initialize and fill the matrix by computing the offset appropriately, e.g. something like
for (int i=0; i<width; i++)
for (int j=0; j<height; j++)
mat[i*height+j] = i+j;
if the matrix has (as you show) dimensions known at compile time, you could either stack allocate it with
{ int matrix [NUM_COLS][NUM_ROWS];
/* do something with matrix */
}
or heap allocate it. I find more readable to make it a struct like
struct matrix_st { int matfield [NUM_COLS][NUM_ROWS]; };
struct matrix_st *p = malloc(sizeof(struct matrix_st));
if (!p) { perror("malloc"); exit(EXIT_FAILURE); };
then fill it appropriately:
for (int i=0; i<NUM_COLS; i++)
for (int j=0; j<NUM_ROWS, j++)
p->matfield[i][j] = i+j;
Remember that malloc returns an uninitialized memory zone so you need to initialize all of it.
A two-dimensional array is not the same as a pointer-to-pointer. Maybe you meant
int (*mat)[MAX_COLS] = malloc(MAX_ROWS * sizeof(*mat));
instead?
Read this tutorial.
A very good & complete tutorial for pointers, you can go directly to Chapter 9, if you have in depth basic knowledge.