I have the following code:
int width = 10;
int height = 9;
float** matrix = (float**) malloc(height*sizeof(float));
for (unsigned int i = 0; i < height; i++)
matrix[i] = (float*) malloc(width*sizeof(float));
//This works
matrix[6][0] = 3.0f;
for (unsigned int j = 0; j < width; j++) {
for (unsigned int i = 0; i < height; i++) {
//This fails on [6][0]
matrix[i][j] = 3.0f;
}
}
I am attempting to create a 2D array and initialise it however a segmentation fault is received when setting the value of matrix[6][0] within the loop. What I'm finding very strange is that no error is thrown when I set matrix[6][0] outside of the loop. It's my understanding that a segmentation fault occurs when illegal memory is accessed but I cannot find any reason why different memory is being accessed within the loop. I have even examined the assembly code to find out what's happening but I cannot find the problem.
Update: The code was part of a CUDA program (C++) but my brain was half thinking about C and half about C++, hence why it became a mess.
float** matrix = (float**)(float*) malloc(height*sizeof(float));
Instead of sizeof(float) use sizeof(float *) . You need to allocate memory for height number of float pointers .
So your code goes like this -
float** matrix = malloc(height*sizeof(float *));
Your code produces seg fault while using loop as you don't correct amount of memory for float pointers and you end up accessing invalid memory , thus getting undefined behaviour.
Note- Don't use cast for malloc . And check retrun of malloc.
Here is the refined code:
int width = 10;
int height = 9;
float **matrix = malloc(height * sizeof (float *));
for (size_t i = 0; i < height; i++)
matrix[i] = malloc(width * sizeof(float));
matrix is "a pointer to pointers to floats", while matrix[i] are "pointers to floats".
Related
I am getting segmentation fault whenever I do this. I wonder if there is a way for me to assign a value to the struct without getting SegFault?
typedef struct _chunk
{
int lo; // lower bound
int hi; // higher bound
} chunk;
chunk_stack = (chunk **)malloc(10 * 10 * sizeof(chunk **));
for (i = 0; i < chunk_per_thread; i++)
{
chunk_stack[myid][i].lo = 0;
chunk_stack[myid][i].hi = 1;
}
Suppose you need to allocate a 2D array of size r * c.You need to first allocate memory for the double pointer which you did but have an error in it i.e., one extra * inside malloc function.
chunk **chunk_stack = (chunk **)malloc(r * sizeof(chunk *));
Then you need to allocate memory for each of the rows separately.
for(int i = 0; i < r; i++){
chunk_stack[i] = (chunk*)malloc(c * sizeof(chunk));
}
I am making a Rubik's Cube that can rotate it's rows and columns. My struct is this:
typedef struct
{
int **cell;
}face;
In this part I am allocating memory for cells. (X is user input)
face faceOne; //There are six of this faces
faceOne.cell = (int **)malloc(x * sizeof(int));
for(int i = 0; i < x; i++)
faceOne.cell[i] = (int *)malloc(x * sizeof(int));
Then I am filling these cells:
for (int i = 0; i < x; i++)
for (int j = 0; j < x; j++)
{
faceOne.cell[i][j] = 0;
printf("%d\n", i);
}
If x is bigger than 3, program crashes with Segmentation fault. How can i solve this?
faceOne.cell = (int **)malloc(x * sizeof(int));
would be
faceOne.cell = malloc(x * sizeof(int*));
Otherwise you are allocating memory for x int variables and then storing int* variable on it. In case sizeof(int) = sizeof(int*) then it wont be any problem but on the system where this is not the case you have earned yoruself a undefined behavior.
To get rid of all this ** while using malloc the best would be to do this
faceOne.cell = malloc(x * sizeof *faceOne.cell);
Similarly,
faceOne.cell[i] = malloc(x * sizeof *faceOne.cell[i] );
And also don't cast the return value of malloc. This suppresses many warning generated by compiler and is unnecessary.
Apart from all this, have you ever wondered what would happen if malloc did fail and returned NULL? Then you are working on a NULL and dereferencing it(Undefined behavior). Check the return value of malloc, in case it returns NULL handle the error and move on.
You want here
faceOne.cell = (int **)malloc(x * sizeof(int));
to allocate x pointers to int, so this should be
faceOne.cell = malloc(x * sizeof(int *));
It seems on your system the size of a pointer (to int) is bigger than the size of an int ...
Also in C don't cast malloc() (see this)
My goal is to dynamically reallocate memory for a two dimensional int array in C. I know there are already several questions about that topic, but unfortunately my code does not run properly and i don't know what is going wrong.
First i am allocating memory:
int n = 10;
int m = 4;
int** twoDimArray;
twoDimArray = (int**)malloc(n * sizeof(int*));
for(int i = 0; i < n; i++) {
twoDimArray[i] = (int*)malloc(m * sizeof(int));
}
And initializing the array with integer numbers:
for(int i = 0; i < n; i++) {
for(j = 0; j < 4; j++) {
twoDimArray[i][j] = i * j;
}
}
Then i use realloc() to reallocate memory dynamically:
int plus = 10;
int newArraySize = n + plus;
twoDimArray = (int**)realloc(twoDimArray, newArraySize * sizeof(int));
I am expecting my aray twoDimArray to be accessible at [10][0] now, but when running
printf("twoDimArray[10][0] = %d\n", twoDimArray[10][0]);
i get an "EXC_BAD_ACCESS" runtime error.
Probably i am missing something rather simple, but since i am new to C and can't figure out my mistake. Any help is appreciated.
reallocating the array of pointers is necessary, but then you have only n values that point to something valid. You need to allocate the rest of the sub-arrays because the newly allocated memory points to unallocated/invalid areas. The error is not from accessing the pointer, but from dereferencing it.
You need to add something like:
for(int i = n; i < n+plus; i++) {
twoDimArray[i] = malloc(m * sizeof(int));
}
(same goes for deallocation: first deallocate the arrays in a loop, then deallocate the array of pointers)
Aside:
assigning the result of realloc directly to the original variable can be a problem if realloc returns NULL. Even if it's rare in those resizing cases (Under what circumstances can malloc return NULL?), you should copy the result in a temp variable, check for NULL, and free the old pointer if reallocation failed.
Do I cast the result of malloc?
I'm trying to code a function that would ask the user to enter matrix. It prompts the number of lines, the number of columns and then prompts the value for each element of the matrix:
#include <stdio.h>
#include <stdlib.h>
void enterMatrix(float ** matrix, int nbLines, int nbColumns){
for (int i = 0; i < nbLines * nbColumns; i++){
printf("i = %d? ", i);
scanf("%f", matrix[i]);
}
}
int main(void){
int nbLines, nbColumns;
printf("nbLines? "); scanf("%d", &nbLines);
printf("nbColumns? "); scanf("%d", &nbColumns);
float *matrix[nbL * nbCol];
enterMatrix(matrix, nbLines, nbColumns);
}
Everything works fine until I enter a value for i = 0 and then press enter, which leads to a segmentation fault.
Any idea as to what could be wrong?
Your problem is because
float *matrice[nbL * nbCol];
defines an array of uninitialised pointers (i.e. an array of float *), not an array of float. This is then passed to enterMatrix() as a pointer to pointer to float (i.e. a float **). The scanf() calls then read to matrix[i] which is an uninitialised pointer. The result is undefined behaviour.
One fix would be to change the definition of matrice in main() to
float matrice[nbL * nbCol];
and change the function to (I've used comment to highlight changes)
void enterMatrix(float *matrix, int nbLines, int nbColumns) /* note type of matrix */
{
for (int i = 0; i < nbLines * nbColumns; i++)
{
printf("i = %d? ", i);
scanf("%f", &matrix[i]); /* note what the second argument is */
}
}
You need to allocate memory dinamically since you do not know at compile time which values the variables nbLines and nbColumns will hold.
So you need to first declare the pointer to the matrix:
float **matrix;
Then start allocate memory according to user input:
matrix = (float **)malloc(nbLines * sizeof(float *));
for (int i = 0; i < nbLines; i++)
{
matrix[i] = (float *)malloc(nbColums * sizeof(float ));
}
The segmentation fault error happens becouse you are not allocating memory for your matrix, but just a [nbL * nbCol] pointers to float
You don't allocate enough memory for your array, thus you invoke Undefined Behavior, since you go out of bounds, causing a Segmentation fault.
You could declare it like a 2D array, like this:
/* TODO: Check if allocation succeeded. (check for NULL pointer) */
float** matrix;
matrix = malloc(nbLines * sizeof(float*));
for(int i = 0 ; i < N ; i++)
matrix[i] = malloc(nbColumns * sizeof(float));
I have other approaches for allocating a 2D array dynamically here.
Please note: Do I cast the result of malloc? No!
Also do not forget to free().
You could emulate a 2D array with an 1D array, like this:
void enterMatrix(float* matrix, int nbLines, int nbColumns){
for (int i = 0; i < nbLines ; i++) {
for (int j = 0; j < nbColumns; j++) {
scanf("%f", matrix[i + nbColumns * j]);
}
}
float matrix[nbLines * nbColumns];
You're creating a variable length array, and, of pointers.
While other answers are perfectly valid, if you really want a 2D array, you just need to change the declaration:
float matrix[nbLines][nbColumns];
to declare a 2D variable length array of floats.
Now the hard point is to pass this VLA to a function and preserve dimensions.
For that you could use C99 way of passing VLAs (note that dimensions must be located before the VLA itself). Reference: Passing a multidimensional variable length array to a function:
void enterMatrix(int nbLines, int nbColumns, float matrix[][nbColumns] ){
for (int i = 0; i < nbLines; i++){
for (int j = 0; j < nbColumns; j++)
{
scanf("%f", &matrix[i][j]);
}
}
}
call as follows:
enterMatrix(nbLines, nbColumns, matrix);
I want to declare the correct pointer and allocate memory for a two-dimensional array, and correctly pass to a function. I am having trouble getting this to work properly and need help.
Here is my code:
double **podrucje;
podrucje=(double **) malloc (sizeof (double *) *123);
for (i=0;i<(123);i++)
{
podrucje[i]=(double *) malloc (sizeof (double) * 11);
}
for (i=0;i<(123);i++)
{
memset (podrucje[i], 0, 10);
}
But this code doesnt work, it messes up whole my program. So i decided to give up on dynamic allocation and use this:
double podrucje[123][11]={0};
But i dont know how to send it and use it in function...
memset works on per byte basis.
double **podrucje = (double **) malloc (sizeof (double *) * 123);
for (i = 0; i < 123; i++)
{
podrucje[i] = (double *) malloc (sizeof (double) * 11);
memset(podrucje[i], 0, sizeof(double) * 11);
}
if you want to pass it just declare it as such
void function(double podrucje[123][11]) {
...
}
You're best off to use malloc, but allocate the whole array on your second line, so it all gets allocated in contiguous memory. So
podrucje = (double*) malloc (sizeof (double) * 123 * 11);
Then the first loop can go away too. And it looks like you're initializing the array to 0 -- in that case, use calloc instead of malloc, eliminating the second loop.
To index into the array, use things like
double myitem = podrucje [11 * row + col];
You should of course use a define or similar to keep the use of 11 consistent, but that's not the point of this answer.
Write the function argument the same way you wrote the variable definition:
void myfunc(double podrucje[123][11])
{
...
}
double podrucje[123][11];
myfunc(podrucje);
Note that the array is passed "by reference" rather than being copied.
In the following code snippet,
podrucje[i]=(double *) malloc (sizeof (double) * 11);
for (i=0;i<(123);i++)
{
memset (podrucje[i], 0, 10);
}
1) You dont need the extra parenthesis against the numbers 123 and 11
The for loop can be as follows,
for (i = 0; i < 123; i++)
2) Instead of using 123 and 11. Try to define a MACRO and use that instead.
Advantage: The code becomes independent of special numbers and is easily maintainable. Especially in the cases of larger code base.
3) If you read the code, podrucje[i] is allocated a memory of 11 doubles But when you memset you are setting it for only 10 doubles while the last one may or may not consist of garbage. Use calloc here, It not only allocated memory but also initializes the same.
podrucje[i]=(double *) calloc(11,sizeof(double));
Also It could be more helpful if you could tell How exactly is it screwing up your code ?
Example, Code Snippet could help more than just stating its screwing up.
It helps others to investigate and solve the issue.
If you have a modern C compiler (C99 would do) you can even declare real 2D matrices with variable sizes. You don't need to fall back to this awful emulation with pointers.
void myfunc(size_t n, size_t m, double podrucje[n][m])
{
...
}
double (*podrucje)[n] = malloc(sizeof(double[n][m]));
myfunc(n, m, podrucje);
For the function you just have to ensure that the declarations of n and m come first, before the matrix.
There are several ways to dynamically allocate memory for an NxM array. Here are two:
You can declare a pointer to an M-element array, and then malloc N instances of it:
double (*podrucje)[11] = malloc(sizeof *podrucje * 123);
As of C89, you don't need to cast the result of malloc, and the practice is discouraged. Also, note that the operand to sizeof is the expression *podrucje; this gives me the same result as sizeof (double) * 11.
You would index this array as podrucje[i][j] like any other 2D array. podrucje[i] implicitly dereferences the pointer (remember that a[i] is equivalent to *(a + i)) so you don't have to do anything funky with it.
You would use it in a function as follows:
void init(double (*podrucje)[11], size_t rows)
{
size_t i, j;
for (i = 0; i < rows; i++)
for (j = 0; j < 11; j++)
podrucje[i][j] = 0.0;
}
which would be called as
init(podrucje, 123);
The drawback to this method is that the function can only operate on Nx11 arrays; if you're using a C99 compiler or a C2011 compiler that supports variable length arrays, you could specify the number of columns as a runtime variable:
void foo(void)
{
size_t rows = 123, cols = 11;
double (*podrucje)[cols] = malloc(sizeof *podrucje * rows);
if (podrucje)
init(cols, podrucje, rows);
...
}
// cols must be declared before it can be used
// in an array declarator
//
void init(size_t cols, double(*podrucje)[cols], size_t rows)
{
size_t i, j;
for (i = 0; i < rows; i++)
for (j = 0; j < cols; j++)
podrucje[i][j] = 0.0;
}
When you're done with the array, deallocate it as follows:
free(podrucje);
The other approach is to allocate each row separately, as follows:
size_t rows = 123, cols = 11;
double **podrucje = malloc(sizeof *podrucje * rows);
if (!podrucje)
{
// malloc failed; handle allocation error here
}
else
{
size_t i;
for (i = 0; i < rows; i++)
{
podrucje[i] = malloc(sizeof *podrucje[i] * cols);
if (!podrucje[i])
{
// malloc failed; handle allocation error here
}
}
}
And you would use it in a function as follows:
void foo()
{
double **podrucje;
// allocate array as above
init(foo, rows, cols);
...
}
void init(double **podrucje, size_t rows, size_t cols)
{
size_t i, j;
for (i = 0; i < rows; i++)
for (j = 0; j < cols; j++)
podrucje[i][j] = 0.0;
}
When you're finished with the array, deallocate it as follows:
for(i = 0; i < rows; i++)
free(podrucje[i]);
free(podrucje);
The first method allocates memory as a single, contiguous block; the second allocates it in a series of smaller, discontinuous chunks. If your array is especially big or your heap especially fragmented, the first method may fail where the second will succeed. If you're working with a compiler that doesn't support variable-length arrays, the first method is much less flexible, because the number of columns must be specified at compile time.
How could the same indexing method work for both forms?
In the first case, each podrucje[i] is an 11-element array of double; indexing it with j works like any other array. In the second case, each podrucje[i] is a pointer to double. Since a[i] is evaluated as *(a + i), array indexing works on pointer expressions just as well as array expressions.
int print_func(char((*p)[26])[10])
{
int i = 0;
for (i=0; i < 26 ; i++) {
fprintf(stderr, "%02d:%.*s\n", i, 10, p[0][i]);
}
return (0);
}
int main(void)
{
int nrow = 26;
int ncol = 10;
char((*p)[26])[10] = (char((*)[26])[10])(0);
char(*q)[10];
char c = 'a';
int i = 0;
p = (char((*)[26])[10])malloc(sizeof(char) * nrow * ncol);
if ((char((*)[26])[10])0 == p) {
return (-1);
}
for (i=0, q=p[0]; i < nrow ; i++) {
memset(q, c, sizeof(char) * ncol);
c++;
q++;
}
for (i=0,q=p[0] ; i < nrow ; i++) {
fprintf(stderr, "%.*s\n", 10, q);
q++;
}
p[0][8][0]='z';
getchar();
print_func(p);
return (0);
}