Attempt to access elements of a 2d struct array failing - c

typedef struct{
unsigned long a;
unsigned long b;
unsigned long c;
} mini_struct;
struct ministruct** build_2Dstruct(unsigned long x, unsigned long y){
double x_squared = pow(2, x);
struct ministruct** temp = (mini_struct**)malloc(x*sizeof(mini_struct*));
for(int i = 0; i < x_squared; i++){
temp[i] = (mini_struct*)malloc(y*sizeof(mini_struct));
for(int j = 0; j < y; j++){
temp[i][j].a = 0;
etc....
}
}
return temp;
}
In the code above I am trying to create a 2D array of ministructs**, with the whole struct being made out of 2^x ministructs*, and each ministruct* has y amount of ministructs.
aka:
x = 2,
y = 2,
[[struct, struct], [struct, struct], [struct, struct], [struct, struct]]
However, for some reason when I try to access the second element or index 1 of the struct inside each struct*, it says there is an error: "expression must be pointer to complete object".
I just do not understand why the code is not allowing me to access each individual element of the elements of the array?
Thanks

You are trying to make an x by y array of structs. So:
// create array of x pointers
mini_struct **temp = malloc(x*sizeof(mini_struct*));
for (int i=0; i<x; i++) {
// to array of y structs
temp[i] = malloc(y*sizeof(mini_struct));
for (int j=0; j < y; j++) {
temp[i][j].a = 0;
... etc.

Question is incomplete so I will be making asumptions.
You seem to be wanting to allocate a 2D array of structs and initialize all members to 0. Here is a possible solution:
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
typedef struct mini_struct{
unsigned long a;
unsigned long b;
unsigned long c;
} mini_struct;
struct mini_struct** build_2Dstruct(unsigned long x, unsigned long y){
double x_squared = pow(x, 2);
mini_struct **temp = (mini_struct **) malloc(x_squared * sizeof(mini_struct*));
for(int i = 0; i < x_squared; i++){
temp[i] = (mini_struct *) calloc(y, sizeof(mini_struct));
}
return temp;
}
int main () {
int x = 3;
int y = 4;
mini_struct **struct2D = build_2Dstruct(x, y);
int x_squared = pow(x,2);
for (int i = 0; i < x_squared; ++i) {
for (int j = 0; j < y; ++j) {
printf("Value of data stored at struct[%d][%d] is: %d\n", i, j, struct2D[i][j]);
}
}
for (int i = 0; i < x_squared; ++i) {
free(struct2D[i]);
}
free(struct2D);
}
As you can see, this contains the whole program, not just the snippet you showed. In this case, a main function would have been useful so that we don't have to guess what you want to do. My solution creates the 2D array with all elements initialized to 0 (you can use calloc to do that, no need for a second for loop).
Another important point is that, because the function returns a newly heap allocated 2D array, you need to free it to avoid a memory leak (end of main function).

You allocate x pointers to mini_struct:
mini_struct **temp = (mini_struct **) malloc(x_squared * sizeof(mini_struct*));
But then when you initialize them:
for(int i = 0; i < x_squared; i++){
temp[i] = (mini_struct *) calloc(y, sizeof(mini_struct));
}
You index temp based on upto x_squared.
Consider if x is 2. You would allocate temp to be an array of two pointers to mini_struct. But then your for loop would attempt to initialize four elements in temp.

Related

C allocate 'array pointer' inside a function

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;
}

Allocating contiguous memory for a 3D array in C

I need to allocate contiguous space for a 3D array. (EDIT:) I GUESS I SHOULD HAVE MADE THIS CLEAR IN THE FIRST PLACE but in the actual production code, I will not know the dimensions of the array until run time. I provided them as constants in my toy code below just to keep things simple. I know the potential problems of insisting on contiguous space, but I just have to have it. I have seen how to do this for a 2D array, but apparently I don't understand how to extend the pattern to 3D. When I call the function to free up the memory, free_3d_arr, I get an error:
lowest lvl
mid lvl
a.out(2248,0x7fff72d37000) malloc: *** error for object 0x7fab1a403310: pointer being freed was not allocated
Would appreciate it if anyone could tell me what the fix is. Code is here:
#include <stdio.h>
#include <stdlib.h>
int ***calloc_3d_arr(int sizes[3]){
int ***a;
int i,j;
a = calloc(sizes[0],sizeof(int**));
a[0] = calloc(sizes[0]*sizes[1],sizeof(int*));
a[0][0] = calloc(sizes[0]*sizes[1]*sizes[2],sizeof(int));
for (j=0; j<sizes[0]; j++) {
a[j] = (int**)(a[0][0]+sizes[1]*sizes[2]*j);
for (i=0; i<sizes[1]; i++) {
a[j][i] = (int*)(a[j]) + sizes[2]*i;
}
}
return a;
}
void free_3d_arr(int ***arr) {
printf("lowest lvl\n");
free(arr[0][0]);
printf("mid lvl\n");
free(arr[0]); // <--- This is a problem line, apparently.
printf("highest lvl\n");
free(arr);
}
int main() {
int ***a;
int sz[] = {5,4,3};
int i,j,k;
a = calloc_3d_arr(sz);
// do stuff with a
free_3d_arr(a);
}
Since you are using C, I would suggest that you use real multidimensional arrays:
int (*a)[sz[1]][sz[2]] = calloc(sz[0], sizeof(*a));
This allocates contiguous storage for your 3D array. Note that the sizes can be dynamic since C99. You access this array exactly as you would with your pointer arrays:
for(int i = 0; i < sz[0]; i++) {
for(int j = 0; j < sz[1]; j++) {
for(int k = 0; k < sz[2]; k++) {
a[i][j][k] = 42;
}
}
}
However, there are no pointer arrays under the hood, the indexing is done by the magic of pointer arithmetic and array-pointer-decay. And since a single calloc() was used to allocate the thing, a single free() suffices to get rid of it:
free(a); //that's it.
You can do something like this:
int ***allocateLinearMemory(int x, int y, int z)
{
int *p = (int*) malloc(x * y * z * sizeof(int));
int ***q = (int***) malloc(x * sizeof(int**));
for (int i = 0; i < x; i++)
{
q[i] = (int**) malloc(y * sizeof(int*));
for (int j = 0; j < y; j++)
{
int idx = x*j + x*y*i;
q[i][j] = &p[idx];
}
}
return q;
}
void deallocateLinearMemory(int x, int ***q)
{
free(q[0][0]);
for(int i = 0; i < x; i++)
{
free(q[i]);
}
free(q);
}
I use it and works fine.

Pass static two dimensional struct array to function by reference

I try to pass a static two dimensional struct as a reference to a function.
But I don't know how to get that done in correct way.
From my understanding, I pass a pointer to the first element of struct test to initfield(). C does know the size of the struct test so I can jump to the specific requested locations of the data. I just don't know how to adress the required data.
Here's my code that hopefully describes what I am looking for.
struct test{
int i;
double d;
};
void initfield(struct test *a, int structsize)
{
int i, j;
for (i = 0; i < structsize; i++)
{
for (j = 0; j < structsize; j++)
{
a[i][j]->i = 1;
a[i][j]->d = 1.0;
}
}
}
int main(void)
{
int i, j;
struct test field[8][8];
initfield(field, 8);
for (i = 0; i < 8; i++)
{
for (j = 0; j < 8; j++)
{
printf("test[%i][%i].i = %i", i, j, test.i);
printf("test[%i][%i].d = %i", i, j, test.d);
}
}
return 0;
}
Update :
I've replaced both printf's with the following :
printf("test[%i][%i].i = %i", i, j, field[i][j].i);
printf("test[%i][%i].d = %lf", i, j, field[i][j].d);
However, I still encounter errors with initfield.
The problem, is actually, in your initfield() code,
void initfield(struct test *a, int structsize)
a is of type struct test *, and later, you're doing
a[i][j]->i = 1;
which expects a to be struct test **
That said,
for (j = 0; j < 8; j++)
{
printf("test[%i][%i].i = %i", i, j, test.i);
printf("test[%i][%i].d = %i", i, j, test.d);
}
is completely wrong. Neither is there any variable called test, nor you can access a 2-D array using structVar.membervar format. Moreover, you are using %d to print a double, which in turn invokes undefined behaviour.
Solution: You can make use of array properties and pointer arithmetic to achieve what you want.
You have to change the loop inside the initfield() code, like
for (i = 0; i < structsize; i++)
{
for (j = 0; j < structsize; j++)
{
((a+(i*structsize))+j)->i = 7; //I changed the value to store, just like that
((a+(i*structsize))+j)->d = 2.0; //I changed the value to store, just like that
}
}
and, in main(), use %f to print the values.
A Live variant
First, structsize is not a good identifier name. It's not the size of the struct, it's the size of one dimension of the array. I'd implement it with parameters x and y, or width and heigth, or any better names for the two dimensions.
field is also a bad name. A field is often used to call a member of a struct. It is very confusing to use the identifier field to call an array of array of structs.
Then to your problem: field is an array of array of structs. In function parameter, this is equivalent to a pointer to a pointer.
The first parameter of initfield should be of type struct test **. Then later in the function, you dereference twice with your [] operators:
a is of type struct test **
a[i] is of type struct test *
a[i][j] is of type struct test
To access the fields of a[i][j], you need the . operator since its a struct test: a[i][j].d. The operator -> would work if a[i][j] was of type struct test *, but it isn't.
In this case it doesn't matter: as other have said, you can't access the second dimension of the array without explicitly calculating with the help of the size of the first dimension. a[i][j] does not work, you need some kind of pointer arithmetic: struct test *p = a + i * structsize + j and use p->i and p->d.
In the main function however, the dimensions of field are know, so field[i][j].d works.
You're assuming C can figure out that a inside refers to a square array with side length structsize, although you clearly say that a has type "pointer to struct test", which is not the same.
You need to do the indexing manually, inside the function:
static void initfield(struct test *a00, size_t sidelength)
{
for(size_t i = 0; i < sidelength; ++i)
{
for(size_t j = 0; j < sidelength; ++j)
{
struct test *aij = a00 + i * sidelength + j;
aij->i = 1;
aij->j = 1.0;
}
}
}
I didn't test the above, but something like that should work. It basically just uses simple pointer arithmetic to compute the address of the 2D array element at (i, j) given the address of the one at (0, 0).
It works fine i din't thought about doing calculation the adresses myself.
Thanks very much!
Here's my final adapted code which just works perfectly!
struct test{
int i;
double d;
};
void initfield(struct test *a00, int structsize)
{
int i, j;
for (i = 0; i < structsize; i++)
{
for (j = 0; j < structsize; j++)
{
struct test *acurrent = a00 + i * structsize + j;
acurrent->i = 1;
acurrent->d = 1.0;
}
}
}
int main(void)
{
int i, j;
struct test field[8][8];
initfield(&field[0][0], 8);
for (i = 0; i < 8; i++)
{
for (j = 0; j < 8; j++)
{
printf("test[%i][%i].i = %i\n", i, j, field[i][j].i);
printf("test[%i][%i].d = %lf\n", i, j, field[i][j].d);
}
}
return 0;
}
Best way to do this:
void initfield(size_t x, size_t y, struct test a[x][y]);
Be aware though, C is strange, the above is still a pointer, equivalent to struct test*. If you wish to have a true array pointer type, you'd have to do:
void initfield(size_t x, size_t y, struct test a[x][y])
{
struct test (*ptr)[x][y] = (void*)a;
or preferably:
struct test (*ptr)[y] = *a;
ptr[i][j] = something; // use the array pointer with sane syntax

int** vs int[const][const] differences

I was writing a code the other day and I found it rather strange, that int** and int[][] does not behave the same way. Can anyone point out the differences between them? Below is my sample code, which fails with a segmentation fault, if I pass constant size 2d array, while it does work fine when I pass a dinamically allocated 2d array.
I am confused mainly because ant int[] array works the same as int*.
#include<stdio.h>
#include<stdlib.h>
void sort_by_first_row(int **t, int n, int m)
{
int i, j;
for(i = m-1 ; i > 0 ; --i)
{
for(j = 0 ; j < i; ++j)
{
if(t[0][j] < t[0][j+1])
{
int k;
for(k = 0 ; k < n ;++k)
{
int swap;
swap = t[k][j];
t[k][j] = t[k][j+1];
t[k][j+1] = swap;
}
}
}
}
}
int main(void) {
int i, j;
/* Working version */
/*int **t;
t =(int**) malloc(3*sizeof(int*));
for(i = 0; i < 3; ++i)
{
t[i] = (int*) malloc(6*sizeof(int));
}*/
/*WRONG*/
int t[3][6];
t[0][0] = 121;
t[0][1] = 85;
t[0][2] = 54;
t[0][3] = 89;
t[0][4] = 879;
t[0][5] = 11;
for( i = 0; i < 6; ++i )
t[1][i] = i+1;
t[2][0] = 2;
t[2][1] = 4;
t[2][2] = 5;
t[2][3] = 3;
t[2][4] = 1;
t[2][5] = 6;
sort_by_first_row(t, 3, 6);
for(i = 0; i < 3; ++i)
{
for(j = 0; j < 6; ++j)
printf("%d ", t[i][j]);
printf("\n");
}
return 0;
}
So based on the below answers I realize, that a multidimensional array is stored continuously in a row major order. After some modification, the below code works:
#include<stdio.h>
#include<stdlib.h>
void sort_by_first_row(int *t, int n, int m)
{
int i, j;
for(i = m-1 ; i > 0 ; --i)
{
for(j = 0 ; j < i; ++j)
{
if(t[j] < t[j+1])
{
int k;
for(k = 0 ; k < n ;++k)
{
int swap;
swap = t[k*m + j];
t[k*m + j] = t[k*m + j+1];
t[k*m + j+1] = swap;
}
}
}
}
}
int main(void) {
int i, j;
/* Working version */
/*int **t;
t =(int**) malloc(3*sizeof(int*));
for(i = 0; i < 3; ++i)
{
t[i] = (int*) malloc(6*sizeof(int));
}*/
/*WRONG*/
int t[3][6];
t[0][0] = 121;
t[0][1] = 85;
t[0][2] = 54;
t[0][3] = 89;
t[0][4] = 879;
t[0][5] = 11;
for( i = 0; i < 6; ++i )
t[1][i] = i+1;
t[2][0] = 2;
t[2][1] = 4;
t[2][2] = 5;
t[2][3] = 3;
t[2][4] = 1;
t[2][5] = 6;
sort_by_first_row(t, 3, 6);
for(i = 0; i < 3; ++i)
{
for(j = 0; j < 6; ++j)
printf("%d ", t[i][j]);
printf("\n");
}
return 0;
}
My new question is this: How to modify the code, so that the procedure works with int[][] and int** also?
Realize that int **t makes t a pointer to a pointer, while int t[3][6] makes t an array of an array. In most cases, when an array is used in an expression, it will become the value of the address of its first member. So, for int t[3][6], when t is passed to a function, the function will actually be getting the value of &t[0], which has type pointer to an array (in this case, int (*)[6]).
The type of what is being pointed at is important for how the pointer behaves when indexed. When a pointer to an object is incremented by 5, it points to the 5th object following the current object. Thus, for int **t, t + 5 will point to the 5th pointer, while for int (*t)[M], t + 5 will point to the 5th array. That is, the result of t + 5 is the same as the result of &t[5].
In your case, you have implemented void sort_by_first_row(int **t, int n, int m), but you are passing it an incompatible pointer. That is, the type of &t[0] (which is what t will become in main) is not the same as what the function wants, a int **t. Thus, when the sorting function starts to use that address, it will think its indexing into pointers, when the underlying structure is an array of arrays.
int** is quite different from int[][]. int** is simply a pointer to a pointer and would appear like the following:
in reality, you can access the entire multidimensional array with simply int* pointing to the first element, and doing simple math from that.
Here is the result of the separate allocations (in your commented code):
However when you allocate a multidimensional array, all of the memory is contiguous, and therefore easy to do simple math to reach the desired element.
int t[3][6];
int *t = (int*) malloc((3 * 6) * sizeof(int)); // <-- similarly
This will result in a contiguous chunk of memory for all elements.
You certainly can use the separate allocations, however you will need to walk the memory differently.
Hope this helps.
int t[3][6] is very nearly the same thing as int t[18]. A single contiguous block of 18 integers is allocated in both cases. The variable t provides the address of the start of this block of integers, just like the one-dimensional case.
Contrast this with the case you have marked as "working", where t gives you the address of a block of 3 pointers, each of which points to a block of memory with 6 integers. It's a totally different animal.
The difference between t[3][6] and t[18] is that the compiler remembers the size of each dimension of the array, and automatically converts 2D indices into 1D offsets. For example, the compiler automatically converts t[1][2] into *(t + 1*6 + 2) (equivalent to t[8] if it were declared as a one-dimensional array).
When you pass a multi-dimensional array to a function, there are two ways to handle it. The first is to declare the function argument as an array with known dimension sizes. The second is to treat your array like a 1D array.
If you are going to declare the size of your array, you would declare your function like this:
void sort_by_first_row(int t[][6], int n)
or this
void sort_by_first_row(int t[3][6])
You either have to declare all array dimension sizes, or you can leave out the first size. In both cases, you access elements of t using t[i][j]; you've given the compiler enough information to do the offset math that converts from 2D notation to a 1D index offset.
If you treat it as a 1D array, you have to pass the array dimensions and then do the offset math yourself.
Here's a full working example, where f and f2 both generate the same output:
void f(int* t, int m, int n)
{
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
std::cout << t[i * n + j] << " ";
std::cout << std::endl;
}
void f2(int t[][6], int m)
{
for (int i = 0; i < m; i++)
for (int j = 0; j < 6; j++)
std::cout << t[i][j] << " ";
std::cout << std::endl;
}
int main()
{
int t[3][6];
int val = 1;
for (int i = 0; i < 3; i++)
{
for (int j = 0; j < 6; j++)
{
t[i][j] = val;
val++;
}
}
f(&(t[0][0]), 3, 6);
f2(t, 3);
return 0;
}
One thing to note is the hack-ish way I had to pass t to f. It's been a while since I wrote in C/C++, but I remember being able to pass t directly. Maybe somebody can fill me in on why my current compiler won't let me.
A int ** is a pointer to a pointer to an int, and can be a pointer to an array of pointers to arrays of ints. A int [][] is a 2-dimensional array of ints. A two-dimensional array is exactly the same as a one-dimensional array in C in one respect: It is fundamentally a pointer to the first object. The only difference is the accessing, a two-dimensional array is accessed with two different strides simultaneously.
Long story short, a int[][] is closer to an int* than an int**.

C and dynamic structure element access

I have this complicated structure thingie:
#include <stdlib.h>
typedef struct {
int x;
int y;
} SUB;
typedef struct {
int a;
SUB *z;
} STRUCT;
#define NUM 5
int main(void)
{
STRUCT *example;
int i;
example = malloc(sizeof(STRUCT));
example->z = malloc(NUM * sizeof(SUB));
for(i = 0; i < NUM; ++i) {
/* how do I access variable in certain struct of array of z's */
}
return 0;
}
example is dynamically allocated structure and z inside the example is dynamically allocated array of SUB structures.
How do I access certain variable in certain element of structure z?
I have been trying something like this: example->z[i].x but it doesnt seem to work.
At the moment I am using this shabby looking workaraound:
SUB *ptr = example->z;
int i;
for(i = 0; i < amount_of_z_structs; ++i) {
/* do something with 'ptr->x' and 'ptr->y' */
ptr += sizeof(SUB);
}
Your problem isn't where you say it is. Your code as posted gives a compile error:
error: request for member ā€˜zā€™ in something not a structure or union
at the line
example.z = malloc(sizeof(STRUCT));
because you meant to write example->z, since example is a pointer to STRUCT, not a STRUCT.
From there on, you can access example->z[i].x exactly as you said. That syntax has always been fine.
For example:
/* your declarations here */
example = malloc(sizeof(STRUCT));
example->z = malloc(NUM * sizeof(SUB));
for(i = 0; i < NUM; ++i) {
example->z[i].x = i;
example->z[i].y = -i;
printf("%d %d\n", example->z[i].x, example->z[i].y);
}
/* output:
0 0
1 -1
2 -2
3 -3
4 -4
*/
When you have pointers pointing to pointers you often end up running into precedence issues. I can't recall if this is one, but you might try (example->b)[i].x.
First of all, your second malloc is wrong; example is a pointer so this:
example.z = malloc(NUM * sizeof(SUB));
should be this:
example->z = malloc(NUM * sizeof(SUB));
Then in your loop you can say things like this:
example->z[i].x = i;
example->z[i].y = i;
You'll also want to have this near the top of your file:
#include <stdlib.h>
Try this:
int my_x = example[3].z[2].x;
The above code will first access the example[3] (the fourth element of the example array).
Once you get that particular element, its contents can be automatically access in the same way as you do with normal objects.
You then access z[2] from that element. Note that, example[3] is an element, so you could use a . to access its members; if its an array, you can access it as an array.
So till now, example[3].z[2] is one element of the SUB array inside one element of the example array.
Now you can simply access the member x using the way shown above.
typedef struct {
int x;
int y;
} SUB;
typedef struct {
int a;
SUB *z;
} STRUCT;
STRUCT *example;
int main() {
example = malloc(sizeof(STRUCT)*10); //array of 10;
int i=0,j=0;
for (;i<10;i++){
example[i].a = i;
example[i].z = malloc(sizeof(SUB)*5);
for (j=0; j<5; j++)
example[i].z[j].x = example[i].z[j].y = j;
}
//access example[3] and access z[2] inside it. And finally access 'x'
int my_x = example[3].z[2].x;
printf("%d",my_x);
for (i=0;i<10;i++){
printf("%d |\n",example[i].a);
//example[i].z = malloc(sizeof(SUB)*5);
for (j=0; j<5; j++)
printf("%d %d\n",example[i].z[j].x,example[i].z[j].y);
free(example[i].z);
}
free(example);
}
In the 'shabby workaround', you wrote:
SUB *ptr = example->z;
int i;
for(i = 0; i < amount_of_z_structs; ++i) {
/* do something with 'ptr->x' and 'ptr->y' */
ptr += sizeof(SUB);
}
The problem here is that C scales pointers by the size of the object pointed to, so when you add 1 to a SUB pointer, the value is advanced by sizeof(SUB). So, you simply need:
SUB *ptr = example->z;
int i;
for (i = 0; i < NUM; ++i) {
ptr->x = ptr->y = 0;
ptr++;
}
Of course, as others have said, you can also do (assuming C99):
for (int i = 0; i < NUM; ++i)
example->z[i].x = example->z[i].y = 0;
#include<stdio.h>
#include<stdlib.h>
#include<conio.h>
#define NUM 5
typedef struct
{
int x;
int y;
}SUB;
typedef struct
{
int a;
SUB* z;
}STRUCT;
void main(void)
{
clrscr();
printf("Sample problem..\n\n");
STRUCT* example;
int i;
example = (STRUCT*)malloc(sizeof(STRUCT));
example->z = (SUB*)malloc(NUM * sizeof(SUB));
for(i = 0; i < NUM; i++)
{
example->z[i].x = i +1;
example->z[i].y = (example->z[i].x)+1;
printf("i = %d: x:%d y:%d\n", i, example->z[i].x, example->z[i].y);
}
}

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