I'm trying to create an array of structs and also a pointer to that array. I don't know how large the array is going to be, so it should be dynamic. My struct would look something like this:
typedef struct _stats_t
{
int hours[24]; int numPostsInHour;
int days[7]; int numPostsInDay;
int weeks[20]; int numPostsInWeek;
int totNumLinesInPosts;
int numPostsAnalyzed;
} stats_t;
... and I need to have multiple of these structs for each file (unknown amount) that I will analyze. I'm not sure how to do this. I don't like the following approach because of the limit of the size of the array:
# define MAX 10
typedef struct _stats_t
{
int hours[24]; int numPostsInHour;
int days[7]; int numPostsInDay;
int weeks[20]; int numPostsInWeek;
int totNumLinesInPosts;
int numPostsAnalyzed;
} stats_t[MAX];
So how would I create this array? Also, would a pointer to this array would look something this?
stats_t stats[];
stats_t *statsPtr = &stats[0];
This is how it is usually done:
size_t n = <number of elements needed>
stats_t *ptr = malloc (n * sizeof (stats_t));
Then, to fill it in,
for (size_t j = 0; j < n; ++j)
{
ptr [j] .hours = whatever
ptr [j] .days = whatever
...
}
The second option of a pointer is good.
If you want to allocate things dynamically, then try:
stats_t* theStatsPointer = (stats_t*) malloc( MAX * sizeof(stats_t) );
as Roland suggests.
Just don't forget to
free(theStatsPointer);
when you're done.
Use malloc():
http://en.wikipedia.org/wiki/Malloc
malloc is your friend here.
stats_t stats[] = (stats_t*)malloc(N * sizeof(stats_t));
stats sort of is a pointer to the array. Or you can use stats[3] syntax as if it were declared explicitly as an array.
Based on your replies to other answers it looks like you need a dynamic data structure like a linked list. Take a look at the queue(3) set of facilities.
Related
I want to use a struct to contain some data and passing them between different functions in my program,this struct has to contain a dynamic 2D array (i need a matrix) the dimensions change depending on program arguments.
So this is my struct :
struct mystruct {
int **my2darray;
}
I have a function that read numbers from a file and has to assign each of them to a cell of the struct array.
I tried doing this :
FILE *fp = fopen(filename, "r");
int rows;
int columns;
struct mystruct *result = malloc(sizeof(struct mystruct));
result->my2darray = malloc(sizeof(int)*rows);
int tmp[rows][columns];
for(int i = 0;i<rows;i++) {
for(int j = 0;j<columns;j++) {
fscanf(fp, "%d", &tmp[i][j]);
}
result->my2darray[i]=malloc(sizeof(int)*columns);
memcpy(result->my2darray[i],tmp[i],sizeof(tmp[i]));
}
But this is giving me a strange result : all the rows are correctly stored except for the first.
(I'm sure that the problem is not in the scanning of file).
While if i change the fourth line of code in this :
result->my2darray = malloc(sizeof(int)*(rows+1));
it works fine.
Now my question is why this happens?
Here's an answer using some "new" features of the language: flexible array members and pointers to VLA.
First of all, please check Correctly allocating multi-dimensional arrays. You'll want a 2D array, not some look-up table.
To allocate such a true 2D array, you can utilize flexible array members:
typedef struct
{
size_t x;
size_t y;
int flex[];
} array2d_t;
It will be allocated as a true array, although "mangled" into a single dimension:
size_t x = 2;
size_t y = 3;
array2d_t* arr2d = malloc( sizeof *arr2d + sizeof(int[x][y]) );
Because the problem with flexible array members is that they can neither be VLA nor 2-dimensional. And although casting it to another integer array type is safe (in regards of aliasing and alignment), the syntax is quite evil:
int(*ptr)[y] = (int(*)[y]) arr2d->flex; // bleh!
It would be possible hide all this evil syntax behind a macro:
#define get_array(arr2d) \
_Generic( (arr2d), \
array2d_t*: (int(*)[(arr2d)->y])(arr2d)->flex )
Read as: if arr2d is a of type array2d_t* then access that pointer to get the flex member, then cast it to an array pointer of appropriate type.
Full example:
#include <stdlib.h>
#include <stdio.h>
typedef struct
{
size_t x;
size_t y;
int flex[];
} array2d_t;
#define get_array(arr2d) \
_Generic( (arr2d), \
array2d_t*: (int(*)[(arr2d)->y])(arr2d)->flex )
int main (void)
{
size_t x = 2;
size_t y = 3;
array2d_t* arr = malloc( sizeof *arr + sizeof(int[x][y]) );
arr->x = x;
arr->y = y;
for(size_t i=0; i<arr->x; i++)
{
for(size_t j=0; j<arr->y; j++)
{
get_array(arr)[i][j] = i+j;
printf("%d ", get_array(arr)[i][j]);
}
printf("\n");
}
free(arr);
return 0;
}
Advantages over pointer-to-pointer:
An actual 2D array that can be allocated/freed with a single function call, and can be passed to functions like memcpy.
For example if you have two array2d_t* pointing at allocated memory, you can copy all the contents with a single memcpy call, without needing to access individual members.
No extra clutter in the struct, just the array.
No cache misses upon array access due to the memory being segmented all over the heap.
The code above never sets rows and columns, so the code has undefined behavior from reading those values.
Assuming you set those values properly, this isn't allocating the proper amount of memory:
result->my2darray = malloc(sizeof(int)*rows);
You're actually allocating space for an array of int instead of an array of int *. If the latter is larger (and it most likely is) then you haven't allocated enough space for the array and you again invoke undefined behavior by writing past the end of allocated memory.
You can allocate the proper amount of space like this:
result->my2darray = malloc(sizeof(int *)*rows);
Or even better, as this doesn't depend on the actual type:
result->my2darray = malloc(sizeof(*result->my2darray)*rows);
Also, there's no need to create a temporary array to read values into. Just read them directly into my2darray:
for(int i = 0;i<rows;i++) {
result->my2darray[i]=malloc(sizeof(int)*columns);
for(int j = 0;j<columns;j++) {
fscanf(fp, "%d", &result->my2darray[i][j]);
}
}
In your provided code example, the variables rows and columns have not been initialized before use, so they can contain anything, but are likely to be equal to 0. Either way, as written, the results will always be unpredictable.
When a 2D array is needed in C, it is useful to encapsulate the memory allocation, and freeing of memory into functions to simplify the task, and improve readability. For example, in your code the following line will create an array of 5 pointers, each pointing to 20 int storage locations: (creating 100 index addressable int locations.)
int main(void)
{
struct mystruct result = {0};
result.my2darray = Create2D(5, 20);
if(result.my2darray)
{
// use result.my2darray
result.my2darray[0][3] = 20;// for simple example, but more likely in a read loop
// then free result.my2darray
free2D(result.my2darray, 5);
}
return 0;
}
Using the following two functions:
int ** Create2D(int c, int r)
{
int **arr;
int y;
arr = calloc(c, sizeof(int *)); //create c pointers (columns)
for(y=0;y<c;y++)
{
arr[y] = calloc(r, sizeof(int)); //create r int locations for each pointer (rows)
}
return arr;
}
void free2D(int **arr, int c)
{
int i;
if(!arr) return;
for(i=0;i<c;i++)
{
if(arr[i])
{
free(arr[i]);
arr[i] = NULL;
}
}
free(arr);
arr = NULL;
}
Keep in mind that what you have created using this technique is actually 5 different pointer locations each pointing to a set of 20 int locations. This is what facilitates the use of array like indexing, i.e. we can say result.my2darray[1][3] represents the second column, forth row element of a 5X20 array, when it is not really an array at all.
int some_array[5][20] = {0};//init all elements to zero
Is what is commonly referred to in C an int array, also allowing access to each element via indexing. In actuality (Even though commonly referred to as an array.) it is not an array. The location of elements in this variable are stored in one contiguous location in memory.
|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0... (~ 82 more)
But C maintains the locations such that they are all indexable as an 2D array.
Hello everyone I got the following struct
struct Test
{
unsigned char* c_string;
unsigned int value;
};
I created a function that creates a new instance of this struct and initialize the attributes with random values like this
struct Test* createNewTest(){
struct Test *NewInstance = (Test * )malloc( sizeof(Test) );
NewInstance->value = rand();
Now I have to create a function that creates n completely initialized instances of my struct.
struct Test** createNewArray(unsigned int n){
};
Can anyone help me to do this? I dont really now how to start here
It's fairly straightforward. First, you'll need to allocate enough storage for n pointers to struct Test:
struct Test **array = malloc(n * sizeof *array);
if (!array) return array;
And then assign each pointer, using the function you have:
for (size_t i = 0; i < n; ++i)
array[i] = createNewTest();
Wrapping it all up, you get
struct Test **createNewArray(size_t n)
{
struct Test **array = malloc(n * sizeof *array);
if (!array) return array;
for (size_t i = 0; i < n; ++i)
array[i] = createNewTest();
return array;
}
Don't forget to write a matching free() function!
Also, consider whether you really want an array of pointers to Test - you may be better off with an array of Test objects instead.
I'm going to try to explain this without giving you the answer to your assignment verbatim ...
First, you need to allocate n times as much memory. malloc can allocate any amount, so you just need to calculate the number.
Second, you need to do the same initialization, but for each entry in the array. You'll need a loop for that.
Hint: in C, a pointer to a single instance and a pointer to a dynamic array are indistinguishable. ptr->value is the same as ptr[0].value, and ptr[1].value is the next element in the array.
I have a function which creates an array, of say, size 5.
Is it possible for the function to accept a pointer (or maybe it needs a pointer to a pointer?) and then point said pointer at an array, so that when the callee then looks at the pointer, it can see all values of the array.
Something along the lines of this (except this will not work):
#define LENGTH 5
void assignArray(int *pointer)
{
int arr[LENGTH] = {0,1,2,3,4};
// Point the pointer at the array, without manually copying each element
pointer = arr;
}
void main()
{
int *pointer;
pointer = malloc(sizeof(int) * LENGTH);
assignArray(pointer);
int i;
for (i = 0 ; i < LENGTH ; i++) printf("%d\n", pointer[i]);
}
C assign array without element by element copy
In C, arrays (compile-time allocated) cannot be assigned. You need to copy the elements from one array to another.
To avoid element-by-element copy, you can copy the whole array all at a time using library function.
I'm not very sure what you want to ask here, but it seems, you need to do memcpy() to achieve your goal.
If you have a secondary array arr to copy from, you can write
memcpy( pointer, arr, ( (sizeof arr[0]) * LENGTH ));
The code to do what you are describing might look like:
#define LENGTH 5
void assignArray(int **pp)
{
static int arr[LENGTH] = {0,1,2,3,4};
// Point the pointer at the array, without manually copying each element
*pp = arr;
}
int main()
{
int *pointer;
assignArray(&pointer);
for (int i = 0 ; i < LENGTH ; i++)
printf("%d\n", pointer[i]);
}
Note that one does not simply point *pp at a non-static local variable arr. That is because int arr[] = .... would go out of scope when assignArray returns.
If you want each call to assignArray to "return" a different array then of course you will have to allocate space and use memcpy each time you want to make a copy of the original array.
int arr[LENGTH] = {0,1,2,3,4}; will be stack allocated, so attempting to return the pointer to any of its elements will give you undefined behaviour as the whole thing will be out of scope when the function returns.
If you want to change what a pointer is pointing to then use 2 levels of indirection ** (i.e. pass a pointer to a pointer). You'll need to allocate the array arr on the heap using malloc or something similar.
As you are trying to do it, it is not possible due to the fact that your local arr is saved to the stack and is cleaned up after the function assignArry finished. As already mentioned you need to memcpy.
This answer will have two parts:
As mentioned in other answers, this is now how you're supposed to do it. A common construct in similar code is:
void assignArray(int *dest, size_t size)
{
int i;
// initialize with some data
for (i=0; i<size; i++)
dest[i] = i;
}
This way you're not wasting space and time with an intermediate buffer.
Second part of this answer is about wrapping arrays in a struct. It's a silly trick, that in a way achieves exactly what you asked, and also something that you probably don't want because of extra data copying.
Example code:
#include <stdio.h>
#include <stdlib.h>
#define LENGTH 5
struct foo { int arr[LENGTH]; };
struct foo assignArray()
{
struct foo bar = { .arr = {0,1,2,3,4} };
/* return the array wrapper in struct on stack */
return bar;
}
int main()
{
struct foo *pointer;
pointer = malloc(sizeof(*pointer));
*pointer = assignArray(); /* this will copy the data, not adjust pointer location */
int i;
for (i = 0 ; i < LENGTH ; i++) printf("%d\n", pointer->arr[i]);
return 0;
}
Is it possible to create an array of arrays in c
Thank you.
It's the same as for example in PHP:
int arrayInArray[10][50];
You read data out of it with:
printf("%d", arrayInArray[3][37]);
I bet you mean Multi Dimensional Array instead of "array of arrays".
Some links for this topic:
http://www.dfstermole.net/OAC/harray2.html
http://webster.cs.ucr.edu/AoA/Windows/HTML/Arraysa2.html
For using an array of arrays with all the power of C you should have some knowledge of dynamic memory handling in c, with the functions malloc, realloc, and free, and some knowledge about pointers. For this example that you ask a possible solution would be this:
#include <stdio.h>
void main(int argc, char* argv[]){
int** myArray; /* This would be a double pointer, because you want a two dimension array.*/
int firstDimension = 10;
int secondDimension = 20;
int i;
myArray = (int**)malloc(firstDimension*sizeof(int*)); This way you initialize the first dimension of the array.
for(i = 0; i < firstDimension; i++){
myArray[i] = (int*)malloc(secondDimension*sizeof(int));
}
/*Once you have the array initialized, you can access in the way myArray[i][j];*/
/*For releasing resources */
for(i = 0; i < firstDimension; i++){
free(myArray[i]);
}
free(myArray);
}
This is the dynamic way, the one that is teached on CS courses.
If you need an array of arrays then you should use structs.
typedef ArrayStruct* ArrayStructPtr;
struct ArrayStruct
{
void* array;//Node array
ArrayStructPtr arrays;//Pointer to sub arrays
};
int main()
{
ArrayStruct* a;//Declare Some Arrays
a=(ArrayStruct*)malloc(sizeof(ArrayStruct)*N);
for(int i=0;i<N;i++)
{
a[i].array=(void*)malloc(sizeof(int)*N);//Malloc the actual array
a[i].arrays=NULL;//Malloc subarrays if needed
}
//add subarray on array 0
ArrayStruck * temp=(ArrayStruct*)malloc(sizeof(ArrayStruct));
temp->array=(void*)malloc(sizeof(char)*MAXNAME*N);
temp->arrays=NULL;
a[0]=arrays=temp;
return 0;
}
What you need is a List Of arrays Where each node of the struct can hold an array and a pointer to another node.
The array type is void* to support int,float,char*.
So each array can have as many subarrays as you want.You can create 3 dimension Arrays if you want!
I made a structure like so:
struct ponto {
int x;
int y;
int z;
};
1) Can I initialize the int's with a default value? int var = value; doesn't seem to work, compiler says "syntax error before '=' token" or something of sorts.
2) I need to work with several of these like in a array of structures, but I only know how many I need after the application starts up, after reading a file. How can I malloc this?
Thanks in advance
EDIT: So many answers, I'm grateful. Sadly I can only mark one
a) You can initalise with
struct pronto p = {1,2,3};
In recent compilers (not sure how portable this is, think it's C99?)
b) You can allocate an array with malloc:
struct pronto *array = malloc(sizeof(struct pronto) * NUMBER);
To initialize your structure members to 0, do:
struct ponto foo = { 0 };
To malloc() an array of the right size, do:
struct ponto *arr = (struct ponto *) malloc(COUNT * sizeof(struct ponto));
Don't forget to free() the array when you're done with it.
struct ponto* create_and_init_ponto(int n)
{
struct ponto* array;
int i;
array = (struct ponto*)malloc( n * sizeof(struct ponto) );
for ( i = 0; i < n; ++i )
{
array[ i ].x = 0;
array[ i ].y = 0;
array[ i ].z = 0;
}
return array;
}
You'e made a structure definition, now you have to create a variable of that structure before you can set the fields:
struct ponto xyz;
xyz.x = 7;
To allocate enough space:
int need_to_have = 24;
struct ponto *pontos = malloc (need_to_have * sizeof(struct ponto));
You cannot have "default" values for structure members. Space is not allocated for a structure definition. You're just creating a new type (like the inbuilt int). When you actually define a variable of type ponto, space will be allocated to it.
You can make an educated guess about how many you will need, allocate space for that many (using malloc) and go ahead. If you find that you're reaching the limit, you can use the realloc function to resize your array.
1) You cannot give a specific structure default values for its elements at the language level, because all variables in C are uninitialized unless you explicitly initialize them (or make them static/external in which case they're zero-initialized). If you design your structs such that all-zeros is a good set of initial values, though, you can always initialize like this:
struct foo myfoo = {0};
The {0} serves as a universal zero-initializer which works for any type.
If you need different defaults, the best way is to use a macro and document that code using your structure must use the macro:
#define FOO_INITIALIZER { 1, 2, 3 }
struct foo myfoo = FOO_INITIALIZER;
2) If you know before you start using any of the struct how many you will need, simply malloc them all once you know the number:
if (count > SIZE_MAX / sizeof *bar) abort();
struct foo *bar = malloc(count * sizeof *bar);
Note the proper idiom for calling malloc and avoiding overflow vulnerabilities.
If you don't know the number you'll need until you start working with them, start out by allocating a decent number, and if you run out, increase the number by a fixed multiple, for example doubling the size is common and easy. You'll want to check for overflows here. Then use realloc.
Question #1: If you need to initialize int with a value:
struct ponto p1;
p1.x = p1.y = p1.z = 3; // initializing with three
Alternatively, if you want to initialize all values to 0, you can use memset like this:
memset(&p1, 0, sizeof(struct ponto));
Question #2: To use malloc:
struct ponto *ps;
ps = (struct ponto *)malloc(N*sizeof(struct ponto));
// where N is your element count.
This will allocate memory to store N elements of type struct ponto. After that, you can initialize its values with:
int initvalue = 3; // assuming you want to initialize points with value 3
for (i=0; i<N; i++) {
ps[i].x = ps[i].y = ps[i].z = initvalue;
}