I'm in a class right now that works with C and one of my assignments requires that I work with a struct that my professor wrote for us. It's actually two structs, with one struct basically containing an array of the first struct.
Here's what they look like:
typedef struct cityStruct
{
unsigned int zip;
char *town
} city;
typedef struct zipTownsStruct
{
int *towns;
city **zips;
city *cities
} zipTowns;
And here's my function for allocating memory for the zipTowns structure:
void getArrs(zipTowns *arrs, int size)
{
arrs->towns = malloc(sizeof(int) * size);
arrs->zips = malloc(sizeof(city **) * size);
arrs->cities = malloc(sizeof(city *) * size);
}
From what I understand, what I'm doing here is allocating space in memory for a certain number of ints, city pointers, and city structures, based on the size variable. I understand that this is basically what an array is.
I'm having trouble with understanding how I can access these arrays and manipulate items in it. Writing this gives me an error:
strcpy(arrs.cities[0]->town, "testTown\0");
You can see what I'm trying to do here. I want to access each "City" in the zipTowns struct by index and insert a value.
How can I access the items in these dynamically allocated array of structures?
Think of x->y as (*x).y.
arrs is not a structure, it's a pointer to a structure, and cities is not a pointer to a pointer to a structure, it's just a pointer to a structure.
Use arrs->cities[0].town instead of arrs.cities[0]->town.
However, you're still not allocating enough room for these structures. This should make it clearer what you're doing with the allocations, and should also give you enough room for your data:
arrs->towns = malloc(sizeof(*arrs->towns) * size);
arrs->zips = malloc(sizeof(*arrs->zips) * size);
arrs->cities = malloc(sizeof(*arrs->cities) * size);
With the second and third, you were only allocating enough room for a pointer to be stored instead of the actual data type.
With this approach, you will be able to access from arrs->cities[0] to arrs->cities[9] and you also will be able to access the members of each city by doing arrs->cities[<number>].<member>.
You also do not need to intentionally null-terminate your strings. This is already done for you. Therefore, you can replace "testTown\0" with "testTown".
Related
I have looked around but have been unable to find a solution to what must be a well asked question.
Here is the code I have:
#include <stdlib.h>
struct my_struct {
int n;
char s[]
};
int main()
{
struct my_struct ms;
ms.s = malloc(sizeof(char*)*50);
}
and here is the error gcc gives me:
error: invalid use of flexible array member
I can get it to compile if i declare the declaration of s inside the struct to be
char* s
and this is probably a superior implementation (pointer arithmetic is faster than arrays, yes?)
but I thought in c a declaration of
char s[]
is the same as
char* s
The way you have it written now , used to be called the "struct hack", until C99 blessed it as a "flexible array member". The reason you're getting an error (probably anyway) is that it needs to be followed by a semicolon:
#include <stdlib.h>
struct my_struct {
int n;
char s[];
};
When you allocate space for this, you want to allocate the size of the struct plus the amount of space you want for the array:
struct my_struct *s = malloc(sizeof(struct my_struct) + 50);
In this case, the flexible array member is an array of char, and sizeof(char)==1, so you don't need to multiply by its size, but just like any other malloc you'd need to if it was an array of some other type:
struct dyn_array {
int size;
int data[];
};
struct dyn_array* my_array = malloc(sizeof(struct dyn_array) + 100 * sizeof(int));
Edit: This gives a different result from changing the member to a pointer. In that case, you (normally) need two separate allocations, one for the struct itself, and one for the "extra" data to be pointed to by the pointer. Using a flexible array member you can allocate all the data in a single block.
You need to decide what it is you are trying to do first.
If you want to have a struct with a pointer to an [independent] array inside, you have to declare it as
struct my_struct {
int n;
char *s;
};
In this case you can create the actual struct object in any way you please (like an automatic variable, for example)
struct my_struct ms;
and then allocate the memory for the array independently
ms.s = malloc(50 * sizeof *ms.s);
In fact, there's no general need to allocate the array memory dynamically
struct my_struct ms;
char s[50];
ms.s = s;
It all depends on what kind of lifetime you need from these objects. If your struct is automatic, then in most cases the array would also be automatic. If the struct object owns the array memory, there's simply no point in doing otherwise. If the struct itself is dynamic, then the array should also normally be dynamic.
Note that in this case you have two independent memory blocks: the struct and the array.
A completely different approach would be to use the "struct hack" idiom. In this case the array becomes an integral part of the struct. Both reside in a single block of memory. In C99 the struct would be declared as
struct my_struct {
int n;
char s[];
};
and to create an object you'd have to allocate the whole thing dynamically
struct my_struct *ms = malloc(sizeof *ms + 50 * sizeof *ms->s);
The size of memory block in this case is calculated to accommodate the struct members and the trailing array of run-time size.
Note that in this case you have no option to create such struct objects as static or automatic objects. Structs with flexible array members at the end can only be allocated dynamically in C.
Your assumption about pointer aritmetics being faster then arrays is absolutely incorrect. Arrays work through pointer arithmetics by definition, so they are basically the same. Moreover, a genuine array (not decayed to a pointer) is generally a bit faster than a pointer object. Pointer value has to be read from memory, while the array's location in memory is "known" (or "calculated") from the array object itself.
The use of an array of unspecified size is only allowed at the end of a structure, and only works in some compilers. It is a non-standard compiler extension. (Although I think I remember C++0x will be allowing this.)
The array will not be a separate allocation for from the structure though. So you need to allocate all of my_struct, not just the array part.
What I do is simply give the array a small but non-zero size. Usually 4 for character arrays and 2 for wchar_t arrays to preserve 32 bit alignment.
Then you can take the declared size of the array into account, when you do the allocating. I often don't on the theory that the slop is smaller than the granularity that the heap manager works in in any case.
Also, I think you should not be using sizeof(char*) in your allocation.
This is what I would do.
struct my_struct {
int nAllocated;
char s[4]; // waste 32 bits to guarantee alignment and room for a null-terminator
};
int main()
{
struct my_struct * pms;
int cb = sizeof(*pms) + sizeof(pms->s[0])*50;
pms = (struct my_struct*) malloc(cb);
pms->nAllocated = (cb - sizoef(*pms) + sizeof(pms->s)) / sizeof(pms->s[0]);
}
I suspect the compiler doesn't know how much space it will need to allocate for s[], should you choose to declare an automatic variable with it.
I concur with what Ben said, declare your struct
struct my_struct {
int n;
char s[1];
};
Also, to clarify his comment about storage, declaring char *s won't put the struct on the stack (since it is dynamically allocated) and allocate s in the heap, what it will do is interpret the first sizeof(char *) bytes of your array as a pointer, so you won't be operating on the data you think you are, and probably will be fatal.
It is vital to remember that although the operations on pointers and arrays may be implemented the same way, they are not the same thing.
Arrays will resolve to pointers, and here you must define s as char *s. The struct basically is a container, and must (IIRC) be fixed size, so having a dynamically sized array inside of it simply isn't possible. Since you're mallocing the memory anyway, this shouldn't make any difference in what you're after.
Basically you're saying, s will indicate a memory location. Note that you can still access this later using notation like s[0].
pointer arithmetic is faster than arrays, yes?
Not at all - they're actually the same. arrays translate to pointer arithmetics at compile-time.
char test[100];
test[40] = 12;
// translates to: (test now indicates the starting address of the array)
*(test+40) = 12;
Working code of storing array inside a structure in a c, and how to store value in the array elements Please leave comment if you have any doubts, i will clarify at my best
Structure Define:
struct process{
int process_id;
int tau;
double alpha;
int* process_time;
};
Memory Allocation for process structure:
struct process* process_mem_aloc = (struct process*) malloc(temp_number_of_process * sizeof(struct process));
Looping through multiple process and for each process updating process_time dyanamic array
int process_count = 0;
int tick_count = 0;
while(process_count < number_of_process){
//Memory allocation for each array of the process, will be containting size equal to number_of_ticks: can hold any value
(process_mem_aloc + process_count)->process_time = (int*) malloc(number_of_ticks* sizeof(int));
reading data from line by line from a file, storing into process_time array and then printing it from the stored value, next while loop is inside the process while loop
while(tick_count < number_of_ticks){
fgets(line, LINE_LENGTH, file);
*((process_mem_aloc + process_count)->process_time + tick_count) = convertToInteger(line);;
printf("tick_count : %d , number_of_ticks %d\n",tick_count,*((process_mem_aloc + process_count)->process_time + tick_count));
tick_count++;
}
tick_count = 0;
the code generated will be identical (array and ptr). Apart from the fact that the array one wont compile that is
and BTW - do it c++ and use vector
I have been trying to figure out how to assign an address to an array of structures in C I have managed to have a limited amount of success. Here is my struct:
struct pdirectory {
uint16_t one;
uint8_t two;
uint8_t three;
uint16_t four;
};
Here is my first attempt at declaring my struct variable:
struct pdirectory *test = (struct pdirectory *)0x00001000;
Which worked fine and placed my struct at the memory location expected.
However I can't successfully assign an address to an array of structures.
I have tired the following code with limited success:
struct pdirectory (*test)[1] = (struct pdirectory *)0x00001000;
Which places my struct at the intended address and I can even access different arrays with the following test code:
void initialise_virtual_manager(){
test[0]->one = 0x3333;
test[0]->four = 0x3333;
test[1]->one = 0x3434;
test[1]->four = 0x3434;
test[2]->one = 0x3535;
test[2]->four = 0x3535;
}
I know my code above is not right. I tired to change the array number in the struct deceleration which offsets the array entries in memory by the struct size (which is not what I am intending to do) I want to declare an array of 1024 strut's . Also with the above code I can access an unlimited amount of array's (which is not what I want, I want the array number to limit the amount of array's I can access). If someone can see what I am doing wrong or has any other suggestion to solve this problem I am having, that would be a big help.
For one, use the proper cast, since you seem to know how to point to an array:
struct pdirectory (*test)[N] = (struct pdirectory (*)[N])0x00001000;
Secondly, test[i] will performs pointer arithmetic in units of the array size. If the array is size N, then the array access will be equivalent to:
*(struct pdirectory(*)[N])((unsigned char*)test + i * sizeof(struct pdirectory) * N)
As you can see the above doesn't index the array that test points to, but treats test as an array.
Your array access should therefore be as follows:
(*test)[0].one = 0x3333;
(*test)[0].four = 0x3333;
First test is dereferenced, and only then the array is indexed.
I have the following struct:
typedef struct dish_t {
const char *name;
const char *cook;
int maxIngredients;
bool tasted;
Ingredient *ingredients[];
}* Dish;
And the following initialization function:
Dish dishCreate(const char* name, const char* cook, int maxIngredients) {
if (name == NULL || cook == NULL || maxIngredients <= 0 )
{
return NULL;
}
Dish newDish=malloc(sizeof(*newDish));
newDish->name=malloc(strlen(name)+1);
newDish->name=name;
newDish->cook=malloc(strlen(cook)+1);
newDish->cook=cook;
newDish->maxIngredients=maxIngredients;
newDish->ingredients=malloc(maxIngredients*sizeof(Ingredient*));
newDish->tasted=false;
return newDish;
}
Ingredient is also a struct.
and on this line
newDish->ingredients=malloc(maxIngredients*sizeof(Ingredient*));
I get the error.
I'm trying to initialize an array of pointers to a struct...what am I doing wrong?
Thanks.
The struct member ingredients is an array of pointers, not a pointer. You cannot assign to an array in C; not to change its contents, and certainly not to change its address. The pattern <array> = malloc(...) makes no sense in C.
Just change the declaration to
Ingredient *ingredients;
Forget about flexible member; that's a memory allocation strategy that won't add to the semantics or quality of your program.
In some kinds of programs, flexible arrays at the end of a structure can be a useful
optimization, since they reduce the number of calls to the allocator: a header structure plus some variable data can be allocated in a single malloc call and released in a single free call.
This doesn't achieve anything semantically interesting in the program, and naive use of this approach can make programs slower and waste memory, since two or more objects with flexible data at the end have to have their own copies of the data in situations where a program that uses a pointer could efficiently share that data.
The problem is that you try to make a flexible array of pointers, and then you try to create this using malloc, which is not what you're supposed to do.
With flexible arrays in a structure, you're supposed to make the initial allocation of the structure longer, to fit the number of elements in the array.
If you're going to use malloc for the array, you might as well use a normal pointer.
To explain what I mean:
typedef struct dish_t {
...
Ingredient ingredients[];
}* Dish;
...
Dish myDish = malloc(sizeof(*myDish) + sizeof(Ingredient) * 10);
The above malloc allocates the structure and enough space for ten entries in the array.
I currently have no code, because I don't know how to do this at all. Could I just, by myself, calculate how many bytes is needed for each lower-level struct and malloc it to it? That's really terrible coding, isn't it. Here's the two structs I'm trying to mash together:
struct property {
int d;
char name [111]; // I just malloc this like I would a normal array, right?
char descr [1025]; // Ditto.
}
struct category {
int d [413]; // The d's of all the child structs sorted highest to lowest.
char name [111];
struct property property [413]; // This. How do I allocate this?
}</code>
Do I have to do struct property* property = (struct property*) malloc(sizeof(struct property) * 413);? Will the malloc of the array within remain intact? How do mallocs in structs behave in general?
You don't have a pointer member inside your structure property so you don't need to malloc any of your structure members.
When you malloc for the structure it will give you enough memory to hold all the structure members including arrays, exceptions are pointer structure members(You don't have any).
Your malloc without the cast would do fine. It allocates contiguous memory for the whole array. The arrays inside the struct's are all allocated along with it, they are proper arrays and not pointers.
Sizeof will give you the size of your entire structure. It properly accounts for the size of arrays and structures.
However, 413 items seems arbitrary. Would a variable sized structure work better for you?
In that case, calculating the size ahead of time to avoid mallocs is a good performance idea. Malloc can be slow, it can require locks, and the heap can fragment over time. This example shows you how to make a "variable length" structure with a pointer instead of an array or a variable length array at the end of your structure:
struct category
{
int cItems; // need this if handling variable # of items now.
int *d; // ptr instead of array
char *name; // ptr again
struct property property[0]; // var length array
}
int cItems = 413; // or whatever
// this is a nifty trick to get the size of a variable length struct:
int cbCategory = (size_t)(&((struct category*)0)->property[cItems]);
int cbD = sizeof(int)*cItems;
int cbName = sizeof(char)*cItems;
struct category *pCategory = (struct category*)malloc(cbCategory + cbD + cbName);
// wire up d:
pCategory->d = (int*)((char*)pCategory + cbCategory);
// or wire d up this way:
pCategory->d = (int*)&pCategory->property[cItems];
// wire up name
pCategory->name = (char*)pCategory->d + cbD;
// or wire up name this way
pCategory->name = (char*)&pCategory->d[cItems];
// set items
pCategory->cItems = cItems;
Note, I assumed that d had 413 elements to it. I could have just as easily left it a an array.
If I create a struct in one class as so
typedef struct
{
int numberOfTiles;
// an array of ints here
int *tileArray;
} CollisionLayer;
is it possible to create an array of ints with an empty [] and set the size on creation? Or how would this array be created? dynamically with a pointer? I will know the size of the array when creating one of these struct "objects", if it is possible to fill in the size of the array on creation, how is the array declared in the struct above?
You will need to initialize the array yourself:
CollisionLayer layer;
layer.numberOfTiles = numberOfTiles;
layer.tileArray = (int*)malloc(sizeof(int) * numberOfTiles);
Or, if you want to create the struct on the heap:
CollisionLayer* pl = (CollisionLayer*)malloc(sizeof(CollisionLayer));
pl->numberOfTiles = numberOfTiles;
pl->tileArray = (int*)malloc(sizeof(int) * numberOfTiles);
// When you are done:
free(pl->tileArray);
free(pl);
The other option would be to hardcode a fixed size limit into CollisionLayer, e.g.:
typedef struct
{
int numberOfTiles;
// an array of ints here
int tileArray[100];
} CollisionLayer;
Of course this will be less desirable in all respects, but it's your only option if you don't want to manage memory manually.
If you don't know the size at compile time, then you must allocate the memory using malloc() at runtime. To use an actual array in C, you must know the size when the code is compiled.
tileArray is a pointer to int. malloc/calloc should be used to create the object to which it will point to. This should happen when creating an object of CollisionLayer.
Defining a struct in which the array [] is empty is not a good idea, refer this. It speaks of C++, bit it should apply for C as well.
VLAs cannot be members of structs so you will need to allocate the memory with malloc when you create the struct object.