I am trying to understand an assignment I have before I have to take a final. I am trying to understand what exactly I am declaring.
So in a given file the typedef struct's are declared as so:
(Struct Declaration)
/** The following two structs must be defined in your <gamename>.c file **/
typedef struct game_position_t *game_position;
/* move struct must code enough information to reverse the move, given the resulting position */
typedef struct move_t *move;
I have then built the structs out as so (yes this has to be separated just because it is interfaced programming):
(Struct Definition)
/** The following two structs must be defined in your <gamename>.c file **/
struct game_position_t {
int mathy;
int numrows;
int *sizes;
};
/* move struct must code enough information to reverse the move, given the resulting position */
struct move_t {
int rownum;
int move_size;
};
Then an example of a functions and declaration of game_position for example is:
(Example Function)
/* return the starting position, NULL if error */
game_position starting_position(int me_first, int argc, char **argv) {
if (argc < 3) {
printf("\n\nToo few arguments, see help below\n\n");
game_help(argv[0]);
return NULL;
}
int mathy;
if (strcmp(argv[2],"search")==0)
mathy = 0;
else if (strcmp(argv[2],"mathy")==0)
mathy = 1;
else {
printf("\n\nSecond argument must be \"search\" or \"mathy\", see help below\n\n");
game_help(argv[0]);
return NULL;
}
int play_default = (argc==3);
if (play_default) printf("\n\nOK, we will play the default game of 7 5 3 1\n\n");
int defaultgame[4] = {7,5,3,1};
game_position result = malloc(sizeof(struct game_position_t)*1);
result->mathy = mathy;
if (result) {
result->numrows = (play_default ? 4 : argc-3);
result->sizes = malloc(sizeof(int)*(result->numrows));
int row;
for (row=0; row<(result->numrows); row++)
(result->sizes)[row] = (play_default ? defaultgame[row] : strlen(argv[row+2]));
}
return result;
}
So my main misunderstanding is when using a struct declaration in this manner, specifically putting the * before the name like this, typedef struct move_t *move;. Is that previous line saying move it a struct pointer or dereferencing move? Continuing from that. When defining them I just use the struct name such as struct move_t. I don't fully understand how they are linking together and in what matter. Then inside the function I just declare game_position, but still need to use a derefencer, 'p->`, to access it fields. So if someone could explain to me when these struct variables are points to structs and when they are the actual struct.
An example of my misunderstanding is that in the Example Function after result was declared. I first thought to use the . operator to access and set it's fields. I then changed it due to compiler errors, but now I want to understand my misunderstanding. And why did I I have to malloc game_position_t and not game_position?
typedef defines a type, so typedef struct move_t *move defines a new type named move, which is a pointer type, pointing to struct move_t. So after this if you define a variable with move ptr, ptr will have a pointer type so that you should use the syntax of accessing members through a pointer. When allocating memory for it, of course you have to specify the exact size of the structure other than the size of a pointer, that's sizeof(struct move_t)
Related
I have a struct initialized on a stack, and i want to write data in memory right after the struct and make a pointer inside a struct point to that data.
I know it is achievable on the stack/heap with uninitialized structure using malloc(sizeof(struct) + additional size) or alloca(). but can i perform initialization of a data after the struct is already initialized on the stack? and can i perform this initialization inside a initializator function?
Simple example:
struct TEST {
wchar_t* itest;
};
void init_struct(struct TEST* test) {
// point to the end of the struct
char* walk_ptr = (char*)test + sizeof(test);
test->itest = (wchar_t*)walk_ptr;
// initialize data after the struct
...
}
int main(void) {
struct TEST test;
init_struct(&test);
return 0;
}
You could do this by embedding the structure inside another structure to reserve memory for the extra data:
int main(void)
{
struct { struct TEST test; wchar_t data[NumberOfElements]; } S;
init_struct(&S.test);
…
}
However, the code in init_struct adds an incorrect size, sizeof(test), to the pointer. You likely wanted to add sizeof (struct Test), or, equivalently, sizeof *test, since you want to get past the struct TEST that test points to, not past a struct TEST * that test is.
However, even adding the correct size of the structure would not guarantee strictly conforming C code, since C implementations may insert padding between structure members. Properly we would add the offset of the data member. To do that, we nwould eed to give the structure a tag and then either make the structure definition visible to init_struct or pass the offset to init_struct. However, it is easier just to pass the address of the extra data:
void init_struct(struct TEST *test, wchar_t *data)
{
test->itest = data;
}
int main(void)
{
struct { struct TEST test; wchar_t data[NumberOfElements]; } S;
init_struct(&S.test, S.data);
…
}
Of course, a pointer can point anywhere, and there is no apparent reason the data should be immediate after the structure, so we can disconnect them:
int main(void)
{
struct TEST test;
wchar_t data[NumberOfElements];
init_struct(&test, data);
…
}
I tried to create a struct that would incude a dynamic array:
typedef struct
{
int idNode;
int* n; //pointer to the int nodes in the dynamically created array of nodes
n = calloc(MAX, sizeof(int)); //dynamic array to store the loser member of the pair
int counter = 0;
}
node;
But I get
error: type name requires a specifier or qualifier
You cannot assign a value to n in the struct declaration. You need to provide a function (something like a constructor ;-) that initializes the structure and assigns a value to its members, including n:
void init_node(node* n)
{
n->idNode = ...;
n->n = calloc(MAX, sizeof(int));
n->counter = 0;
}
Note: you still need to handle errors (e.g. calloc may fail) in the function and propagate errors to its caller.
You can't have statements or initialize an variable inside a structures.
For fix your problem
typedef struct Node {
int idNode;
int* n; //pointer to the int nodes in the dynamically created array of nodes
int counter;
} node;
int main(void)
{
node data = null;
node.n = calloc(sizeof(int), MAX);
node.idNode = 0;
node.counter = 0;
return (0)
}
Now you have initialize your struct
Starting point
I tried to create a struct that would incude a dynamic array
The problem I see in your initial snippet is you mix definition, declaration and use.
From https://www.geeksforgeeks.org/difference-between-definition-and-declaration:
Declaration of a variable is for informing to the compiler the following information: name of the variable, type of value it holds and the initial value if any it takes. i.e., declaration gives details about the properties of a variable. Whereas, Definition of a variable says where the variable gets stored.
Steps to get a basic knowledge of how to do it.
Firstly, you must know how to create a struct.
The next step is how to typedef it.
The next step is how dynamic arrays are declarated, defined, created, stored, modified or destroyed (the life cycle). Pay attention to erros may occur during the execution. The happy path of create things in C is not the only one, there are plenty of errors out there!
The next step is how to insert them into a typedef'd struct.
And the last step is use that typedef struct with a dynamic array inside it. Even you can create multiple dynamic arrays in the struct!
Note: Steps, 1, 2 and 4 may be ordered in other ways depend on the programmer
There is no shortcuts, no trial and error and, of course, you must create test programs to ensure the stuff you want and the stuff you program are the same thing.
n = calloc(MAX, sizeof(int));
int counter = 0;
You cannot use statements to execute inside of the declaration of a structure. You need to initialize n and counter inside of a function after an object of node has been defined.
E.g.:
typedef struct
{
int idNode;
int* n;
int counter;
}
node;
int main (void)
{
node a;
a.n = calloc(MAX, sizeof(int));
a.counter = 0;
}
I have this struct
struct FluxCapacitor{
unsigned char* c_string;
unsigned int value;
};
Now I need to create an instance of this struct. I googled this problem and found that I have to use something like this
typedef struct FluxCapacitor{
unsigned char* c_string
unsigned int value;
};
But I dont really understand the next step with malloc(). Can someone explain it to me?
You do not need malloc() to create an instance of a struct. And I would recommend that you avoid typedefing structures merely to reduce keystrokes. The extra keystrokes would only be saved in declarations and function prototypes (and maybe if you need to cast something), since you don't need the struct keyword elsewhere; the advantage is that when you see struct FluxCapacitor, you know exactly what it is. If you only see FluxCapacitor alone, you don't know if it is a typedef for a struct, or a union, or an integer type or what.
Note that the posted code was missing the semicolon at the end of the declaration. Also, it is unclear why you have unsigned char* c_string;. This may not allow assignment to a string literal. I have changed this in the code below. You can create a single struct like this:
struct FluxCapacitor
{
char *c_string;
unsigned int value;
};
...
struct FluxCapacitor fcap_1;
You can then assign values to the fields of fcap_1:
fcap_1.c_string = "McFly";
fcap_1.value = 42;
Note that you could also use designated initializers at the point of declaration:
struct FluxCapacitor fcap_2 = { .c_string = "Biff",
.value = 1985
};
If you need an array of FluxCapacitor structures, just declare one:
struct FluxCapacitor fcaps[2];
You can assign to the fields of each array member in a loop:
struct FluxCapacitor fcaps[2];
char *somestrings[] = { "McFly", "Biff" };
unsigned somevalues[] = { 42, 1985 };
for (size_t i = 0; i < 2; i++) {
fcaps[i].c_string = somestrings[i];
fcaps[i].value = somevalues[i];
}
Alternatively, you can use designated initializers here too:
struct FluxCapacitor fcaps[2] = { { .c_string = "McFly", .value = 42 },
{ .c_string = "Biff", .value = 1985}
};
Using malloc()
Since OP seems determined to use malloc(), it would be good to first recall that memory allocated with malloc() must later be deallocated with free(). Also note that malloc() can fail to allocate memory, returning a null pointer. Thus the result of a call to malloc() must be checked before attempting to dereference this pointer. The additional complexity should be avoided in favor of the above approaches unless OP has good reason to do manual allocation.
In the code below, the function create_flux_cap() takes a string and an unsigned int as arguments, and returns a pointer to a newly allocated FluxCapacitor structure with the arguments assigned to the appropriate fields. Note that since the FluxCapacitor structure is accessed through a pointer, the arrow operator is used instead of the dot operator.
Inside the function, the return value from the call to malloc() is checked before attempting assignment. If the allocation has failed, no assignment is made and a null pointer is returned to the calling function. Note that in the call to malloc(), the result is not cast, since there is no need for this in C and it needlessly clutters the code. Also observe that an identifier is used instead of an explicit type with the sizeof operator. This is less error-prone, easier to maintain if types change in the future, and is much cleaner code. That is, instead of this:
new_fcap = (struct FluxCapacitor *)malloc(sizeof (struct FluxCapacitor));
use this:
new_fcap = malloc(sizeof *new_fcap);
In main(), the return values from the calls to create_flux_cap() are checked. If an allocation has failed, the program exits with an error message.
The stdlib.h header file has been included for the function prototypes of malloc() and exit(), and also for the macro EXIT_FAILURE.
#include <stdio.h>
#include <stdlib.h>
struct FluxCapacitor
{
char* c_string;
unsigned value;
};
struct FluxCapacitor * create_flux_cap(char *, unsigned);
int main(void)
{
struct FluxCapacitor *fcap_1 = create_flux_cap("McFly", 42);
struct FluxCapacitor *fcap_2 = create_flux_cap("Biff", 1985);
/* Check for allocation errors */
if (fcap_1 == NULL || fcap_2 == NULL) {
fprintf(stderr, "Unable to create FluxCapacitor\n");
exit(EXIT_FAILURE);
}
/* Display contents of structures */
printf("%s, %u\n", fcap_1->c_string, fcap_1->value);
printf("%s, %u\n", fcap_2->c_string, fcap_2->value);
/* Free allocated memory */
free(fcap_1);
free(fcap_2);
return 0;
}
struct FluxCapacitor * create_flux_cap(char *str, unsigned val)
{
struct FluxCapacitor *new_fcap;
new_fcap = malloc(sizeof *new_fcap);
if (new_fcap != NULL) {
new_fcap->c_string = str;
new_fcap->value = val;
}
return new_fcap;
}
You need malloc for dynamic allocation of memory.In your case, both the types char and int are known to the compiler, it means the compiler can know the exact memory requirement at compile time.
For e.g. you can create a struct object like in the main function
#include<stdio.h>
#include<stdlib.h>
struct FluxCapacitor{
unsigned char* c_string;
unsigned int value;
};
int main() {
FluxCapacitor x;
x.c_string = "This is x capacitor"
x.value = 10
}
The x is of value type. You can make a copy and pass around this value. Also, observe we are using . notation to access its member variables.
But this doesn't happen at all time. We are not aware of future FluxCapacitor requirement and so above program will need more memory as while it is running and by using the malloc we can ask the compiler to provide us requested memory. This is a good place to use malloc, what malloc does is, it returns us a pointer to a piece of memory of the requested size. It is dynamic memory allocation.
Here's a simple example: let suppose if you need struct declaration of FluxCapacitor but don't know how many you will need, then use malloc
#include<stdio.h>
#include<stdlib.h>
typedef struct FluxCapacitor {
unsigned char* c_string;
int value;;
} flux;
// typedef is used to have the alias for the struct FluxCapacitor as flux
int main() {
flux *a = malloc(sizeof(flux)); // piece of memory requested
a -> c_string = "Hello World"; // Pointer notation
a -> value = 5;
free(a); // you need to handle freeing of memory
return 0;
}
.
I want to define a new data type consisting of an array with a size inputted by the user. For example if the user inputs 128, then my program should make a new type which is basically an array of 16 bytes. This structure's definition needs to be global since I am going to use that type thereafter in my program. It is necessary to have a dynamic size for this structure because I will have a HUGE database populated by that type of variables in the end.
The code I have right now is:
struct user_defined_integer;
.
.
.
void def_type(int num_bits)
{
extern struct user_defined_integer
{
int val[num_bits/sizeof(int)];
};
return;
}
(which is not working)
The closest thing to my question, I have found, is in here:
I need to make a global array in C with a size inputted by the user
(Which is not helpful)
Is there a way to do this, so that my structure is recognized in the whole file?
When doing:
extern struct user_defined_integer
{
int val[num_bits/sizeof(int)];
};
You should get the warning:
warning: useless storage class specifier in empty declaration
because you have an empty declaration. extern does not apply to user_defined_integer, but rather the variable that comes after it. Secondly, this won't work anyway because a struct that contains a variable length array can't have any linkage.
error: object with variably modified type must have no linkage
Even so, variable length arrays allocate storage at the point of declaration. You should instead opt for dynamic memory.
#include <stdlib.h>
typedef struct
{
int num_bits;
int* val;
} user_defined_integer;
void set_val(user_defined_integer* udi, int num_bits)
{
udi->num_bits = num_bits;
udi->val = malloc(num_bits/sizeof(int));
}
What you need is a VLA member, as asked about here. Basically, you declare a struct with a size field and one element's worth of storage as last member, and over-allocate it.
Imported from that question :
typedef struct Bitmapset {
int nwords;
uint32 words[1];
} Bitmapset;
Bitmapset *allocate(int n) {
Bitmapset *p = malloc(offsetof(Bitmapset, words) + n * sizeof *p->words);
p->nwords = n;
return p;
}
I want to define a new data type consisting of an array with a size inputted by the user. For example if the user inputs 128, then my program should make a new type which is basically an array of 16 bytes.
This is not possible in C, because C types are a compile-time thing and don't exist at all at run-time.
However, with a C99 conforming compiler, you might use flexible array member. You'll need a struct containing some members and ending with an array without any given dimension, e.g.
struct my_flex_st {
unsigned size;
int arr[]; // of size elements
};
Here is a way to allocate it:
struct my_flex_st *make_flex(unsigned siz) {
struct my_flex_st* ptr
= malloc(sizeof(struct my_flex_st) + siz * sizeof(int));
if (!ptr) { perror("malloc my_flex_st"); exit(EXIT_FAILURE); };
ptr->size = siz;
memset (ptr->arr, 0, siz*sizeof(int));
return ptr;
}
Don't forget to free it once you don't use it anymore.
Of course, you'll need to use pointers in your code. If you really want to have a global variable, declare it as e.g.
extern struct my_flex_st* my_glob_ptr;
Try this method-
#include<stdio.h>
#include<stdlib.h>
#include<limits.h>
struct user_defined_integer
{
int *val;
}user_int;
void memory_allocate(int num_bit)
{
int result;
result = (num_bit+CHAR_BIT-1)/CHAR_BIT; // since 8 bit =1 byte
user_int.val=malloc(result*sizeof(int));
if(user_int.val == NULL){
printf("Failed to allocate memory\n");
return ;
}
else
printf("Allocated %d bytes for val\n",result);
}
int main()
{
int num_bit;
printf("Enter the number of bits\n");
scanf("%d",&num_bit);
memory_allocate(num_bit);
// do your stuff here
free(user_int.val); // free the memory at the end;
return 0;
}
Is there an easy explanation for what this error means?
request for member '*******' in something not a structure or union
I've encountered it several times in the time that I've been learning C, but I haven't got a clue as to what it means.
It also happens if you're trying to access an instance when you have a pointer, and vice versa:
struct foo
{
int x, y, z;
};
struct foo a, *b = &a;
b.x = 12; /* This will generate the error, should be b->x or (*b).x */
As pointed out in a comment, this can be made excruciating if someone goes and typedefs a pointer, i.e. includes the * in a typedef, like so:
typedef struct foo* Foo;
Because then you get code that looks like it's dealing with instances, when in fact it's dealing with pointers:
Foo a_foo = get_a_brand_new_foo();
a_foo->field = FANTASTIC_VALUE;
Note how the above looks as if it should be written a_foo.field, but that would fail since Foo is a pointer to struct. I strongly recommend against typedef:ed pointers in C. Pointers are important, don't hide your asterisks. Let them shine.
You are trying to access a member of a structure, but in something that is not a structure. For example:
struct {
int a;
int b;
} foo;
int fum;
fum.d = 5;
It may also happen in the following case:
eg. if we consider the push function of a stack:
typedef struct stack
{
int a[20];
int head;
}stack;
void push(stack **s)
{
int data;
printf("Enter data:");
scanf("%d",&(*s->a[++*s->head])); /* this is where the error is*/
}
main()
{
stack *s;
s=(stack *)calloc(1,sizeof(stack));
s->head=-1;
push(&s);
return 0;
}
The error is in the push function and in the commented line. The pointer s has to be included within the parentheses. The correct code:
scanf("%d",&( (*s)->a[++(*s)->head]));
I have enumerated possibly all cases where this error may occur in code and its comments below. Please add to it, if you come across more cases.
#include<stdio.h>
#include<malloc.h>
typedef struct AStruct TypedefedStruct;
struct AStruct
{
int member;
};
void main()
{
/* Case 1
============================================================================
Use (->) operator to access structure member with structure pointer, instead
of dot (.) operator.
*/
struct AStruct *aStructObjPtr = (struct AStruct *)malloc(sizeof(struct AStruct));
//aStructObjPtr.member = 1; //Error: request for member ‘member’ in something not
//a structure or union.
//It should be as below.
aStructObjPtr->member = 1;
printf("%d",aStructObjPtr->member); //1
/* Case 2
============================================================================
We can use dot (.) operator with struct variable to access its members, but
not with with struct pointer. But we have to ensure we dont forget to wrap
pointer variable inside brackets.
*/
//*aStructObjPtr.member = 2; //Error, should be as below.
(*aStructObjPtr).member = 2;
printf("%d",(*aStructObjPtr).member); //2
/* Case 3
=============================================================================
Use (->) operator to access structure member with typedefed structure pointer,
instead of dot (.) operator.
*/
TypedefedStruct *typedefStructObjPtr = (TypedefedStruct *)malloc(sizeof(TypedefedStruct));
//typedefStructObjPtr.member=3; //Error, should be as below.
typedefStructObjPtr->member=3;
printf("%d",typedefStructObjPtr->member); //3
/* Case 4
============================================================================
We can use dot (.) operator with struct variable to access its members, but
not with with struct pointer. But we have to ensure we dont forget to wrap
pointer variable inside brackets.
*/
//*typedefStructObjPtr.member = 4; //Error, should be as below.
(*typedefStructObjPtr).member=4;
printf("%d",(*typedefStructObjPtr).member); //4
/* Case 5
============================================================================
We have to be extra carefull when dealing with pointer to pointers to
ensure that we follow all above rules.
We need to be double carefull while putting brackets around pointers.
*/
//5.1. Access via struct_ptrptr and ->
struct AStruct **aStructObjPtrPtr = &aStructObjPtr;
//*aStructObjPtrPtr->member = 5; //Error, should be as below.
(*aStructObjPtrPtr)->member = 5;
printf("%d",(*aStructObjPtrPtr)->member); //5
//5.2. Access via struct_ptrptr and .
//**aStructObjPtrPtr.member = 6; //Error, should be as below.
(**aStructObjPtrPtr).member = 6;
printf("%d",(**aStructObjPtrPtr).member); //6
//5.3. Access via typedefed_strct_ptrptr and ->
TypedefedStruct **typedefStructObjPtrPtr = &typedefStructObjPtr;
//*typedefStructObjPtrPtr->member = 7; //Error, should be as below.
(*typedefStructObjPtrPtr)->member = 7;
printf("%d",(*typedefStructObjPtrPtr)->member); //7
//5.4. Access via typedefed_strct_ptrptr and .
//**typedefStructObjPtrPtr->member = 8; //Error, should be as below.
(**typedefStructObjPtrPtr).member = 8;
printf("%d",(**typedefStructObjPtrPtr).member); //8
//5.5. All cases 5.1 to 5.4 will fail if you include incorrect number of *
// Below are examples of such usage of incorrect number *, correspnding
// to int values assigned to them
//(aStructObjPtrPtr)->member = 5; //Error
//(*aStructObjPtrPtr).member = 6; //Error
//(typedefStructObjPtrPtr)->member = 7; //Error
//(*typedefStructObjPtrPtr).member = 8; //Error
}
The underlying ideas are straight:
Use . with structure variable. (Cases 2 and 4)
Use -> with pointer to structure. (Cases 1 and 3)
If you reach structure variable or pointer to structure variable by following pointer, then wrap the pointer inside bracket: (*ptr). and (*ptr)-> vs *ptr. and *ptr-> (All cases except case 1)
If you are reaching by following pointers, ensure you have correctly reached pointer to struct or struct whichever is desired. (Case 5, especially 5.5)
It may means that you forgot include a header file that define this struct/union.
For example:
foo.h file:
typedef union
{
struct
{
uint8_t FIFO_BYTES_AVAILABLE : 4;
uint8_t STATE : 3;
uint8_t CHIP_RDY : 1;
};
uint8_t status;
} RF_CHIP_STATUS_t;
RF_CHIP_STATUS_t getStatus();
main.c file:
.
.
.
if (getStatus().CHIP_RDY) /* This will generate the error, you must add the #include "foo.h" */
.
.
.
can also appear if:
struct foo { int x, int y, int z }foo;
foo.x=12
instead of
struct foo { int x; int y; int z; }foo;
foo.x=12
I saw this when I was trying to access the members.
My struct was this:
struct test {
int a;
int b;
};
struct test testvar;
Normally we access structure members as
testvar.a;
testvar.b;
I mistook testvar to be a pointer and did this.
testvar->a;
That's when I saw this error.
request for member ‘a’ in something not a structure or union
My ridiculous experience is that I incorrectly put '.' instead of ','.
printf("%c". ch);