I need some ideas on my array of struct implementation. This is what I have in my structs. I plan on making SHAPES an array because I will have multiple SHAPES. Each shape will have multiple x and y coordinates. Because of this I'm not sure if should make a linked list or not. The purpose of the start and finish is I will eventually be running a search algorithm after I get my shapes right.
struct START
{
int x;
int y;
};
struct END
{
int x;
int y;
};
struct SHAPES
{
int x [100];
int y [100];
int *link;
};
struct DISTANCE
{
int distance_traveled;
int distance_to_finish;
};
I was reading this and was wondering if I needed to malloc or calloc my structs as I create them. If so why and if not why? Malloc and calloc are always confusing to me.
How do you make an array of structs in C?
int main(int argc, char *argv[])
{
//Is it better to make these pointers?
struct START start;
struct END end;
struct SHAPES shapes[100];
while(fgets(line, 80, fp) != NULL)
{
//I have a very annoying issue here, I don't know the size of the lines.
//If there are more values I want to increment the x[] and y[] array but NOT the
//shapes array. I can only think of one way to deal with this by putting a bunch of
//variables in the sscanf. I discovered putting sscanf on the next line didn't do what
//I was hoping for.
sscanf(line, "%d%*c %d%*c ", &shapes[i].x[i] , &shapes[i].y[i] );
printf(" line is: %s \n", line);
sscanf(line, "%d%*c %d%*c ", &x1_main , &y1_main );
printf(" line is: %s \n", line);
printf(" shapes[i].x[i] is: %d \n", shapes[i].x[i]);
printf(" shapes[i].y[i] is: %d \n", shapes[i].y[i]);
printf(" x1_main is: %d \n", x1_main);
printf(" y1_main is: %d \n", y1_main);
i++;
memset(line, 0, 80);
}
}
This is what my file looks like. Adding the %*c seemed to handle the commas and semicolons appropriately.
10, 4
22, 37
22, 8; 2, 0; 3, 6; 7, 8; 5, 10; 25, 2
1, 2
I got that idea from here.
https://www.daniweb.com/software-development/c/threads/334515/reading-comma-separated-values-with-fscanf
First of all, you might want to consider something like this:
struct point {
int x;
int y;
};
so you can use a struct point data structure (array) instead of two separate data structures for x and y. Using it like this should also speed up access to the points, since their coordinates are next to each other in memory. Otherwise you will have x somewhere in the x array and y somewhere in the y array.
The choice of the data structure to store the points depends on your usage. If you need to address points directly, a linked list may be a bad choice. If you always access all points in a linear order, it is fine. However, consider that a singly linked list will add 8 bytes per point for the next pointer. A doubly linked list will use another 8 bytes for prev (assuming 64-bit arch that is; sizeof(pointer) in general). I assume, that you create x[100] and y[100] to make sure you have enough space. You might be better off using a dynamic array (the ADT) e.g. glib's GArray after all. It will grow as big as you need it without you doing anything.
For malloc vs calloc: it doesn't really matter. A call to calloc is basically a malloc followed by a
memset(ptr, 0, sizeof(mallocd area);
i.e. the memory is zeroed; cf manpage calloc. If you initialize the memory directly you may not need to do this.
A struct with no pointer members
If your struct has no pointer members, such as:
typedef struct {
int a;
int b;
} DEMO;
Then you can simply declare an array instance of a typedef struct like this:
DEMO demo[10];// instance of array of 10 DEMO
Example, struct with Pointer members
If you have a pointer in the list of members:
#define SIZE_STR 20
typedef struct {
int a;
int b;
char *str;//pointer, will require memory allocation
} DEMO;
DEMO demo[10];// instance of array of 10 DEMO
int main(void)
{
int i;
for(i=0;i<10;i++)//create memory for each instance of char * in array of DEMO
{
demo[i].str = malloc(SIZE_STR);
}
return 0;
}
Don't forget to free() all instances of malloc'ed memory.
Dynamically allocate array of struct
If you need to dynamically allocate memory for a struct:
typedef struct {
int a;
int b;
} DEMO;
DEMO demo, *pDemo;// create a pointer to DEMO
In a function, main() for example:
int main(void)
{
pDemo = &demo;
pDemo = malloc(sizeof(DEMO)*10);//provides an array of 10 DEMO
return 0;
}
Again, don't forget to free() all instances of malloc'ed memory.
Related
hey I am trying to create a program in which I am trying store elements from one array to another with the use of a pointer to pointer but the problem is that is caused undefined behavior I believe that the problem is that I do not pass the elements in members with a proper way
I know it is a vague way of doing this but It is in only for practising reasons
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
typedef struct student{
char *name;
int *number;
}T;
int main(void) {
char array[10][100]={"araaaa","bbgt","gffkghgh"};
T arr[10][100];
T *p;
T **p1;
p=&arr[0][0];
p1=&p;
int i=0;
for(i = 0 ; i < 3 ; i++)
{ p=arr[i];
strcpy((*p1)->name,array[i]);
}
/*******print_elements*************/
for(i = 0 ; i < 3 ; i++)
{ p=arr[i];
printf("\n the elements are %s",(*p1)-> name);
}
return 0;
}
When you do this:
strcpy ((*p1)->name, array[i]);
(*p1)->name is an uninitialised pointer. What happens, therefore, is in the lap of the gods.
The easiest fix is to modify your student structure such that name is a buffer, rather than a pointer. At the same time, change number to an int, rather than a pointer to an int::
typedef struct student{
char name [100];
int number;
} T;
If you want to keep name as a pointer then you have to allocate some memory before you store your string in it. This should work:
(*p1)->name = strdup (array[i]);
Don't forget to free the memory when done.
T is made of of two pointers, this first one points to a string of characters in memory.
arr is a 2D array that is allocated to store a total of 1000 T structures.
arr[i] would reference a 1D array of T structures within arr
*p1 would essentially be arr[i], since dereferencing p1 gives you p, which was just set to arr[i]. So, that is not a pointer to a T structure, but to an array of T structures. Forcing the cast will likely give you a reference to the first T structure in that row, however.
->name This value is never set. You allocated an array, but "name" is a pointer to memory, not an array of characters, so '->name' is undefined.
I think you need to change arr to be a single dimension array. You aren't using 90% of it.
And, you need to initialize every T struct in that array. You can use malloc or strdup, and then remember to free them all. Or, set the struct to use an array instead.
How can i create N elements of this struct after the input N?
typedef struct cat{
int code;
int age;
float weight;
enum {kibbles,canned_food,tuna_fish}food;
} cats;
int n,i;
printf("Insert a number: ");
scanf("%d",&n);
for(i=0;i<n;i++){
....
}
I want to create N cats (named cat1,cat2 etc..)
Make a function to input a single struct cat
struct cat inputsinglecat(void);
After you know how many cats you need, get the amount of memory required
struct cat *memcat;
memcat = malloc(n * sizeof *memcat);
if (memcat == NULL) exit(EXIT_FAILURE);
Then, to enter cats, use a loop and the function defined above
for (int k = 0; k < n; k++) {
memcat[k] = inputsinglecat();
}
Don't forget to release the memory when you don't need it anymore
free(memcat);
You can either do it statically, by creating an array of cats:
cats myCatsArray[10];
or dynamically, using malloc or calloc (the latter defined as void *calloc(size_t nitems, size_t size)):
cats *myCatsArray = calloc( 10, sizeof (cats)):
Just avoid the static definition as a local variable of a function, in order to avoid to occupy to much memory in the stack. In case of dynamic allocation you have to remember to free() the structs as soon as you don't need them anymore.
After allocating all the N cats you need, you are able to populate their fields according to your requirements.
In both cases you can access an element (lets say the sixth one) in this way
int myCode = myCatsArray[5].code;
Today I was working on a problem of moving an array of smaller structs directly into an array of larger structs (arrayNew) (essentially upgrading the smaller structs to store more information). The smaller structs needed to be read from a HDD in one single read operation into the array of new 'upgraded' larger structs, a function would be called to do the 'upgrading'. Also all the new fields in the structs that were read from the hard drive would be set to '0'.
Other more simple solutions that I tried were:
Creating a local array of the old structures (arrayOld), loading the structures from the HDD into it then simply looping through the empty array of the new structures (arrayNew) and manually moving each structs contents from arrayOld into arrayNew. (e.g. arrayNew[i].x = arrayOld[i].x; )
The problem with this is that in my case the arrays I was working with were very large and too large for the stack ( about 1mb for each array) causing a segmentation fault the instant the upgrading function was called.
Another viable solution was to create a dynamic array of the old structures (arrayDy) and load the old structures into arrayDy and then again manually moving each structs contents from arrayDy into arrayNew. (e.g. arrayNew[i].y = arrayDy[i].y; ) This addressed the issue of running out of stack memory.
After implementing the second solution. I decided to experiment and develop a solution that uses no dynamically allocated memory and loads the array of old structures from the HHD directly into the larger array of larger structs arrayNew in one read operation and manipulate the contents of arrayNew in memory to pad out the missing values that are there due to the array being bigger.
I will post my solution below in a scaled down version of what I implemented, using the following structs for my example:
typedef struct INNER_STRUCT_ {
int i_item1;
int i_item2;
char i_item3;
} INNER_STRUCT;
typedef struct SMALL_STRUCT_ {
int item1;
char item2;
INNER_STRUCT item3;
} SMALL_STRUCT;
typedef struct BIG_STRUCT_ {
int item1;
char item2;
INNER_STRUCT item3;
INNER_STRUCT item4;
} BIG_STRUCT;
Yes, this is possible - you can use union for that. C99 standard makes a special guarantee that can be used to implement your requirement:
6.5.2.3-5: One special guarantee is made in order to simplify the use of unions: if a union contains several structures that share a common initial sequence (see below), and if the union object currently contains one of these structures, it is permitted to inspect the common initial part of any of them anywhere that a declaration of the complete type of the union is visible.
Your structA_ and structB_ do share a common initial sequence, so creating a union and accessing the structs through it would do the trick:
union {
structA a;
structB b;
} u;
memset(&u.b, 0, sizeof(structB)); // Zero out the bigger structB
loadFromHdd(&u.a); // Load structA part into the union
// At this point, u.b is valid, with its structA portion filled in
// and structB part zeroed out.
Note that you cannot do it to an array (unless, of course, you make an array of unions). Each structA needs to be loaded individually into the union, from which it could then be read as structB.
The method I propose and used as a solution basically loads the smaller structs for the HDD ( a file in this case) into the array of new larger structs and then rearranges the block of memory so that each field can be accessed properly. The code to illustrate this is below, and is an mcve.
#include <stdio.h>
#include <string.h>
typedef struct INNER_STRUCT_ {
int i_item1;
int i_item2;
char i_item3;
} INNER_STRUCT;
typedef struct SMALL_STRUCT_ {
int item1;
char item2;
INNER_STRUCT item3;
} SMALL_STRUCT;
typedef struct BIG_STRUCT_ {
int item1;
char item2;
INNER_STRUCT item3;
INNER_STRUCT item4;
/*
Note that the big struct is exactly the same as the small
struct with one extra field - Key to this method working
is the fact that the extension to the struct is appended
at the end, in an array of the structs will be placed one
after the other in memory with no gaps*/
} BIG_STRUCT;
void printSmallStruct (SMALL_STRUCT *inStruct, int count) {
// Print everything inside given small struct
printf("\n\n Small struct %d, item1: %d \n",count,inStruct->item1);
printf(" Small struct %d, item2: %c \n",count,inStruct->item2);
printf(" Small struct %d, item3.i_item1: %d \n",count,inStruct->item3.i_item1);
printf(" Small struct %d, item3.i_item2: %d \n",count,inStruct->item3.i_item2);
printf(" Small struct %d, item3.i_item3: %c \n",count,inStruct->item3.i_item3);
}
void printBigStruct (BIG_STRUCT *inStruct, int count) {
// Print everything inside given big struct
printf("\n\n Big struct %d, item1: %d \n",count,inStruct->item1);
printf(" Big struct %d, item2: %c \n",count,inStruct->item2);
printf(" Big struct %d, item3.i_item1: %d \n",count,inStruct->item3.i_item1);
printf(" Big struct %d, item3.i_item2: %d \n",count,inStruct->item3.i_item2);
printf(" Big struct %d, item3.i_item3: %c \n",count,inStruct->item3.i_item3);
printf(" Big struct %d, item4.i_item1: %d \n",count,inStruct->item4.i_item1);
printf(" Big struct %d, item4.i_item1: %d \n",count,inStruct->item4.i_item2);
printf(" Big struct %d, item4.i_item1: %c \n",count,inStruct->item4.i_item3);
}
int main() {
SMALL_STRUCT smallStructArray[5]; // The array of small structs that we will write to a file then read
BIG_STRUCT loadedBigStructArray[5]; // The large array of structs that we will read the data from the file into
int i; // Counter that we will use
FILE *pfile; // pointer to our file stream
void *secondary_ptr; // void pointer that we will use to 'chop' memory into the size we want
/* Fill the array of structs (smallStructArray) */
for (i = 0; i < 5; i++) {
/* We fill each field with different data do we can ID that the right data is in the right fields */
smallStructArray[i].item1 = 111;
smallStructArray[i].item2 = 'S';
INNER_STRUCT* temp = &smallStructArray[i].item3;
temp->i_item1 = 777;
temp->i_item2 = 999;
temp->i_item3 = 'I';
}
/* Write the contents of smallStructArray to binary file then display it */
pfile = fopen("test.dat","wb");
if (pfile!=NULL){
for (i = 0; i < 5; i++) {
fwrite(&smallStructArray[i],sizeof(SMALL_STRUCT),1,pfile);
}
fclose(pfile);
}
else{
printf("Unable to open file!");
return 1;
}
for (i = 0; i < 5; i++) {
printSmallStruct(&smallStructArray[i],i);
}
/* Clear array of big structs using memset */
memset(&loadedBigStructArray[0],0,sizeof(loadedBigStructArray));
/* Here we read from the smallStructArray that was aved to file into the loadedBigStructArray */
pfile = fopen("test.dat","rb");
if (pfile !=NULL){
/*
He we pass fread the following: size_t fread(void *args1, size_t args2, size_t args3, FILE *args4)
args1 - a pointer to the beginning of a block of memory, in our case the beginning of the
array loadedBigStructArray.
args2 - the size of the ammout of bytes we wish to read, in our case the size of a SMALL_STRUCT,
the size one of the elments in the array saved to the file.
args3 - the ammount of elements to read, in our case five (which is the number of elements the
array saved to the file has.
args4 - a pointer to a FILE that specifies our input stream.
Essentially what fread will do here is read a block of bytes the size of the array we saved to
the file (smallStructArray) into the array in memory loadedBigStructArray from the
beggining of loadedBigStructArray. Fig 1 illustrates what this will look like in memory.
*/
fread(&loadedBigStructArray,sizeof(SMALL_STRUCT),5,pfile);
fclose(pfile);
}
else{
printf("Unable to open file!");
return 1;
}
/*
Due to the way the array on the file has been read into the array in memory, if we try
to access the data in loadedBigStructArray only the first 5 values will be valid, due to
the memory not being in the order we want. We need to re-arrange the data in loadedBigStructArray
*/
/*
Here we use a void pointer to point to the beggining of the loadedBigStructArray.
we will use this pointer to 'chop' the data loadedBigStructArray into SMALL_STRUCT
sized 'chunks' we can read from.
Due to the way pointers and arrays work in C we can cast the void pointer to any type we want
and get a chunk of memory that size begginnig from the pointer and its off set.
E.g. : int temp = ((int *)void_ptr)[i];
This example above will give us an integer 'temp' that was taken from memory beggining from position
void_ptr in memory and its offset i. ((int *)void_ptr) casts the pointer to type int and [i] dereferances
the pointer to location i.
*/
secondary_ptr = &loadedBigStructArray;
/*
Not we are going through the array backwards so that we can rearange the data with out overwriting
data in a location that has data which we havent moved yet. As the bottom end of the loadedBigStructArray
is essentially empty we can shift data down that way.
*/
for (i = 5; i > -1; i=i-1) {
SMALL_STRUCT temp = ((SMALL_STRUCT *)secondary_ptr)[i]; // dereference pointer to SMALL_STRUCT [i] inside loadedBigStructArray call it 'temp'
/*
Now that we have dereferenced a pointer a given SMALL_STRUCT inside loadedBigStructArray called 'temp'
we can use temp to move the data inside temp to its corresponding position in loadedBigStructArray
which rearragnes the data.
*/
loadedBigStructArray[i].item1 = temp.item1;
loadedBigStructArray[i].item2 = temp.item2;
loadedBigStructArray[i].item3.i_item1 = temp.item3.i_item1;
loadedBigStructArray[i].item3.i_item2 = temp.item3.i_item2;
loadedBigStructArray[i].item3.i_item3 = temp.item3.i_item3;
/* We then fill the new field to be blank */
loadedBigStructArray[i].item4.i_item1 = 0;
loadedBigStructArray[i].item4.i_item2 = 0;
loadedBigStructArray[i].item4.i_item3 = '0';
}
/* Print our new structures */
for (i = 0; i < 5; i++) {
printBigStruct(&loadedBigStructArray[i],i);
}
return 0;
}
Visualization of technique:
When fread does the single read operation of the array saved on disk into the array in memory due to it being smaller it will take up the first potion of the array in memory but the 'bottom' section could be anything, if we try to access the data in the new array with the current handles we have on the data we will either get inaccurate information or a bad piece of memory. We have to rearrange this data before we can use any of our handles on the structs in the array.
I'm currently creating an array of structs, and when i initialize the array, it is starting with 8 elements in the struct, instead of 1. Why is it doing this? If more code is needed (but i doubt it as they are all seperate functions, i can post it if asked)
This is the relevant bits of code:
typedef struct {
int valid;
int row, col;
} Point;
typedef struct {
Point point;
int number;
char name;
char param[20];
char type[20];
} Agent;
int main(int argc, char **argv)
{
int steps;
int listSize = 0;
Agent *agentList = (Agent *) calloc(1, sizeof(Agent));
printf("%d size of agentList when initialised\n", sizeof(agentList));
if (argc != 4) {
error(SHOW_USAGE);
}
sscanf(argv[2], "%d", &steps);
if ((steps < 1)) {
error(BAD_STEPS);
}
readMap(argv[1]);
agentList = readAgentFile(argv[3], agentList);
print_agents(agentList);
return 0;
printf("%d size of agentList when initialised\n", sizeof(agentList));
This will give you the size of the agentList pointer which will most likely be four or eight in current systems.
Similarly, sizeof(*agentList) would simply give you the size of the structure pointed to by agentList but that's a single element of the array rather than the entire array.
There is no way to get from a pointer to the size of the array pointed at by it, simply because a pointer points at a single thing, not an array. By that, I mean it may well point to the first element of an array but that's not the same thing as pointing at an array. The distinction is subtle but important.
If you want to know the size of an array where the only thing you have is a pointer to it, you will have to remember it separately.
This here is not giving you the size of the array, but giving you the size of the pointer:
printf("%d size of agentList when initialised\n", sizeof(agentList));
It's not possible to determine the size of the array from the pointer, you need to store the size of the array separately and carry it around with the array.
This is why you will notice many functions in C will have both a pointer and a size argument.
The following code causes a SIGSEGV, but only while debugging.
#include <stdio.h>
#include <stdlib.h>
typedef struct enemy_desc
{
int type;
int x;
int y;
}enemy;
int main()
{
enemy **enemies;
enemies=(enemy **)malloc(sizeof(enemy *)*16);
enemies[0]->type=23;
printf("%i",enemies[0]->type);
return 0;
}
You are only creating space for 16 pointers to enemy, but are not creating the actual enemy objects that you're attempting to use.
Here is an example where I create an enemy object to the first pointer in the array.
#include <iostream>
typedef struct enemy_desc
{
int type;
int x;
int y;
}enemy;
using namespace std;
int main(int argc, char **argv)
{
enemy **enemies;
enemies=(enemy **)malloc(sizeof(enemy *)*16);
memset(enemies, 0, sizeof(enemy*)*16);
enemies[0] = (enemy *) malloc(sizeof(enemy));
memset(enemies[0], 0, sizeof(enemy));
enemies[0]->type=23;
printf("type: %i x: %i y: %i\n\n",enemies[0]->type, enemies[0]->x, enemies[0]->y);
enemies[0]->x = 10;
enemies[0]->y = 25;
enemies[0]->type= 7;
printf("type: %i x: %i y: %i\n\n",enemies[0]->type, enemies[0]->x, enemies[0]->y);
free(enemies[0]);
free(enemies);
return 0;
}
You have allocated memory for 16 enemy * pointers, but you have not allocated room for the 16 enemy structs themselves. There are two ways to fix this. One is to add a loop that allocates each of the 16 enemy structs one by one:
int main()
{
enemy **enemies;
int i;
enemies = (enemy **) malloc(sizeof(enemy *) * 16);
for (i = 0; i < 16; ++i) {
enemies[i] = (enemy *) malloc(sizeof(enemy));
}
enemies[0]->type = 23;
printf("%i",enemies[0]->type);
return 0;
}
The other is to remove one level of indirection. If you declare enemy *enemies then you can allocate the 16 structs at once and forgo a loop. If there's no need for the double indirection this would be my preferred solution:
int main()
{
enemy *enemies;
enemies = (enemy *) malloc(sizeof(enemy) * 16);
enemies[0].type=23;
printf("%i",enemies[0].type);
return 0;
}
Notice that the -> operator switches to ..
You need to create the struct that the pointers point to. The reason why it only gives a SEGV while debugging will be that there's some initialisation done during debugging to cause this sort of thing to segfault; whatever random data's in enemies[0] when you're not debugging is getting dereferenced and is just happening not to cause a segfault.
Your code probably wants to read like this:
int main()
{
enemy *enemies;
enemies=(enemy *)malloc(sizeof(enemy)*16);
enemies[0].type=23;
printf("%i",enemies[0].type);
return 0;
}
Well it should crash right away... because you initialize the array of (array of pointers). And then you DEREFERENCE the first item (enemies[0]) which should give you any random pointer. You try to access that random memory area to write in the value 23.
It should be along the lines of this:
enemies = (enemy **)malloc(sizeof(enemy *) * 16);
for (int i = 0; i < 16; i++) {
enemies[i] = (enemy *)malloc(sizof(enemy));
}
... before you access it.
Looks like you want to allocate an array of pointer to struct, but you're trying to access the struct without allocating space for them. You should do:
enemy **enemies;
enemies=(enemy **)malloc(sizeof(enemy *)*16);
for(i=0;i<16;i++) // allocate space for the structs
enemies[i] = (enemy *)malloc(sizeof(enemy))
enemies[0]->type=23; // now access type field of the first struct obj in array.
On my system (x86, Debian GNU/Linux), the code always segfaults.
Crash backtrace:
signal SIGSEGV, Segmentation fault.
0x08048413 in main () at en.c:16
16 enemies[0]->type=23;
(gdb)
The assignment cited (enemies[0]->type=23;) is the problem. You only allocate memory for the array enemies, which is an array of pointers. You then access the structure that enemies[0] is supposed to point to, but you have not allocated memory for the structure, and enemies[0] is an uninitialized pointer, hence the segfault.
In cases like this, a debugger is your friend :-).