Modifying a structure array through a pointer passed to a function - c

I am trying to pass a structure array pointer and a pointer to a structure array pointer into a function and then have it modified rather than using a return.
This example code is indeed pointless, its just a learning example for me.
Basically I want to create array[0]...array[1]..array[2] and so on and have a pointer that points to these while using a different index...such as array_ref[2] points to array[0] and array_ref[3] points to array[1].
The code below compiles, but immediately crashes. Any suggestions?
typedef struct unit_class_struct {
char *name;
char *last_name;
} person;
int setName(person * array, person ***array_ref) {
array[0].name = strdup("Bob");
array[1].name = strdup("Joseph");
array[0].last_name = strdup("Robert");
array[1].last_name = strdup("Clark");
*array_ref[2] = &array[0];
*array_ref[3] = &array[1];
return 1;
}
int main()
{
person *array;
person **array_r;
array = calloc (5, sizeof(person));
array_r = calloc (5, sizeof(person));
setName(array,&array_r);
printf("First name is %s %s\n", array_r[2]->name, array_r[2]->last_name);
printf("Second name is %s %s\n", array_r[3]->name, array_r[3]->last_name);
return 0;
}

You are allocating an array of structures but declaring an array of pointers.
Calloc() and malloc() return a void * object that can be assigned to everything. So, even with -Wall your program will compile with no warnings.
However, when it runs, it tries to deal with what you said was an array of pointers, in fact, an array of pointers to pointers to pointers. None of these pointers are ever created. All you start with are one-level pointers to an array of objects. Depending on what you really want, you may need something more like...
array_ref[2] = &array[0];
array_ref[3] = &array[1];
return 1;
}
int main()
{
person *array, *array_r_real;
person **array_r;
array = calloc(5, sizeof(person));
array_r_real = calloc(5, sizeof(person));
array_r = calloc(5, sizeof(struct unit_class_struct *));
int i;
for (i = 0; i < 5; ++i)
array_r[i] = &array_r_real[i];
setName(array, array_r);

Related

Accessing members of dynamically allocated array of pointers to structs

I need to have a global dynamic array of pointers, in which I will store my structs, beacuse later I will need to iterate through this array to list all the stored information, I also need to be able to read the name, age and job variables from the console, and store them in a person_t in the iterator array.
#include <stdio.h>
#include <stdlib.h>
typedef struct Person
{
char name[30];
int age;
char job[30];
} person_t;
person_t **iterator;
int capacity = 10;
int size = 0;
int main()
{
int i;
*iterator = (person_t *)malloc(capacity * sizeof(person_t));
for (i = 0; i < capacity; ++i)
{
person_t p;
p.age = i;
*iterator[i] = p;
}
return 0;
}
I get no errors/warnings compiling this code (gcc -ansi -pedantic -Wall -Wextra), but when I try to run it, I get a Segmentation fault immediately.
When you do this:
*iterator = (person_t *)malloc(capacity * sizeof(person_t));
You're deferencing iterator, however as a file-scope pointer variable it's initialized to NULL. Attempting to dereference a NULL pointer invokes undefined behavior.
I suspect what you really want is an array of structs, not an array of pointers to structs. That being the case, define iterator as:
person_t *iterator;
Then you allocate memory for it like this:
iterator = malloc(capacity * sizeof(person_t));
Then assign to array elements like this:
iterator[i] = p;
Your stated purpose is to create a "global dynamic array of pointers, in which I will store my structs". The following modification of your code (see comments) will do this:
person_t p[10] = {0};
int main()
{
int i;
// with declaration: person_t **iterator = NULL;,
//following is all that is needed to create an array of pointers:
iterator = malloc(capacity * sizeof(person_t *));//no need to cast return of malloc
for (i = 0; i < capacity; ++i)
{
//person_t p;//moved to scope that will exist outside of main()
p[i].age = i;
iterator[i] = &p[i];//assign the address of the object to the pointer
//iterator[i] is the ith pointer in a collection of
//pointers to be assigned to point to
//instances of struct person_t
}
//Once all fields are populated (to-do), the following will display the results:
for (i = 0; i < capacity; ++i)
{
printf("%d) Name: %s Age: %d Job: %s\n", i, iterator[i]->name,iterator[i]->age,iterator[i]->job);
}
return 0;
}
you are not allocating memory correctly
First you need to allocate memory for a pointer which can store capacity number of address i.e done through iterator = malloc(capacity * sizeof(person_t*)); and then you need to allocate memory for holding each structure element i.e iterator[i] = malloc(sizeof(person_t));
all the malloc'ed memory should be free'd once we are done with it.
Also, have not done the error check for malloc's , that is left as an exercise for you.
int main()
{
int i;
// test data
char *names[] = {"ABC", "DEF"};
char *jobs[] = {"Accountant", "Security"};
int ages[] = {50, 60};
// first allocate memory for iterator , which can hold pointers to store iterator poniters
iterator = malloc(capacity * sizeof(person_t*));
for (i = 0; i < capacity; ++i)
{
// now allocate memory for individual iterator
iterator[i] = malloc(sizeof(person_t));
strcpy(iterator[i]->name,names[i]);
iterator[i]->age = ages[i];
strcpy(iterator[i]->job, jobs[i]);
}
for (i = 0; i < capacity; ++i)
{
printf("name = %s ", iterator[i]->name);
printf("Age = %d ", iterator[i]->age);
printf("Job = %s\n", iterator[i]->job);
}
return 0;
}

Accessing Element in Array Inside Struct Dynamically Allocated

So I have an array of structs with every struct containing a dynamic array of strings.
typedef struct _test {
int total;
char *myarray[];
} Test;
This how I allocated enough memory
Test *mystruct = (Test *)malloc(size);
Above is how my struct is formated
mystruct[x].myarray[index] = strdup(name);
index++;
mystruct[x].total = index;
when trying to access a string in that array i.e. :
printf("%s", mystruct[0].myarray[0]);
Nothing prints, thanks for any help!
The following is correct. I'll explain below.
typedef struct TEST {
int total;
char *myarray[];
} Test;
int StackOveflowTest(void)
{
int index;
Test *mystruct = malloc(sizeof(Test)+10*sizeof(char *));
for (index=0; index<10; index++)
mystruct->myarray[index] = strdup("hello world");
mystruct->total = index;
for (index=0; index<10; index++)
printf("%s\n", mystruct->myarray[index]);
return 0;
}
Actually, you declare an "array of pointers" myarray, however that array has zero elements. This "trick" is used to have the ability that at the end of the struct you have an array of variable size when malloc'ing the array:
Test *mystruct = malloc(sizeof(Test)+10*sizeof(char *));
This allocates the size of the struct and adds room for 10 array elements.
Since you did not allocate this flexible part of the struct, you wrote into "nothing" and are lucky the program did not abort (or it did, which is why there was no output).
P.s.: don't forget to free the memory when you are done:
for (index=0; index<10; index++)
free(mystruct->myarray[index]);
free(mystruct);

Garbage value even after initializing members of dynamically allocated struct array

I have a dynamically allocated array of structures, 'buff'. Each element is a structure that has a few integer variables and a pointer 'buffer_ptr' which points to another dynamically allocated array of structures. The size of both arrays is given as command line input.
int buffer_size;
int user_num;
struct tuple
{
char userID[5];
char topic[16];
int weight;
};
struct buff_ctrl
{
struct tuple* buffer_ptr;
int in;
int out;
int numItems;
int done;
};
The arrays are created and initialized in main() as follows:
int main(int argc, char* argv[])
{
void *status;
pthread_t mapThd;
if(argc != 4)
{
printf("Input format: ./combiner <buffer_size> <number_of_users> <input_file>\n");
return -1;
}
buffer_size = atoi(argv[1]);
user_num = atoi(argv[2]);
struct buff_ctrl *buff = (struct buff_ctrl*)malloc(user_num * sizeof(struct buff_ctrl));
for(int i=0; i<user_num; i++)
{
struct buff_ctrl* curr_buff = (buff + (i*sizeof(struct buff_ctrl)));
struct tuple *ptr = (struct tuple*)malloc(buffer_size * sizeof(struct tuple));
curr_buff->buffer_ptr = ptr;//points to another array
curr_buff->in = 8;
curr_buff->out = 4;
curr_buff->numItems = 7;
curr_buff->done = 0;
printf("%p\n",curr_buff);
}
Then, I need to pass the 'buff' pointer as an argument when creating thread using pthread_create:
pthread_create(&mapThd, NULL, mapper, (void*)buff);
pthread_join(mapThd, &status);
free(buff);
/*end of main*/
My function pointer is as follows:
void* mapper(void *buff)
{
struct buff_ctrl* arr = (struct buff_ctrl *)buff;
struct buff_ctrl* temp_ptr;
printf("######################################################\n");
for(int k=0; k<user_num; k++)
{
/*Printing just to check values */
temp_ptr = arr + (k*sizeof(struct buff_ctrl));
printf("buffer ptr = %p\n", temp_ptr->buffer_ptr);
printf("in = %d\n", temp_ptr->in);
printf("out = %d\n", temp_ptr->out);
printf("numItems = %d\n", temp_ptr->numItems);
}
printf("######################################################\n");
pthread_exit((void*)buff);
}
But, when I print the values of 'buffer_ptr' from the created thread (only one), for ODD number of user_num, there is always ONE element of the array 'buff' which gives garbage value after pthread_create statement! When the values are checked in main itself after removing calls to pthread, it runs fine.
This line
struct buff_ctrl* curr_buff = (buff + (i*sizeof(struct buff_ctrl)));
should be
struct buff_ctrl* curr_buff = buff + i;
buff + i is pointer arithmetic and the compiler already takes the size of the
object pointed to by buff into consideration. By doing i*sizeof(struct buff_ctrl) you are assigning
a pointer that may be after the allocated memory.
As general suggestion:
Don't cast malloc. And instead of using sizeof(<type>), use sizeof *variable, this is more safe, because
it's easier to make mistakes when writing sizeof(<type>).
So:
struct buff_ctrl *buff = malloc(user_num * sizeof *buff);
...
struct tuple *ptr = malloc(buffer_size * sizeof *ptr);
And you don't need to declare a separate pointer, you can do:
for(int i=0; i<user_num; i++)
{
buff[i].buffer_ptr = malloc(buffer_size * sizeof *buff[i].buffer_ptr);
buff[i].in = 8;
buff[i].out = 4;
buff[i].numItems = 7;
buff[i].done = 0;
}
Also you should always check for the return value of malloc. If it returns
NULL, you cannot access that memory.
This is wrong:
struct buff_ctrl* curr_buff = (buff + (i*sizeof(struct buff_ctrl)));
When you do pointer arithmetic, it operates in units of the size of what the pointer points to, so you don't need to multiply by sizeof. As a result, you're effectively multiplying twice and accessing outside the array bounds.
Just treat buff as an array, rather than dealing with pointers.
for(int i=0; i<user_num; i++)
{
struct tuple *ptr = malloc(buffer_size * sizeof(struct tuple));
buff[i].buffer_ptr = ptr;//points to another array
buff[i].in = 8;
buff[i].out = 4;
buff[i].numItems = 7;
buff[i].done = 0;
}
Also, see Do I cast the result of malloc?
You have a fundamental error.
Pointer arithmetics works by adding the offset in multiples of the pointer type, so adding the offset yourself will not work as you apparently expect it to.
If it was a char * pointer then you would need to add the offset manually, increments would be multiplied by one. But in your case increments by n are multiplied by the size of the pointer base type.
There are times when doing pointer arithmetics with the addition notation makes sense, but most of the time it's much clearer to write index notation instead.

What is wrong with my dynamically allocated array of pointers to structs?

I have the following program that references array elements through a double pointer.
typedef struct {
int num1;
int num2;
int num3;
} DATA_SET;
typedef struct {
int structID;
DATA_SET *data_set_array; // Pointer to an array of DATA_SET structs
} MY_STRUCT;
int main () {
MY_STRUCT *Struct1;
DATA_SET **DataSetArray; // Array of pointers
Struct1 = malloc(sizeof(MY_STRUCT));
Struct1->data_set_array = malloc(sizeof(DATA_SET*)) //Allocate mem for the pointer to array of DATA_SETs
DataSetArray = malloc(sizeof(DATA_SET*) * 2) // Allocate mem for an array of 2 DATA_SET pointers
DataSetArray[0] = malloc(sizeof(DATA_SET)) // Allocate mem for the actual DATA_SET struct
DataSetArray[0]->num1 = 1;
DataSetArray[0]->num2 = 2;
DataSetArray[0]->num3 = 3;
DataSetArray[1] = malloc(sizeof(DATA_SET)) // Allocate mem for the actual DATA_SET struct
DataSetArray[1]->num1 = 1;
DataSetArray[1]->num2 = 2;
DataSetArray[1]->num3 = 3;
memcpy(Struct1->data_set_array, *DataSetArray, sizeof(DATA_SET*); //Copy data set array into Struct1
When I print all the data out in Struct1, i get:
Struct1->data_set_array[0].num1 = 1
Struct1->data_set_array[0].num2 = 2
Struct1->data_set_array[0].num3 = 3
Struct1->data_set_array[1].num1 = 50 //This should be 1
Struct1->data_set_array[1].num2 = 50 //This should be 2
Struct1->data_set_array[1].num3 = 65 //This should be 3
Seems to be misuse/data corruption for the 2nd element in the array.
I know there's probably different ways to do this, but I wanted to get familiar with referencing the array indices via double pointers. Am I allocating memory properly? I have a feeling the memcpy is incorrect.
Struct1->data_set_array = malloc(sizeof(DATA_SET*)
/* ^ wrong */
DataSetArray = malloc(sizeof(DATA_SET*) * 2)
/* ^ wrong */
You need to do sizeof(DATA_SET) because that is the actual size of a struct, not a pointer to a DATA_SET struct.
You're making the mistake of allocating only the size of a pointer times 2, which in many cases can be smaller than you'd hope for.
memcpy(Struct1->data_set_array, *DataSetArray, sizeof(DATA_SET*);
This is also wrong. You must copy the sizeof(DATA_SET). Keep in mind that sizeof(DATA_SET) is not equal to sizeof(DATA_SET*). A pointer's size in bytes is the same regardless of the type of pointer.
What is wrong with my dynamically allocated array of pointers to structs?
DataSetArray is an array of pointers to struct and there is not anything wrong with it.
The problem is 2 other things:
You can't use memcpy on an array of pointers to struct. The memory isn't consecutive.
Further, data_set_array is not an array of pointers to struct so you are trying to copy between incompatible types.
I guess you basically want to create a Double Dimension array. So basically you can do the code like this
struct abc {
int i;
};
struct abc **arr;
arr = (struct abc **) malloc(sizeof(struct abc) * 2);
arr[0] = (struct abc*)malloc(sizeof(struct abc));
arr[0]->i = 100;
arr[1] = (struct abc*)malloc(sizeof(struct abc));
arr[1]->i = 200;
One thing that you need to remember is that the outer array is just a pointer, so it doesn't need to be sizeof(abc) , it could very well be sizeof(int*) / sizeof (void*) etc.
so basically the above code could be written as
struct abc **arr;
arr = (struct abc **) malloc(sizeof(int *) * 2);
arr[0] = (struct abc*)malloc(sizeof(struct abc));
arr[0]->i = 100;
arr[1] = (struct abc*)malloc(sizeof(struct abc));
arr[1]->i = 200;
Hope this helps in underlying memory management you want to do with your code

What is the way of correct way in using calloc for an array?

I want to create a pointer to an array of pointers (with 10 pointers in the array), then I want to give a pointer a value.
This is what I have so far:
char **arraypointer = calloc (10, sizeof (char*));
How will I give value to this array?
I've tried:
arraypointer[0] = "string"
But I get a seg fault.
Edit:
I want to create a pointer that points to an array of pointers. Each of these pointers will have a struct property. How is it possible to access the struct property for this pointer? I have no code to post because I am still trying to figure out how it should look. The struct for this pointer will contain a string which is a char *string and an int number. I am thinking of it working like this:
arraypointer[0]->string = "this";
arraypointer[0]->number = 3;
Update:
struct mystruct* pointer = calloc(N, sizeof(struct mystruct));
pointer[0].string = "this"; //notice . instead of ->
^
pointer[0].number = 3;
^
Original answer:
char **arraypointer = calloc (10, sizeof (char*));
arraypointer[0] = "string";
These two lines are perfectly fine. You'd get a segfault(Undefined behavior, actually) if you tried to do something like
arraypointer[0][3] = 's';
later.
Also, you'd get in trouble if you did
strcpy(arraypointer[0], "string");
because you haven't actually allocated any memory that arraypointer[i] point to
You have to type :
char **arraypointer = (char**) calloc (10, sizeof (char*));
As calloc is meant for void*
//READING STRINGS USING POINTERS (CALLOC)
#include<stdio.h>
#include<stdlib.h>
int main() {
int n;
printf("Enter number of Strings : \n");
scanf("%d", &n);
char** a = (char**) calloc(n, sizeof(char*));
int i=0;
printf("Enter Strings : \n");
for (i=0;i<n;i++) {
a[i]= (char*) malloc(100*sizeof(char));
scanf("%s", a[i]);
}
i=0;
printf("**************\n");
for (i;i<n;i++) {
printf("%s\n", a[i]);
}
}

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