CODEWITHSUNDEEP
c
Ryan Tee
Ryan Tee

Reputation: 59

Some questions about memory allocation in C

So I have defined a structure struct process with name as of its attributs. I also a global variable struct process *table[3].

So inside one of my functions, I did:

struct process *new_element =  table[0];
new_element = malloc(sizeof(struct process));
new_element->name = some_name;
...

After that, when I try to access the name with table[0]->name it shows me an memory access error.

But if I use malloc on table[0] it works:

table[0] = malloc(sizeof(struct process));
struct process *new_element = table[0];
new_element->name = some_name;
...

I dont understand why the first case failed, when I point new_element towards table[0] then use malloc on it, shouldnt it allocates memory on table[0]?

Upvotes: 1

Views: 55

Answers (2)

Luis Romero
Luis Romero

Reputation: 81

When yo do this:

struct process *new_element =  table[0]; // this make variable points to memory table[0]
new_element = malloc(sizeof(struct process)); // in this point you don't creating memory for table[0] you are creating memory and assigning th

at to new_element new_element->name = some_name; // here is an new memory

After that, when yo try table[0]->name. here is not memory table[0] is undefined

table[0] = malloc(sizeof(struct process)); // Here you create a memory for table[0]
struct process *new_element = table[0]; // here you point to the memory wich table[0] points to
new_element->name = some_name; // Here is not undefined and can be accesible

Upvotes: 0

Acorn
Acorn

Reputation: 26066

struct process *new_element =  table[0];

This copies the pointer (which is a value, similar to an integer) into a new variable called new_element. When you modify it later (assigning it what malloc() returns), table[0] is not updated.

shouldnt it allocates memory on table[0]?

No, it does not. The address is saved to new_element, but table[0] is not modified. If you want to do it that way, you would need to copy the address later on:

table[0] = new_element;

This doesn`t copy the element, just the pointer (the address of where it is in memory)!

Upvotes: 3

Related Questions