Thread: struct question

Code:

typedef struct
{
int a =1;
}
test_struct;


test_struct *test_obj = NULL;

int main ()
{

}

how do i change *test_obj to a value and not null in main i am confused

"test_obj" is a pointer that is not pointing to anything.

If you want to use the pointer to create a new instance of the struct, you need to allocate memory for it.

Code:

typedef struct{
int a;
}test_struct;

int main(){
test_struct.a = *value;
return 0;

is this what do you need?

Originally Posted by Giwrgows x

Code:

typedef struct{
int a;
}test_struct;

int main(){
test_struct.a = *value;
return 0;

is this what do you need?

no its not i need test_obj to equal to something but not null

If you want to point to a structure that already exists, just assign its address to the pointer:

Code:

test_struct test;
test_struct *ptr = &test;

If you want to create a new instance of the struct, you need to allocate memory for it:

Code:

test_struct *test = malloc(sizeof(*test));  // be sure to free your memory when you're done with it

Code:

typedef struct{
int a;
}test_struct, *test_pnt;
 
 
test_struct mystruct = {1};

 
int main ()
{
 test_pnt myptr = &mystruct;    
 myptr->a = 2;     // change it to two            
}

Typedef allow the shortcut of assigning pointer declaration

Typedef allow the shortcut of assigning pointer declaration[/QUOTE]

Typing test_struct mystruct has the same number of keystrokes as struct test mystruct

Also, please don't hide a pointer using typedef for absolutely no reason at all; it's confusing!