Error in initializing members of a structure - c

Question

Following is a program for initializing members of a structure in c
struct stack
{
int *a;
int top;
int size;
}s;
void init()
{
const int size =10; /*s.size=10;*/ /*const int s.size=10*/
s.a=(int*)malloc(size*sizeof(int));
s.top=0;
}
int main()
{
init();
printf("Initialization done !\n");
return 0;
}
Q1 : In init method instead of const int size=10 when I wrote s.size=10 , I got an error "size was not declared in scope " , but I have already declared size in stack struct .I was able to initialize top in the same manner Then why the error ?
Q2 : In init method I am getting correct output with const int size=10 . I am confused , in this statement how are we able to access size member of the struct stack without using a struct variable , shouldn't it be const int s.size=10?

Answer 1

s.size=10 has no problem. The problem is when you allocate memory for s.a, there's no variable named size, you should change it to:
s.a = malloc(s.size * sizeof(int));
You seem to be confused about the variable size and the member size in the structure struct stack s, they are not related except having the same name.

Answer 2

Yes as size is a structure variable, you have to access with the structure variable and initialize it.
if you initialize size =10 it will take as a new variable. because init function will be stored in a separate stack and the scope of variable size will be inside the init function only.
Then while allocating the memory you should allocate for the s.size variable.
s.a = malloc(s.size * sizeof(int));

Answer 3

I think, your confusion is because you've used the same variable name size twice,
as a structure member variable
as a local variable in void init().
Please be noted, these two are seperate variables.
The size member variable in struct stack is a member of the structure. You need to access a member variable by either . or -> operator [Yes, even if the structure is global].
OTOH, the int size in void init() is a normal variable of type int.
Without having a variable of type struct stack, there does not exist a size, which belongs to struct stack. Likewise, there's no way you can access size, which is a member variable in struct stack directly [without using a structure variable of type struct stack] anywhere.
To summarize
Answer 1:
The error is not for replacing const int size=10 with s.size = 10. It's rather from the next line,
s.a= malloc(size*sizeof(int));
^
|
Where, no size variable is present when const int size=10 is removed.
Answer 2
const int size=10 declares and defines a new variable called size. It's not the same as the s.size [member of struct stack]. That's why using
s.a= malloc(size*sizeof(int));
^
|
is valid, as a variable named size is in scope.

Answer 4

Note: the following method of defining/declaring a struct is depreciated
struct stack
{
int *a;
int top;
int size;
}s;
The preferred method is:
// declare a struct type, named stack
struct stack
{
int *a;
int top;
int size;
};
struct stack s; // declare an instance of the 'struct stack' type
// certain parameters to the compile command can force
// the requirement that all functions (other than main)
// have a prototype so:
void init ( void );
void init()
{
s.size =10;
// get memory allocation for 10 integers
if( NULL == (s.a=(int*)malloc(s.size*sizeof(int)) ) )
{ // then, malloc failed
perror( "malloc failed" );
exit( EXIT_FAILURE );
}
s.top=0;
} // end function: init