//findreferencesize.C
#include<iostream>
using namespace std;
using namespace std;
struct student
{
static int i;
static int j;
}s;
int *x = new int(11);
int* &j = x;
cout << "x: " << x << endl;
cout << "&x: " << &x << endl;
cout << "j: " << j << endl;
cout << "&j: " << &j << endl;
delete(x);
x = NULL;
cout << "x: " << x << endl;
cout << "&x: " << &x << endl;
cout << "j: " << j << endl;
cout << "&j: " << &j << endl;
cout << "size " << sizeof(&j) << endl;
cout << "size " << sizeof(x) << endl;
cout << "size " << sizeof(s) << endl;
cout << "size " << sizeof(s.i) << endl;
static int q;
cout << "size " << sizeof(q) << endl;
return 0;
}
result:
x: 0x7f9359c000e0
&x: 0x7fff52cbcae0
j: 0x7f9359c000e0
&j: 0x7fff52cbcae0
x: 0
&x: 0x7fff52cbcae0
j: 0
&j: 0x7fff52cbcae0
size 8
size 8
size 1
size 4
size 4
//findsize.C
#include<iostream>
using namespace std;
struct bstree{
static int info;
bstree *left;
bstree *right;
}obj;
int main()
{
cout << "sizeof int is :" << sizeof(int) << endl;
cout << "sizeof struct member pointer: " << sizeof(obj.left) << endl;
cout << sizeof( bstree) << endl;
return 0;
}
result:
sizeof int is :4
sizeof struct member pointer: 8
16
//nonstatic.C
#include<iostream>
using namespace std;
struct bstree{
int info;
bstree *left;
bstree *right;
}obj;
int main()
{
cout << "sizeof int is :" << sizeof(int) << endl;
cout << "sizeof struct member pointer: " << sizeof(obj.left) << endl;
cout << sizeof( bstree) << endl;
cout << "static in size: ";
static int x;
cout << sizeof(x) << endl;
return 0;
}
result:
sizeof int is :4
sizeof struct member pointer: 8
24
static in size: 4
NOTE:C Program does allow storage type in structure, Above programs are in C++.
below is the code written in C.
//structsize.c
#include <stdio.h>
#include <stdlib.h>
struct node {
static int val;
};
int main()
{
struct node obj;
printf("%d", sizeof(obj));
return 0;
}
{
cout << "sizeof int is :" << sizeof(int) << endl;
cout << "sizeof struct member pointer: " << sizeof(obj.left) << endl;
cout << sizeof( bstree) << endl;
cout << "static in size: ";
static int x;
cout << sizeof(x) << endl;
return 0;
}
result:
sizeof int is :4
sizeof struct member pointer: 8
24
static in size: 4
NOTE:C Program does allow storage type in structure, Above programs are in C++.
below is the code written in C.
//structsize.c
#include <stdio.h>
#include <stdlib.h>
struct node {
static int val;
};
int main()
{
struct node obj;
printf("%d", sizeof(obj));
return 0;
}
result:
type name does not allow storage class to be specified.
//notallowed_initialization.c
#include<stdio.h>
int main()
{
struct node
{
char name[] = "C skills";
int rank = 200;
};
struct node *ptr;
printf("%d ", ptr->rank);
printf("%s", ptr->name);
getchar();
return 0;
}
result:
Compiliation error
Reason:
When we declare a structure or union, we actually declare a new data type suitable for our purpose. So we cannot initialize values as it is not a variable declaration but a data type declaration.
//staticinstruct.c
#include<stdio.h>
struct node
{
int m;
static int n;
};
int main()
{
printf("%d", sizeof(struct node));
return 0;
}
return 0;
}
result:
Compiliation error
Reason:
In C, struct and union types cannot have static members. In C++, struct types are allowed to have static members, but union cannot have static members in C++ also.
//sizeofstrucunion.c
#include<stdio.h>
int main()
{
struct {
short s[5];
union {
float y;
long z;
}u;
} t;
printf("%d\n", sizeof(t.u));
printf("%d\n", sizeof(t));
printf("%d", sizeof(long));
return 0;
}
result:
8
24
8
Reason: long is 8 bytes. short is 2 bytes. Hence short s[5] is 5 *2 which is 10, but here size is allocated based on alignment or padding for short s[5] in terms of long z which is 8bytes .
i.e multiples of 8. When we declare a union, memory allocated for union is equal to memory required for largest member of it, and all members share this same memory space. Hence for short s[5] with padding the size allocated 16 bytes, thus totaling to 16 + 8 =24.
//structasmember.c
#include<stdio.h>
int main()
{
struct node
{
int x;
struct node next;
};
struct node temp;
temp.x = 10;
temp.next = temp;
printf("%d", temp.next.x);
return 0;
}
result:structasmember.c|48|error: field has incomplete type 'struct node'|
Compiliation error
A structure cannot contain a member of its own type because if this is allowed then it becomes impossible for compiler to know size of such struct. Although a pointer of same type can be a member because pointers of all types are of same size and compiler can calculate size of struct
