Structure
struct student
{
int rollno;
char name[10];
float marks;
};
main Function
void main()
{
struct student **s;
int n;
Total size of memory to be dynamically allocated
printf("Enter total number of students:\n");
scanf("%d",&n);
printf("Size of **s: %ld\n", sizeof(**s));
Why the size of *s and s is 8 here ?
printf("Size of *s: %ld\n", sizeof(*s));
printf("Size of s: %ld\n", sizeof(s));
s = (struct student **)malloc(sizeof(struct student *)* n);
Allocating memory dynamically to array of pointers
for(int i = 0; i<n; i++)
s[i]=malloc(sizeof(struct student));
User defined data
for(int i = 0; i<n; i++)
{
printf("Enter the roll no:\n");
scanf("%d",&s[i]->rollno);
printf("Enter name:\n");
scanf("%s",s[i]->name);
printf("Enter marks:\n");
scanf("%f",&s[i]->marks);
}
printf("Size of **s:%ld\n",sizeof(**s));
for(int i = 0; i<n; i++)
printf("%d %s %f\n", s[i]->rollno, s[i]->name, s[i]->marks);
free(s);
}
Pointer types are as big as they need to be to represent an address on the host system. On x86-64 platforms, all pointer types tend to be 64 bits, or 8 8-bit bytes, wide. The expressions s and *s both have pointer type (struct student ** and struct student *, respectively), so on your platform they’re both 8 bytes wide.
The sizes of pointer types will be different on different platforms, and different pointer types on the same platform may have different sizes. The only requirements are:
char * and void * have the same size and alignment;sizeof (const int *) == sizeof (int *))struct pointer types have the same size and alignment;union pointer types have the same size and alignment.