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Why use an asterisk after a type?
#include<stdio.h>
struct Node{
int data;
struct Node* next;
};
Here I used an asterisk after Node. What is it used for? What if I don't put any asterisk after Node (both Node's are structures)?
Is *ptr and ptr* same?
2 answers
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Here I used asterisk after Node. Actually, why asterisk used for? What if I don't put any asterisk after Node (Both Node are structure).
It's a pointer.
A pointer, like its name implies, points to an actual object in memory. (More technically, it holds an address to the memory location).
In your Node example, it is necessary because using a plain object is impossible.
Think of a struct like a box. The box has compartments, which hold the items in it. By necessity, the compartment must be smaller than the box itself.
Now, imagine trying to stuff a Node into another Node. This is like trying to stuff a box into a compartment of a box of the same size -- it just doesn't fit.
However, if we use a pointer, we get around that problem. We slip in a piece of paper into the box that says "the next box is over there", so we don't try to stuff boxes in boxes.
Is
*ptrandptr*same?
It's not.
The basic syntax to declare a pointer is type * var_name. The most likely confusion you have is which one is the type, and which one is the variable name.
In struct Node * next, struct Node is the type, and next is the variable name.
In int *ptr;, int is the type, and ptr is the variable name.
Note that spaces are completely irrelevant. int* ptr is the same as int *ptr is the same as int * ptr.
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The * for variables and not mathematical operators are the pointers.
Assigning a pointer goes this way:
char *text; // string
Here, we assign a pointer named text and its type is a char, but alternatively, this is a string.
As for your question, int *ptr is basically the same as int* ptr, only being a pointer of an int, and nothing else. Your given struct Node* next is the same as struct Node *next, and I don't think ptr* even exists, unless it's made as some union or struct. struct *Node wouldn't work likely.
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