typedef struct {
int Key_value;
Node *link;
}Node;
Is the above declaration valid? or should I use
typedef struct _node{
int Key_value;
_node *link;
}Node;
No, it's not valid (as you would have noticed if you tried to compile it).
The typedef alias isn't introduced until the after the typedef, so you can't use it inside itself.
The second one isn't valid either (in C), _node is not a valid type name. You must use struct _node for the self-reference.
I tend to use pre-declaration and split it:
typedef struct Node Node;
struct Node {
int Key_Value;
Node *link;
};
The thing is that you can actually give the same name to both the structure and the typedef:
typedef struct Node {
int Key_value;
struct Node *link;
} Node;
Here I have added something which would have caused your code to not compile in a C compiler: The typedef isn't created until after the structure is defined. This means we must use the structure name in the link member declaration.
This can either be solved by giving the structure a name, as above, or by declaring the typedef first:
typedef struct Node *Node;
struct Node {
int Key_value;
Node *link;
};
Also note that in C++ you don't need to use the typedef keyword, or the struct keyword when declaring the link member:
// C++ version
struct Node {
int Key_value;
Node *link;
};
Structure (and class) names can be used as types in C++.
The first declaration is invalid, because Node is not known when you declare link as a structure member. The reason is that a declaration name is visible only after the declarator (simply put, that is after a comma, equal sign, or semi-colon). So, typedef being a declaration like any other, the name Node is only visible after the final semi-colon that ends the declaration statement.
Thus, you must use the second form (the first won't even compile). However, if you're on C, note that you should prepend the struct keyword to _node, like this:
typedef struct _node {
int Key_value;
struct _node *link;
} Node;
This is not necessary if you're on C++.
Both are invalid. Here's one valid way, for C and C++:
struct Node
{
int Key_value;
struct Node *link;
};
// if C, you can also do this
typedef struct Node Node;
The main point is that whatever the type of Link is, it must be something that's already been declared. The line struct X { .... declares that struct X is a type (but does not define it yet, but that's OK).
In C you should do:
typedef struct _node {
int Key_value;
struct _node *link;
} Node;
However, if you are using C++, it's simpler to omit the typedef at all:
struct Node {
int Key_Value;
Node* link;
}
Related
struct Element{
Element() {}
int data = NULL;
struct Element* right, *left;
};
or
struct Element{
Element() {}
int data = NULL;
Element* right, *left;
};
I was working with binary trees and I was looking up on an example. In the example, Element* right was struct Element* right. What are the differences between these and which one would be better for writing data structures?
I was looking up from this website:
https://www.geeksforgeeks.org/binary-tree-set-1-introduction/
In C, struct keyword must be used for declaring structure variables, but it is optional(in most cases) in C++.
Consider the following examples:
struct Foo
{
int data;
Foo* temp; // Error in C, struct must be there. Works in C++
};
int main()
{
Foo a; // Error in C, struct must be there. Works in C++
return 0;
}
Example 2
struct Foo
{
int data;
struct Foo* temp; // Works in both C and C++
};
int main()
{
struct Foo a; // Works in both C and C++
return 0;
}
In the above examples, temp is a data member that is a pointer to non-const Foo.
Additionally, i would recommend using some good C++ book to learn C++.
In C++, defining a class also defines a type with the same name so using struct Element or just Element means the same thing.
// The typedef below is not needed in C++ but in C to not have to use "struct Element":
typedef struct Element Element;
struct Element {
Element* prev;
Element* next;
};
You rarely have to use struct Element (other than in the definition) in C++.
There is however one situation where you do need it and that is when you need to disambiguate between a type and a function with the same name:
struct Element {};
void Element() {}
int main() {
Element x; // error, "struct Element" needed
}
I ran into this bit of code for a c++ linked list implementation.
struct node
{
int info;
struct node *next;
}*start;
What does this mean to have *start rather than just start??
What happens when it is later used like this? What does s mean it is not referenced anywhere else in the function?
struct node *temp, *s;
temp = new(struct node);
Fragment
struct node
{
int info;
struct node *next;
}*start;
is equivalent to
struct node
{
int info;
struct node *next;
};
struct node *start;
So in your first fragment, you define a structure named node and a variable named start of type struct node * within one statement. That's all.
Note that in C++ (unlike in C), you could also write
struct node
{
int info;
node *next;
};
node *start;
i.e. you can omit the struct-keyword when defining variables of type struct node.
I am always confused what is why we are using typedef struct node for creating node rather same thing can be easily implemented using struct node only...
That's not the way it is done.
it is usually
struct node { ... };
versus
typedef struct { ... } node;
//here node is a alias for an anonymous struct or...
typedef struct node_ { ... } node; //note the two names are different
In C++ that doesn't make any particular difference: node is in any case a type.
But it is different in C, where node is a tag for the type struct node (not just node) in the first case, and just the type node in the second case
Libraries defining types that have to work the same for both C and C++ usually adopt the second form, so that they can in any other expression to mention just node, instead of struct node, like in
node* first() { ... }
instead of
struct node* first() { ... }
I came across the code below and I'm a little confused as to its purpose.
struct bob{
int myNum;
struct bob * next;
};
static struct bob_stuff{
int theNum;
struct bob *lists;
} bob;
I know the second struct is static and being intialized as a bob struct, but why would you do that? But I'm not really sure why'd you have 2 structs.
I think that it is the names of the structures that confuse you.
It is a definition of a single linked list.
The first structure
struct bob{
int myNum;
struct bob * next;
};
defines a node of the list. That it would be more clear I will rewrite it with different names
struct Node{
int value;
struct Node * next;
};
The second structure simply defines a head of the list and the number of nodes in the list (as I thik)
static struct bob_stuff{
int theNum;
struct bob *lists;
} bob;
So it can be rewritten like
static struct SingleLinkedList{
int nodes_count;
struct Node *head;
} list;
This abstract list is used as a container for some Bob stuff.:)
That looks like "module" (or "class", if you will) state for maintaining a bunch of singly-linked lists of integers.
Of course the naming is terrible, it should be e.g.
struct list_node {
int myNum;
struct list_node *next;
};
static struct {
int theNum;
struct list_node *lists;
} listState;
Note that the name ("struct tag") bob_stuff is pointless and confusing, and should probably be removed. If you want a static, thus local, variable of struct type, chances are the tag won't be useful anyway.
If I create a structure in C++ like this:
typedef struct node {
int item;
int occurrency;
};
I know that a structure is allocated in memory using successive spaces, but what is the name of the structure (node in this example)? A simple way to give a name to the structure?
In C++ you don't have to use typedef to name a structure type:
struct node {
int item;
int occurrency;
};
is enough.
A pointer to an instance of that struct would be defined as node* mypointer;
E.g: You want to allocate a new instance with new:
node* mypointer = new node;
In C
struct node {
int item;
int occurrency;
};
is a tag, and by itself, it doesn't represent a type.
That is why you cannot do
node n;
You have to do
struct node n;
So, to give it a "type name", many C programmers use a typedef
typedef struct node {
int item;
int occurrency;
} node;
That way you can do
node n;
Instead of
struct node n;
Also, you can omit the tag and do the following
typedef struct {
int item;
int occurrency;
} node;
However, in C++ this all changes, the typedef syntax is no longer needed. In C++ classes and structs are considered to be user-defined types by default, so you can just use the following
struct node {
int item;
int occurrency;
};
And declare nodes like this
node n;
node is the name of the type. You can have multiple objects of that type:
struct node {
int item;
int occurrency;
};
node a;
node b;
In this example, both a and b have the same type (==node), which means that they have the same layout in memory. There's both an a.item and a b.item.