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
}
Related
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.
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;
}
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.
As an example, consider a simple data structure like a linked list. In C, it might look like:
struct Node
{
struct Node *next;
void *data;
};
void *getLastItem(struct Node*);
...
I'd like to have the same struct and functions, but with better type checking by declaring the type of the data field, which will always be a pointer to something. An example use:
Node<Thing*> list = getListOfThings();
Thing *t = list->data;
t = getLastItem(list);
...
But I don't want to generate an implementation for every type of pointer, as happens with a normal template. In other words, I want something more like a generic or parametric type from Java, ML, and other languages. I just tried the code below as a test. The untyped C-like part would eventually go in a implementation file, while the template and function declarations would be in the header file. I'm assuming they would be optimized away and I'd be left with machine code that is about the same as the C version, except it would be type-checked.
But I'm not great with C++... Is there a way to improve this, or use more idiomatic C++, perhaps template specialization?
#include <stdio.h>
struct NodeImpl
{
NodeImpl *next;
void *data;
};
void *getLastItemImpl(NodeImpl *list)
{
printf("getLastItem, non-template implementation.\n");
return 0; // not implemented yet
}
template <typename T>
struct Node
{
Node<T> *next;
T data;
};
template <typename T>
T getLastItem(Node<T> *list)
{
return (T)getLastItemImpl((NodeImpl*)list);
}
struct A { };
struct B { };
int main()
{
Node<A*> *as = new Node<A*>;
A *a = getLastItem(as);
Node<B*> *bs = new Node<B*>;
B *b = getLastItem(bs);
}
This is exactly what Boost.PointerContainer does, check its implementation. Basically what it does is implement the specialization for void*, and have any other implementation forward to it static_casting the parameters in and out.
struct Node
{
struct Node *next;
void *data;
};
void *getLastItem(struct Node*);
...
This is common for C, but not for C++. In C++ it usually looks like this:
template<typename T>
struct Node
{
struct Node *next;
T data;
};
T& getLastItem(const Node&);
...
Note the important difference -- the C version has another level of indirection in order to share implementations, while the C++ version need not do this. This means the C version has another n dynamic memory allocations, where n is the number of items in the list. Given that each allocation usually requires obtaining a global lock, often has at least 16 bytes of overhead per allocation, as well as all the overhead the memory manager brings to the party, the advantage of the C++ version is not insignificant, particularly when you include things like cache locality in the considerations.
Put another way, for Node<int>, the C++ version stores an int, while the C version stores an int *, along with a dynamic allocation for the int.
This of course discounting that a linked list is a horrendous data structure 90% of the time.
If you must use a linked list, and if you must use dynamic allocation for the data members, then your idea of "replace the pointers with void*s" is not unreasonable. However, if you have access to a C++11 compiler (VS2010, recent GCC versions, etc.), you should put in an assert that you depend on T being a pointer type, using std::is_pointer and static_assert, and you should use static_cast rather than C-style casts in your interface methods. The C style cast would let someone do Node<SomeTypeBiggerThanVoidPtr>, and it would compile, but explode at runtime.
As the other answers and comments said, use std::forward_list, or another existing library. If you refuse, this is more like I would do:
#include <stdio.h>
struct NodeImpl
{
NodeImpl *next;
void *data;
public:
// we have pointers, so fulfill the rule of three
NodeImpl() : next(NULL), data(NULL) {}
~NodeImpl() {}
NodeImpl& operator=(const NodeImpl& b) {next = b.next; data = b.data; return *this;}
// This function now a member. Also, I defined it.
void* getLastItem()
{
if (next)
return next->getLastItem();
return data;
}
void* getData() {return data;}
void setData(void* d) {data = d;}
};
// the template _inherits_ from the impl
template <typename T>
struct Node : public NodeImpl
{
Node<T> operator=(const Node<T>& b) {NodeImpl::operator=(b);}
// we "redefine" the members, but they're really just wrappers
T* getLastItem()
{ return static_cast<T*>(NodeImpl::getLastItem());}
T* getData() {return static_cast<T*>(NodeImpl::getData());}
void setData(T* d) {NodeImpl::setData(static_cast<void*>(d));}
//or, if you prefer directness...
operator T*() {return static_cast<T*>(NodeImpl::getData());}
Node<T> operator=(T* d) {NodeImpl::setData(static_cast<void*>(d));}
};
struct A { };
struct B { };
int main()
{
Node<A> as; //why were these heap allocated? The root can be on the stack
A *a = as.getLastItem();
Node<B> bs; //also, we want a each node to point to a B, not a B*
B *b = bs.getLastItem();
B* newB = new B;
bs = newB; //set the data member
newB = bs; //read the data member
}
http://ideone.com/xseYk
Keep in mind that this object doesn't encapsulate next or data really, so you have to manage all of that yourself.
I have a class something like that:
template <class T>
class bag
{
public:
private:
typedef struct{void* prev; struct{T item; unsigned int count;} body; void* next;}* node;
typedef struct{
node operator->() { return current; }
operator(){;} // <- i can not do that, right?
private:
node current;
} iterator;
//...
};
So, how to write a constructor for the bag::iterator?
Make some nice name for it :-)
typedef struct NoName1 {void* prev; NoName1(){}; struct NoName2{T item; unsigned int count; NoName2() {}} body; void* next;}* node;
EDIT: LOL sorry, wrote it for the wrong one, but the principle is the same :-)
There is no way to write a constructor for bag::iterator because iterator is a typedef name, which are prohibited from being used as constructor names:
14882:2003 12.1/3
a typedef-name that names a class shall not be used as the identifier in the declarator for a constructor declaration.
There is even an example in the standard, although in a different paragraph, 7.1.3/5:
typedef struct {
S(); //error: requires a return type because S is
// an ordinary member function, not a constructor
} S;
You will have to give that struct a name, if you want a user-defined constructor. The typedef struct { } name; style of programming is usually discouraged by C++ style guides anyway, in favor of struct name { };.