When and why should we use the 'struct' keyword when declaring a class pointer variable in C++?
I've seen this in embedded environments so I suspect that this is some kind of hold over from C. I've seen plenty of explanations on when to use the 'struct' keyword when declaring a struct object as it relates to namespaces in C (here), but I wasn't able to find anyone talking about why one might use it when declaring a class pointer variable.
Example, in CFoo.h:
class CFoo
{
public:
int doStuff();
};
inline Foo::doStuff()
{
return 7;
}
And later in a different class:
void CBar::interesting()
{
struct CFoo *pCFoo;
// Go on to do something interesting with pCFoo...
}
There's rarely a reason to do this: it's a fallover from C and in this case the programmer is simply being sentimental - perhaps it's there as a quest for readability. That said, it can be used in place of forward declarations.
In some instances you might need to disambiguate, but that's not the case here. One example where disambiguation would be necessary is
class foo{};
int main()
{
int foo;
class foo* pf1;
struct foo* pf2;
}
Note that you can use class and struct interchangeably. You can use typename too which can be important when working with templates. The following is valid C++:
class foo{};
int main()
{
class foo* pf1;
struct foo* pf2;
typename foo* pf3;
}
There are two reasons to do this.
The first one is if we are going to introduce a new type in the scope using an elaborated name. That is in this definition
void CBar::interesting()
{
struct CFoo *pCFoo;
// Go on to do something interesting with pCFoo...
}
the new type struct CFoo is introduced in the scope provided that it is not yet declared. The pointer may point to an incomplete type because pointers themselves are complete types.
The second one is when a name of a class is hidden by a declaration of a function or a variable. In this case we again need to use an elaborated type name.
Here are some examples
#include <iostream>
void CFoo( const class CFoo * c ) { std::cout << ( const void * )c << '\n'; }
class CFoo
{
public:
int doStuff();
};
int main()
{
class CFoo c1;
return 0;
}
Or
#include <iostream>
class CFoo
{
public:
int doStuff();
};
void CFoo( void ) { std::cout << "I am hidding the class CGoo!\n"; }
int main()
{
class CFoo c1;
return 0;
}
In C, two different styles are the most common:
typedef struct { ... } s; with variables declared as s name;.
struct s { ... }; with variables declared as struct s name;
In C++ you don't need to typedef to omit the struct keyword, so the former style is far more in line with the C++ type system and classes, making it the most common style in C++.
But then there are not many cases in C++ when you actually want to use struct instead of class in the first place - structs are essentially classes with all members public by default.
The reason for this may be as simple as not having to include a header file whose contents aren't needed other than for announcing that CFoo names a type. That's often done with a forward declaration:
class CFoo;
void f(CFoo*);
but it can also be done on the fly:
void f(struct CFoo*);
I'm working on a project with a pre-made .hpp file with all the declarations and stuff.
A struct is declared in the private part of the class, along with some private members.
I need to create an array with the type of the struct in my .cpp file.
//.hpp
private:
struct foo
{
std::string a;
unsigned int b;
};
std::string* x;
unsigned int y;
//.cpp
unsigned int returny()
{
return y; // No errors
}
foo newArray[10]; // Compile time error; unknown type name
Why is it that I can return y, which is also private, but not make an array out of the struct foo?
How can I fix this? (I'm in an introductory C++ class... so hopefully there's a simple solution)
There are couple of issues.
You can't use a type that's defined in the private section of class like you are trying.
The nested type can be used by specifying the appropriate scope.
EnclosingClass::foo newArray[10];
But this will work only if foo is defined in the public section of EnclosingClass.
you should define the struct int the outside of the class like this
struct Foo
{
std::string a;
unsigned int b;
};
class A {
private:
Foo foo;
...
}
I was confused about how I would create two structs that reference each other. I couldn't find any question like it asked before.
So, I know that this will compile just fine:
struct MyStruct {
MyStruct* p;
};
But, for instance,
struct A {
B* pBstruct;
};
struct B {
A* pAstruct;
};
This won't compile.
You need a forward declaration to make the type known:
struct B;
struct A {
B* pBstruct;
};
struct B {
A* pAstruct;
};
So, I know I asked it, but I figured it out and thought that answering it might help others.
Just like making recursive functions, structures must also be prototyped.
So, consider making recursive functions:
int fnc_a();
int fnc_b();
int fnc_a()
{
fnc_b();
}
int fnc_b()
{
fnc_a();
}
The first two lines identify as the prototypes for the functions so that they can be used before their actual definition/declaration is provided.
Now consider making two recursive structs again:
struct A;
struct B;
struct A {
B* pBstruct;
};
struct B {
A* pAstruct;
};
These first two lines declare the existence of the two structs so that they can be used before they are declared. (Don't bother commenting on the fact I only need the B struct prototype - I do recognize that)
One final note, don't forget to use struct 'pointer' variables and not struct variables. This will not compile because it would create an infinitely large structure:
struct A;
struct B;
struct A {
B pBstruct;
};
struct B {
A pAstruct;
};
I have the following situation:
typedef void (*F_POINTER)(ClassName*);
class ClassName {
public:
F_POINTER f;
}
This happens because an instance of ClassName needs to pass a pointer itself to the client. However, If I write things in this order, I get complains about ClassName not declared, whatnot. However, if I switch them around, I get complains about F_POINTER not being declared when I declare the instance in the class.
So maybe I'm missing something simple here, but how would I accomplish this?
Forward declaration:
class ClassName;
typedef (*F_POINTER)(ClassName*);
class ClassName {
public:
F_POINTER f;
}
Or shorter:
typedef (*F_POINTER)(class ClassName*); // implicit declaration;
This seems to be working under GCC 4.5.2 (mingw32):
typedef void (*F_POINTER)(class ClassName);
class ClassName {};
void f (ClassName) {}
int main ()
{
F_POINTER x = f;
}
I wouldn't trust my life on whether it is legal C++ or not, not even the pointer version described by other people. Forward declarations and pointers do not solve everything, for example it is not possible to declare methods that throw pointers of incomplete types, you need full definition for that. This might be similarly illegal, just accepted by compilers.
Forward declare ClassName. The type can be incomplete to use a pointer to it.
class ClassName;
typedef (*F_POINTER)(ClassName*);
class ClassName { ... };
Alternatively, move the typedef inside the class. Also consider using boost::function/std::function instead of function pointers.
Either you can forward declare it as,
class ClassName;
typedef void (*F_POINTER)(ClassName*); // assume that return is `void`
before the F_POINTER or you can use the template trick to avoid such hassels for every class:
template<typename RETURN, typename ARGUMENT>
struct FuncPtr { typedef RETURN (*F_POINTER)(ARGUMENT); };
Usage:
class ClassName {
public:
FuncPtr<void,ClassName*>::F_POINTER f;
};
Can a struct have a constructor in C++?
I have been trying to solve this problem but I am not getting the syntax.
In C++ the only difference between a class and a struct is that members and base classes are private by default in classes, whereas they are public by default in structs.
So structs can have constructors, and the syntax is the same as for classes.
struct TestStruct {
int id;
TestStruct() : id(42)
{
}
};
All the above answers technically answer the asker's question, but just thought I'd point out a case where you might encounter problems.
If you declare your struct like this:
typedef struct{
int x;
foo(){};
} foo;
You will have problems trying to declare a constructor. This is of course because you haven't actually declared a struct named "foo", you've created an anonymous struct and assigned it the alias "foo". This also means you will not be able to use "foo" with a scoping operator in a cpp file:
foo.h:
typedef struct{
int x;
void myFunc(int y);
} foo;
foo.cpp:
//<-- This will not work because the struct "foo" was never declared.
void foo::myFunc(int y)
{
//do something...
}
To fix this, you must either do this:
struct foo{
int x;
foo(){};
};
or this:
typedef struct foo{
int x;
foo(){};
} foo;
Where the latter creates a struct called "foo" and gives it the alias "foo" so you don't have to use the struct keyword when referencing it.
As the other answers mention, a struct is basically treated as a class in C++. This allows you to have a constructor which can be used to initialize the struct with default values. Below, the constructor takes sz and b as arguments, and initializes the other variables to some default values.
struct blocknode
{
unsigned int bsize;
bool free;
unsigned char *bptr;
blocknode *next;
blocknode *prev;
blocknode(unsigned int sz, unsigned char *b, bool f = true,
blocknode *p = 0, blocknode *n = 0) :
bsize(sz), free(f), bptr(b), prev(p), next(n) {}
};
Usage:
unsigned char *bptr = new unsigned char[1024];
blocknode *fblock = new blocknode(1024, btpr);
Class, Structure and Union is described in below table in short.
Yes, but if you have your structure in a union then you cannot. It is the same as a class.
struct Example
{
unsigned int mTest;
Example()
{
}
};
Unions will not allow constructors in the structs. You can make a constructor on the union though. This question relates to non-trivial constructors in unions.
In c++ struct and c++ class have only one difference by default struct members are public and class members are private.
/*Here, C++ program constructor in struct*/
#include <iostream>
using namespace std;
struct hello
{
public: //by default also it is public
hello();
~hello();
};
hello::hello()
{
cout<<"calling constructor...!"<<endl;
}
hello::~hello()
{
cout<<"calling destructor...!"<<endl;
}
int main()
{
hello obj; //creating a hello obj, calling hello constructor and destructor
return 0;
}
Yes. A structure is just like a class, but defaults to public:, in the class definition and when inheriting:
struct Foo
{
int bar;
Foo(void) :
bar(0)
{
}
}
Considering your other question, I would suggest you read through some tutorials. They will answer your questions faster and more complete than we will.
Note that there is one interesting difference (at least with the MS C++ compiler):
If you have a plain vanilla struct like this
struct MyStruct {
int id;
double x;
double y;
} MYSTRUCT;
then somewhere else you might initialize an array of such objects like this:
MYSTRUCT _pointList[] = {
{ 1, 1.0, 1.0 },
{ 2, 1.0, 2.0 },
{ 3, 2.0, 1.0 }
};
however, as soon as you add a user-defined constructor to MyStruct such as the ones discussed above, you'd get an error like this:
'MyStruct' : Types with user defined constructors are not aggregate
<file and line> : error C2552: '_pointList' : non-aggregates cannot
be initialized with initializer list.
So that's at least one other difference between a struct and a class. This kind of initialization may not be good OO practice, but it appears all over the place in the legacy WinSDK c++ code that I support. Just so you know...
One more example but using this keyword when setting value in constructor:
#include <iostream>
using namespace std;
struct Node {
int value;
Node(int value) {
this->value = value;
}
void print()
{
cout << this->value << endl;
}
};
int main() {
Node n = Node(10);
n.print();
return 0;
}
Compiled with GCC 8.1.0.
struct HaveSome
{
int fun;
HaveSome()
{
fun = 69;
}
};
I'd rather initialize inside the constructor so I don't need to keep the order.
Syntax is as same as of class in C++. If you aware of creating constructor in c++ then it is same in struct.
struct Date
{
int day;
Date(int d)
{
day = d;
}
void printDay()
{
cout << "day " << day << endl;
}
};
Struct can have all things as class in c++. As earlier said difference is only that by default C++ member have private access but in struct it is public.But as per programming consideration Use the struct keyword for data-only structures. Use the class keyword for objects that have both data and functions.
Yes structures and classes in C++ are the same except that structures members are public by default whereas classes members are private by default. Anything you can do in a class you should be able to do in a structure.
struct Foo
{
Foo()
{
// Initialize Foo
}
};
Yes it possible to have constructor in structure here is one example:
#include<iostream.h>
struct a {
int x;
a(){x=100;}
};
int main() {
struct a a1;
getch();
}
In C++ both struct & class are equal except struct'sdefault member access specifier is public & class has private.
The reason for having struct in C++ is C++ is a superset of C and must have backward compatible with legacy C types.
For example if the language user tries to include some C header file legacy-c.h in his C++ code & it contains struct Test {int x,y};. Members of struct Test should be accessible as like C.
In C++, we can declare/define the structure just like class and have the constructors/destructors for the Structures and have variables/functions defined in it.
The only difference is the default scope of the variables/functions defined.
Other than the above difference, mostly you should be able to imitate the functionality of class using structs.