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*);
Suppose I have nested classes as follows defined in a header file:
class ClassA
{
private:
class ClassB
{
private:
int member_b;
public:
void function_name();
};
};
In order to give a definition to the function "function_name()" in an external .cpp file, I have to access it like this:
void ClassA::ClassB::function_name()
{
std::cout << member_b;
return;
}
For the sake of this example, please do not ask why I'm using nested classes; I have a reason for doing so in my actual project. However, my question is this; is it possible to somehow shorten the ClassA::ClassB::function_name() in the implementation file to something like short::function_name() while still keeping the classes nested? I don't think that typedefs or new namespace definitions can help me here, but maybe I'm wrong.
Qualified type names allow you to define a typedef to represent a qualified class name. You can then use the typedef with the :: (scope resolution) operator to refer to a nested class or class member, as shown in the following example:
class outside
{
public:
class nested
{
public:
static int x;
static int y;
int f();
int g();
};
};
int outside::nested::x = 5;
int outside::nested::f() { return 0; };
typedef outside::nested outnest; // define a typedef
int outnest::y = 10; // use typedef with ::
int outnest::g() { return 0; };
However, using a typedef to represent a nested class name hides information and may make the code harder to understand.
Source : https://www.ibm.com/support/knowledgecenter/en/SSPSQF_9.0.0/com.ibm.xlcpp111.aix.doc/language_ref/cplr061.html
Have you tried using aliases?
// C++11
using fmtfl = std::ios_base::fmtflags;
// C++03 equivalent:
// typedef std::ios_base::fmtflags fmtfl;
fmtfl fl_orig = std::cout.flags();
fmtfl fl_hex = (fl_orig & ~std::cout.basefield) | std::cout.showbase | std::cout.hex;
// ...
std::cout.flags(fl_hex);
Code source: https://msdn.microsoft.com/en-us/library/dn467695.aspx
I'm not very familiar with structs.
But I created this:
in test.h:
class test {
public:
struct Astruct
{
int age;
int weight;
};
struct Astruct& MethodOne();
};
and in test.cpp:
#include "test.h"
test::test() {}
struct Astruct& test::MethodOne() {
Astruct testStruct;
// code to fill in testStruct
return testStruct;
}
The goal of this code above is that I'm able to return a struct with MethodOne.
But on the line of
struct Astruct & test::MethodOne(){
it says :error: declaration is incompatible with what's in the header file.
I don't understand this. If I'd replace the struct with an int return type there wouldn't be an error?
What's wrong here?
And a second error I'm getting when I return testStruct: Error: a reference of type "Astruct&" (not const qualified) cannot be initialized with a value of type "test::Astruct"
Your code has several errors (missing ; etc.). A class is not different from a struct in C++. The only difference is the default access specifier which is private in a class and public in a struct (members and inheritance).
Structs are typically used to indicate that it actually is just a data structure with no logic or methods. Imho they are nice to encapsulate input and output of methods. If you want to have it for a member function it could look like this:
class Foo{
public:
struct BarIn {}; // need ; here
struct BarOut {}; // and here
BarOut bar(const BarIn& b){return BarOut();}
};
int main() {
Foo::BarOut result = Foo().bar(Foo::BarIn());
}
Note that I had to write Foo:BarOut and Foo::BarIn, because the structs are declared inside the class. Also there is no need to write struct when you declare a variable of a type that is declared as struct (because there really is no difference between an instance of a class or struct).
Last but not least you should never return a reference (or pointer) to a local variable:
struct Astruct & test::MethodOne(){
Astruct testStruct;
return testStruct; // testStruct is destroyed here
} // and the returned ref is invalid
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.
I would like to be able to achieve something like this:
class Zot
{
namespace A
{
static int x;
static int y;
}
}
I am working with a legacy system that uses code generation heavily off a DB schema, and certain fields are exposed as methods/variables in the class definition. I need to add a few extra static variables to these classes and would like to guarantee no clashes with the existing names.
The best I have come up with is to use another struct to wrap the statics as if it were a namespace:
class Zot
{
struct A
{
static int x;
static int y;
}
}
Is there a better way?
Update:
An extra requirement is to be able to access these from a template elsewhere
e.g.
template<class T>
class B
{
void foo() { return T::A::x; }
};
So putting them in a separate class won't work
Really the inner struct is your best bet. Another possibility would be to use a typedef to bring in a class of statics. This works well for code generation in that you can separate the extras from the generated code:
In the generated file that doesn't care at all what's in Zot_statics:
class Zot_statics;
class Zot
{
public:
typedef Zot_statics A;
int x; // This is ok
};
In a hand-maintained header for when you need to access x and y:
class Zot_statics
{
public:
static int x;
static int y;
};
In a hand-maintained cpp file:
int Zot_statics::x;
int Zot_statics::y;
And your template should work just fine with Zot::X referring to the instance variable X on Zot, and Zot::A::x refering to the static variable.