[#PasserBy has found that my question is a duplicate. The question can be closed, thanks.]
How can I get a named namespace with internal linkage? That is, how can I get a named namespace invisible to external source files? I want this:
static namespace N {
int foo() {return 10;}
int bar() {return 20;}
}
However, unfortunately, C++ does not recognize static namespace.
Enclose the named namespace within an unnamed namespace:
namespace {
namespace N {
int foo() {return 10;}
int bar() {return 20;}
}
}
int sum()
{
return N::foo() + N::bar();
}
This works because an unnamed namespace automatically exports its members (the sole member in this case being the namespace N) to the surrounding scope—without exposing the members to other source files.
Related
Quote from the standard:
A using-directive specifies that the names in the nominated namespace
can be used in the scope in which the using-directive appears after
the using-directive. During unqualified name lookup (3.4.1), the names
appear as if they were declared in the nearest enclosing namespace
which contains both the using-directive and the nominated namespace.
Look at this code:
namespace A {
int fn() { return 1; }
}
namespace Inner {
int fn() { return 2; }
namespace B {
using namespace A;
int z = fn();
}
}
Here, before I knew the exact rules of namespaces, I had expected that z will be initialized to 1, as I written using namespace A, so expected that A::fn() will be used. But it is not the case, z will be initialized to 2, as Inner::fn() is called because of the rule I quoted.
What is the rationale behind this behavior: "as if they were declared in the nearest enclosing namespace which contains both the using-directive and the nominated namespace"?
What would be the cons, if using namespace worked as applying using declarations for everything in that namespace?
Note: this is the related issue that motivated me to ask this question.
A desirable property of a namespace system is that of what I call incremental API compatibility. That is, if I add a symbol to a namespace, then any previously working program should keep working and mean the same thing.
Now, plain C++ with overloads is not incrementally API compatible:
int foo(long x) { return 1; }
int main()
{
foo(0);
}
Now I add the overload int foo(int x) { return 2; } and the program silently changes meaning.
Anyway, when C++ people designed the namespace system they wanted that when incrementing an external API, previously working code should not change the namespace from where the symbol is chosen. From your example, the previous working code would be something like:
namespace A {
//no fn here, yet
}
namespace Inner {
int fn() { return 2; }
namespace B {
using namespace A;
int z = fn();
}
}
And z is easily initialized to 2. Now augmenting namespace A with a symbol named fn will not change the meaning of that working code.
The opposite case does not really apply:
namespace A {
int fn() { return 1; }
}
namespace Inner {
// no fn here
namespace B {
using namespace A;
int z = fn();
}
}
Here z is initialized to 1. Of course, if I add fn to Inner it will change the meaning of the program, but Inner is not an external API: actually, when Inner was written initially, A::fn did already exist (it was being called!), so there is no excuse for being unaware of the clash.
A somewhat practical example
Imagine this C++98 program:
#include <iostream>
namespace A {
int move = 0;
void foo()
{
using namespace std;
cout << move << endl;
return 0;
}
}
int main()
{
A::foo();
return 0;
}
Now, if I compile this with C++11, everything works fine thanks to this using rule. If using namespace std worked as applying using declarations for everything in that namespace, then this program would try to print function std::move instead of A::move.
that this snippet of code actually do?
#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;
void test();
namespace {
static struct StaticStruct {
StaticStruct() {
test();
}
} TheStaticSupport;
}
int main(void) {
return 0;
}
void test() {
printf("testing function\n");
}
why does the test function actually get called? and why use the "anonymous" namespace? I found this piece of code in an open source project...
This:
static struct StaticStruct {
StaticStruct() {
test();
}
} TheStaticSupport;
Is equivalent to this:
struct StaticStruct {
StaticStruct() {
test();
}
};
static StaticStruct TheStaticSupport;
It defines a type named StaticStruct and an instance of the type named TheStaticSupport with internal linkage (though, since it is declared in an unnamed namespace, the static is redundant).
The constructor for TheStaticSupport is called before main() is entered, to construct the object. This calls the test() function.
The anonymous namespace gives the contained objects internal linkage, as their fully qualified name can never be known to anyone outside the translation unit. It's the sophisticated man's version of the old static in C.
Note that you do declare a global object of type StaticStruct, and its constructor (which runs before main() is called) calls test().
Is it legal to define the implementation of a function or class members outside the unnamed (anonymous) namespace they've been defined inside.
My compiler accepts it but I want to be sure it's legal
e.g.
////////////////
// foo.cpp
namespace {
struct X
{
void foo(int x);
};
}
// Is this legal?
void X::foo(int x)
{
}
The reason is I would like to avoid the unnecessary indentation imposed by our uncrustify formatting
It is really no different than the following, which is totally legit:
namespace ns {
struct s {
void f();
};
}
using namespace ns;
void s::f() { }
The names from the named namespace are brought into the global namespace and thus definitions can be provided for them there. The only difference with the unnamed namespace is that it has no name (really, it has some unique, unutterable name) and the using namespace is implicit.
#include <iostream>
using namespace std;
namespace
{
struct X
{
X(int);
int x_;
};
}
X::X(int x) : x_(x) {}
int main()
{
X x(5);
cout << x.x_ << "\n";
return 0;
}
Compiles (and runs) under gcc 4.6.0
Yes. It's perfectly legal. The difference an unnamed namespace makes is that, the content of the namespace is available only within the file it's declared. If it's in .h file then, it will be added to all the subsequent .cpp file and every .cpp file will have a unique copy of the contents of the namespace.
In fact it's a better way to declare a global static variable. Now see, what kind of difference it will make:
namespace {
struct X
{
void foo(int x);
};
int i; // declare this global variable
}
If you put this code inside a header file then, wherever that header file is #includeed, all those .cpp file will have a different copy of int i; inside them. Changing the value of i in one .cpp file doesn't affect the other .cpp file.
Moreover, it will not give linker error for multiple definitions, since it's in unnamed namespace.
Edit: To evaluate it more, define the namespace as following:
// x.h
namespace
{
struct X
{
void foo(int x)
{
static int c; // static inside the function
cout<<"a = "<<(c++)<<endl;
}
};
}
Now #include this header file in 2 different .cpp files. In both of them try to call foo() with object of X. It will print:
a = 0
a = 0
Which means that X::foo() in both the .cpp files are different. If you give the namespace a name and repeat the same thing, it will output
a = 0
a = 1
Thus unnamed namespace creates a different copy for each translation unit.
It's legal in your scenario, but in one scenario, it will cause compiler ambiguous error.
In case have another class in X.h:
// X.h
struct X
{
void foo(int x) { }
};
And in foo.cpp, need to use the X defined in X.h
////////////////
// foo.cpp
#include "X.h"
namespace {
struct X
{
void foo(int x);
};
// use the X declared in anonymous namespace
void test()
{
X x;
x.foo(3);
}
}
// reference to 'X' is ambiguous
void X::foo(int x)
{
}
void otherFunction()
{
// Use struct X in X.h
::X x;
x.foo(3);
}
if leave the implementation code out of the anonymous namespace, compiler complains ambiguous. If move the implementation code inside anonymous namespace, it works fine.
I am starting to learn C++ and Qt, but sometimes the simplest code that I paste from a book results in errors.
I'm using g++4.4.2 on Ubuntu 10.04 with QtCreator IDE. Is there a difference between the g++ compiler syntax and other compilers? For example when I try to access static members something always goes wrong.
#include <iostream>
using namespace std;
class A
{
public:
static int x;
static int getX() {return x;}
};
int main()
{
int A::x = 100; // error: invalid use of qualified-name 'A::x'
cout<<A::getX(); // error: : undefined reference to 'A::x'
return 0;
}
I think it's exactly the same as declared here and here (isn't it?). So what's wrong with the above code?
You've declared the static members fine, but not defined them anywhere.
Basically what you've said "there exists some static member", but never set aside some memory for it, you need:
int A::x = 100;
Somewhere outside the class and not inside main.
Section [9.4.2]
Static Data Members
The declaration of a static data member in its class definition is not a definition and may be of an incomplete type other than cv-qualified void. The definition for a static data member shall appear in a namespace scope enclosing the member’s class definition. In the definition at namespace scope, the name of the static data member shall be qualified by its class name using the :: operator
You need to define the static member variable of the class outside the class as static member variables require declaration as well as definition.
#include <iostream>
using namespace std;
class A
{
public:
static int x;
static int getX() {return x;}
};
int A::x; // STATIC MEMBER VARIABLE x DEFINITION
int main()
{
A::x = 100; // REMOVE int FROM HERE
cout<<A::getX();
return 0;
}
Try:
#include <iostream>
using namespace std;
class A
{
public:
// This declares it.
static int x;
static int getX(){return x;}
};
// Now you need an create the object so
// This must be done in once source file (at file scope level)
int A::x = 100;
int main()
{
A::x = 200;
// Notice no 'int' keyword before A::x on this line. You can modify A::x
cout<<A::getX(); // Should work
return 0;
}
The definition of static member variables must live at file scope, i.e. outside all functions, etc.
Try this example:
#include<iostream>
using namespace std;
class check
{
static int a;
public:
void change();
} ;
int check::a=10;
void check::change()
{
a++;
cout<<a<<"\n";
}
int main()
{
int i,j;
check c;
check b;
c.change();
b.change();
return 0;
}
Now you have worked out how to use static class members I will advise you that you should generally use them only in the following circumstances:
For use in templates. So in your example you could have GetX() in different classes and in a template somewhere you would use
template< typename T >
int func()
{
return T::GetX();
}
although obviously more elaborate. But here your static function being in a class serves a purpose.
Where the function needs access to the class, i.e. to private members. You could make it a friend but you may as well make it static. Often the case in callbacks.
The rest of the time you can probably use compilation-unit level functions and variables which has the advantage of taking your members out of the header (particularly if they are private). The less implementation detail you give the better.
You can use the inline keyword since c++ 17 in front of static members to avoid a definition outside of class scope. Your code should now look like this:
#include <iostream>
using namespace std;
class A
{
public:
inline static int x;
static int getX() {return x;}
};
int main()
{
A::x = 100; //Works now
cout<<A::getX()<<'\n';
return 0;
}
Case 1: static variable
As we all know, defining a static variable inside a class which will throw compilation error. E.g. below
class Stats
{
public:
static int AtkStats[3];
*static int a =20;* // Error: defining a value for static variable
};
int Stats::AtkStats[3] = {10, 0, 0};
Output:
error: ISO C++ forbids in-class initialization of non-const static member 'Stats::a'
Case 2: const static variable
For const static variable, we can define a value either inside a class or Outside class.
class Stats
{
public:
static const int AtkStats[3];
static const int a =20; // Success: defining a value for a const static
};
const int Stats::AtkStats[3] = {10, 0, 0};
const int Stats::a = 20; // we can define outside also
Output:
Compilation success.
Consider the following snippet:
void Foo() // 1
{
}
namespace
{
void Foo() // 2
{
}
}
int main()
{
Foo(); // Ambiguous.
::Foo(); // Calls the Foo in the global namespace (Foo #1).
// I'm trying to call the `Foo` that's defined in the anonymous namespace (Foo #2).
}
How can I refer to something inside an anonymous namespace in this case?
You can't. The standard contains the following section (§7.3.1.1, C++03):
An unnamed-namespace-definition behaves as if it were replaced by
namespace unique { /* empty body */ }
using namespace unique;
namespace unique { namespace-body }
where all occurrences of unique in a
translation unit are replaced by the
same identifier and this identifier
differs from all other identifiers in the entire program.
Thus you have no way to refer to that unique name.
You could however technically use something like the following instead:
int i;
namespace helper {
namespace {
int i;
int j;
}
}
using namespace helper;
void f() {
j++; // works
i++; // still ambigous
::i++; // access to global namespace
helper::i++; // access to unnamed namespace
}
While Georg gives standard-complient, correct, right, and respectable answer, I'd like to offer my hacky one - use another namespace within the anonymous namespace:
#include <iostream>
using namespace std;
namespace
{
namespace inner
{
int cout = 42;
}
}
int main()
{
cout << inner::cout << endl;
return 0;
}
The only solution I can think of that doesn't modify the existing namespace arrangement is to delegate main to a function in the anonymous namespace. (main itself is required to be a global function (§3.6.1/1), so it cannot be in an anonymous namespace.)
void Foo() // 1
{
}
namespace
{
void Foo() // 2
{
}
}
namespace { // re-open same anonymous namespace
int do_main()
{
Foo(); // Calls local, anonymous namespace (Foo #2).
::Foo(); // Calls the Foo in the global namespace (Foo #1).
return 0; // return not optional
}
}
int main() {
return do_main();
}
The only real way is to put the code you want to access that namespace within the namespace itself. There's no way to resolve to the unnamed namespace otherwise, since it has no identifier you can give it to solve the ambiguous resolution problem.
If your code is inside the namespace{} block itself, the local name gets priority over the global one, so a Foo() will call the Foo() within your namespace, and a ::Foo() will call the namespace at global scope.
Just rename the local namespace function.