In C++, if I have
namespace myNamespace {
boost::whatever();
}
will the compiler look for whatever() in the boost namespace or in myNamespace::boost?
The question is not about whether it will look for boost::whatever, but where it will find boost itself.
If you have a nested namespace definition with the name boost, it will hide the global boost namespace at its point of declaration. The global boost namespace will be visible up until the point where myNamespace::boost is declared (§3.3.2/2).
[ Note: a name from an outer scope remains visible up to the point of declaration of the name that hides it. [ Example:
const int i = 2;
{ int i[i]; }
declares a block-scope array of two integers. — end example ] — end note ]
So if you're calling boost::whatever() before the nested name myNamespace::boost is created, it will look for the global boost namespace. You can qualify the name with ::boost if you always want it to find boost in the global namespace.
First of all I think you mean this:
namespace myNamespace {
using boost::whatever; // NOT boost::whatever();
}
Remember that C++ namespaces are mainly to avoid naming clashes not a design mechanism, which means when you are using a method of other namespaces in another one, you just call the original one and there is no copy in new namespace scope. So, when you call myNamespace::whatever(), compiler would choose boost::whatever() not myNamespace::boost::whatever() because there is no boost namespace declared inside myNamespace.
For more details on namespaces you could see:
http://en.cppreference.com/w/cpp/language/namespace
Assuming you want to call a function using some qualification, you may end up using a local namespace rather than a global namespace:
#include <iostream>
namespace A { void f() { std::cout << "::A::f()\n"; } }
namespace B {
namespace A { void f() { std::cout << "::B::A::f()\n"; } }
void g() { A::f(); }
void h() { ::A::f(); }
}
int main() {
B::g();
B::h();
}
If you want to make sure you are picking up a specific namespace you'll need to use absolute qualification as in ::A::f().
Related
I read these 2 pieces of codes and their descriptions and thought that they were clashing with each other.
Namespace definition can appear only in a global scope.
void f()
{
namespace space1{
}
}
// So this is not allowed as space1 is local to f()
Namespaces can be nested
namespace namespace1{
int i;
namespace namespace2{
int j;
}
}
In this case won't namespace2 be local to namespace1 thus resulting in an error?
Because your quote is wrong. A more correct description would be:
Namespace definitions are only allowed at namespace scope, including
the global scope.
https://en.cppreference.com/w/cpp/language/namespace
I have some C++ libraries that are written using nested namespaces. These libraries use many mathematical functions that, for readability, are better off read without explicitly specifying namespaces. Right now, a simplified version of the code would look like this.
namespace Base::Sibling1 {
float cos(float x) { return cosf(x); }
};
namespace Base::Sibling2 {
using namespace Sibling1;
float f(float x) { return cos(x); }
};
We wanted to move to use flatter namespaces mostly to make it easier to extend the library with sibling code. We had hoped that a simple change like this would work:
namespace Sibling1 {
float cos(float x) { return cosf(x); }
};
namespace Sibling2 {
using namespace Sibling1;
float f(float x) { return cos(x); }
};
Instead this fails now since Sibling2::f() calls cos(x) that is now ambiguous.
I have two questions
why is it ambiguous now and not int the first version?
is it possible to obtain the behavior we had before without listing all functions explicitly using using Sibling1::cos?
The ambiguity is due to how using directives work.
[namespace.udir] (emphasis mine)
2 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
([basic.lookup.unqual]), the names appear as if they were declared in
the nearest enclosing namespace which contains both the
using-directive and the nominated namespace. [ Note: In this context,
“contains” means “contains directly or indirectly”. — end note ]
3 A using-directive does not add any members to the declarative
region in which it appears. [ Example:
namespace A {
int i;
namespace B {
namespace C {
int i;
}
using namespace A::B::C;
void f1() {
i = 5; // OK, C::i visible in B and hides A::i
}
}
namespace D {
using namespace B;
using namespace C;
void f2() {
i = 5; // ambiguous, B::C::i or A::i?
}
}
void f3() {
i = 5; // uses A::i
}
}
void f4() {
i = 5; // error: neither i is visible
}
— end example ]
So given your structure of namespaces, according to bit I made bold in paragraph 2, when you write
using namespace Sibling1;
It kinda translates to this
namespace /* Enclosing */ {
using Sibling1::cos;
namespace Sibling2 {
float f(float x) { return cos(x); }
};
}
The namespace I marked as enclosing is either Base or the global namespace. The "kinda" bit (according to paragraph 3) is that it's not an actual declaration being added. I.e. if something named cos already exists in /* Enclosing */, it's not a re-declaration. This is by design, because using directives can potentially bring a lot of names in, and so it shouldn't cause an error when the names they bring are not actually used.
In your case however, the name brought in is used.
When /* Enclosing */ is Base, it matches only one declaration, the one in Sibling1, as-if it was declared in Base.
When /* Enclosing */ is the global namespace, it matches the actual declaration there too, presumably the one brought in by math.h (you seem to be using that). So you get an ambiguity (the name refers to two potential entities).
So on the whole, compilers that reject this code are behaving as expected. While I understand your plight, I don't think there's really a problem for you to solve here. If client code finds Base::Sibling1::cos too verbose, it itself can employ a using directive of the form using namespace Base;. Or using a namespace alias to shorten things namespace sib1 = Base::Sibling1;.
The problem cannot be reproduced as you describe it. Moreover, there is no reason that the flattening alone introduces ambiguities, if you use expose in the siblings the same names you introduced in Base. So, the root cause is probably in some parts of the code you are not showing.
The second version does not define cosf(), so you probably are using some namespace. If in that namespae there is also a cos() you create an ambiguity between the two overloads. I could reproduce this by using namespace std:
#include <iostream>
#include <cmath>
using namespace std;
namespace Sibling1 {
float cos(float x) { return cosf(x); }
}
namespace Sibling2 {
using namespace Sibling1;
float f(float x) { return cos(x); }
}
Online demo: the compilation error indicates which are the candidates behind the ambiguity. If you now comment the using namespace out, the ambiguity goes away, and the code compiles. Online proof.
Namespaces are meant to avoid naming conflicts and keep control of ambiguities:
Creating a namespace and systematically using namespace everywhere defeats the whole purpose.
On the other side, long using lists are painful to maintain, especially if you have to repeat them in several sibling namespaces. This is probably why the Base and the nesting were created in the first place.
In the first version, you do not show what's in Base. It is possible that some more tailored using clauses are used therein instead of full namespaces: if selectively injecting the really required functions in the Base namespace, they are made available within the siblings avoiding ambiguity that can be introduced by injecting the name of unnecessary functions.
Could somebody tell me what is the difference between
using namespace android;
....
and
namespace android {
....
}
I found that almost all .cpp files in Android source code use the second one.
Also, If I want to include some files that use the type of second one in my own project, do I need to use namespace android{...} too?
Because if I do not, compiler would report error when I call methods of included files. Or do I need to add any prefix before method call?
namespace android {
extern int i; // declare here but define somewhere
void foo ();
}
-- is used for scoping variables and functions inside a particular name. While using/calling those variables/functions, use scope resolution operator ::. e.g.
int main ()
{
android::foo();
}
There is no restriction for putting all namespace declarations in a single body instance. Multiple namespace android bodies spread across several files, is possible and also recommended sometimes. e.g.
// x.cpp
namespace android {
void somefunc_1 ();
}
// y.cpp
namespace android {
void somefunc_2 ();
}
Now, sometimes you may find using :: operator inconvenient if used frequently, which makes the names unnecessarily longer. At that time using namespace directive can be used.
This using directive can be used in function scope / namespace scope / global scope; But not allowed in class scope: Why "using namespace X;" is not allowed inside class/struct level?).
int main ()
{
using namespace android;
foo(); // ok
}
void bar ()
{
foo(); // error! 'foo' is not visible; must access as 'android::foo()'
}
BTW, Had using namespace android; declared globally (i.e. above main()), then foo() can be accessed without :: in Bar() also.
My answer is probably only helpful if you are more experienced with Java. I'm guessing since you are doing android stuff that this is the case.
The following means that you are declaring a class called MyClass in the namespace android. The qualified name of the class would be android::MyClass.
namespace android {
class MyClass {...};
}
It can be thought of similarly to the Java code:
package android;
public class MyClass {...}
The following means that you can use classes, functions etc. defined in the android namespace without having to use their qualified name (assuming they have been included).
using namespace android;
This
#include <path/to/MyClass.h>
using namespace android;
can be thought of similarly to the Java code:
import android.MyClass;
I'm learning C++ now. What are the complete legal entities that can be put in a namespace?
Legal entities here means valid members of a namespace
Oh, this is a real question. I'm coming from .net and I have the .net mindset.
Any code can be put inside namespace.
However main() function must be at global namespace. It cannot be put inside user-defined namespace.
namespace userns
{
int main()
{
return 0;
}
}
This program wouldn't compile link : http://www.ideone.com/k6SPc
Its because userns::main() will not be considered entry-point of the program; it became just like any other user function, not the standard main(). To compile it successfull, you've to add main() at global namespace:
namespace userns
{
int main()
{
return 0;
}
}
int main()
{
return 0;
}
This will compile link now : http://www.ideone.com/76Ynu
Anything can be put in a namespace (which is legal for C++, of course).
Actually, everything is in some namespace - the global namespace, if not specified.
Everything can be put in namespace except few "entities", which will not compile.
(1) Globally overloaded operator new and operator delete
namespace N
{
void* operator new (size_t size) // error
{ ... }
}
(2) Definition of the constructs which are declared in outer scope of the namespace; for example you have a class A declared globally then you cannot define its method inside your namespace N. In the same way, if you have method declared in a namespace N then you cannot put its definition inside namespace N::Nested (i.e. Nested is a namespace inside N).
//file
struct A {
void foo ();
static int i;
};
namespace N
{
int A::i = 0; // error
void A::foo() // error
{}
}
Demo: this is not allowed.
I remember at least these 2 restrictions from my experience. Don't know about specs.
I expected to be able to use a namespace alias in a class declaration but get a compiler syntax error.
struct MyClass
{
namespace abc = a_big_namespace;
void fn() {
abc::test();
}
};
The only way I can get it to work is to put the alias in every function.
void fn() {
namespace abc = a_big_namespace;
abc::test();
}
Additionally I would like to be able to use the alias for function parameters. I haven't found a work-around for that.
void fn(abc::testType tt) {
abc::test(tt);
}
Is there a way to do what I want?
EDIT: my solution
I found that I didn't need unnamed namespace for my particular problem and can simply do this:
namespace myspace
{
namespace abc = a_big_namespace;
struct MyClass
{
void fn(abc::testType tt) {
abc::test(tt);
}
};
}
To switch to the other library, which is what my alias namespace refers to I just change the alias. This method even allows me to have the same class in a single file twice, each time refering to a different library.
Thanks for all your help.
Namespace alias in the class definition is illegal, as specified by the language specification.
Its allowed in only in namespace scope or function scope.
You can make alias at namespace scope. But that will be make permanent alias which can be used from other files as well. But the solution is simple : you can use unnamed namespace to prevent alias (and therefore all symbols from the big namespace) from being visible from other files. This is how it can be done:
//MyFile.cpp
namespace myspace
{
namespace //this is unnamed namespace
{
namespace abc = a_big_namespace;
}
struct MyClass
{
void fn()
{
abc::test(); //don't worry, this will work!
}
};
}
//OtherFile.cpp
myspace::abc::test(); //error - that means, prevention worked.
The alias is not visible from other files. When compiling OtherFile.cpp, GCC (4.5.0) says,
'myspace::abc' has not been declared
That proves the alias abc is visible only in MyFile.cpp. Thanks to unnamed namespace.
Demo : http://www.ideone.com/2zyNI (though it doesn't demonstrate OtherFile concept. I cannot have more than one file at ideone.com)
The scope of a namespace alias is a code block.
So you can put it in any code block.
BUT, you can't put it inside a class, because that will mean it's a member of the class.
A namespace alias can't be a member.
More about namespace aliases:
What is the scope of a namespace alias in C++?
Namespaces
It works if you declare the alias outside of the struct.
You can of course also put the alias outside the class:
namespace abc = a_big_namespace;
struct MyClass {
void fn()
{ abc::test(); }
};