Is Warning 1573 ("Name of generic function template declared in namespace associated with type") really relevant when the namespace is an
anonymous namespace? Most of the helper functions I have for tests go in unnamed namespace and it breaks the above rule.
Example:
namespace
{
template <typename T>
T template_func(T arg)
{
return arg;
}
class foo {};
}
int main()
{
return template_func(0);
}
How do I get around in the above, to satisfy the rule?
As state in their example, you might use extra namespace, something like:
namespace
{
template< class T >
T template_func(T arg) { return arg; }
namespace X
{
class foo{};
}
using X::foo;
}
int main()
{
return template_func(0);
}
I'd guess that it isn't aimed at polluting global namespace but to avoid class picking up weird matches from generic templates belonging to the same namespace.
It goes along the cpp core guideline T.47: Avoid highly visible unconstrained templates with common names
Specifically:
Reason :
An unconstrained template argument is a perfect match for anything so such a template can be preferred over more specific types that require minor conversions. This is particularly annoying/dangerous when ADL is used. Common names make this problem more likely.
Note:
If an unconstrained template is defined in the same namespace as a type, that unconstrained template can be found by ADL (as happened in the example). That is, it is highly visible.
As for how to bypass it, Jarod42 was 1st with an example in his answer.
I have a template function:
namespace Example
{
template<class T>
T Foo() { return T(0); };
}
I would like to use a using statement or similar so that I do not have to prefix the function name with it's namespace when calling it i.e.
template<class T> using Foo = Example::Foo<T>;
However this does not work.
I do not want to use the following approach as it would include everything form the namespace Example:
using namespace Example;
Is there a nice C++ 11 way to create a shortened alias to a function in a namespace?
As for any symbol, you can do using Example::Foo;. This can be used either in namespace scope, or in function scope (this is actually present in C++98, it is not new in C++11). The approach you were trying to do works only for types, while Foo is a function.
I am writing a C++ class which represents an arithmetic type (a c++ wrapper around mpfr), and I'd like to support some functions found in <cmath> (I'll take std::sqrt as an example).
So I have the following class:
namespace ns
{
class MyClass
{
/* ... */
public:
friend MyClass sqrt(const MyClass& mc);
};
}
And I can use it this way:
MyClass c;
/* ... */
MyClass d = ns::sqrt(c);
MyClass e = sqrt(c); // Apparently I don't have to specify ns::
But I cannot use it this way:
MyClass f = std::sqrt(c);
Compiler (g++ (Debian 4.7.2-5)) error is: "no matching function for call to sqrt(ns::MyClass&)".
This is normal, but it's a problem to me. I need this to be valid, because MyClass is supposed to be used into existing template functions (that I'm not supposed to modify). For example:
template <typename T>
void func(T a)
{
/* ... */
T c = std::sqrt(a);
/* ... */
}
int main()
{
func<float>(3);
func<MyClass>(MyClass(3));
/* ... */
}
The following piece of code actually resolve my problem:
namespace std
{
using ns::sqrt;
}
But adding things into the std namespace seems very unnatural to me. I am afraid to run into unexpected troubles later, doing this.
Is it safe ? If not, why ?
Is there a better alternative ?
Adding stuff to namespace std is prohibited by the standard. The correct way to tackle this problem is normally seen with swap (that is available in the std namespace, but that can be specialized by user-defined types to work more efficiently): when a template function needs to use e.g. sqrt, it will do
using std::sqrt;
a=sqrt(b);
This way, for "regular" types it will use std::sqrt ("taken in" by the using statement), while for your type your overload will prevail due to Koenig lookup (which, incidentally, is the reason of the behavior you observed at // Apparently I don't have to specify ns::).
It’s not safe, because it’s not legal (§17.6.4.2.1):
The behavior of a C++ program is undefined if it adds declarations or definitions to namespace std or to a namespace within namespace std unless otherwise specified. A program may add a template specialization for any standard library template to namespace std only if the declaration depends on a user-defined type and the specialization meets the standard library requirements for the original template and is not explicitly prohibited.
So you may add specialisations for your own types. You may not add overloads (or indeed anything else).
Your current code is the correct way of doing this.
The alternative, within generic functions, is to do this:
template <typename T>
void func(T a)
{
using std::sqrt;
/* ... */
T c = sqrt(a);
/* ... */
}
And defining additional things into the std-namespace is not safe in general (and not strictly legal).
Not so good an idea to put it in std namespace.
Since you have your own namespace, you could import sqrt into your namespace and add specialized sqrt functions:
namespace ns {
using std::sqrt;
MyClass sqrt(const MyClass &)
}
ns::sqrt(...);
A quick one for the gurus: C++11 allows unnamed namespaces to be declared inline. This seems redundant to me; things declared in an unnamed namespace are already used as if they were declared in the enclosing namespace.
So my question is this: what does it mean to say
inline namespace /*anonymous*/ {
// stuff
}
and how is it different from the traditional
namespace /*anonymous*/ {
// stuff
}
that we know and love from C++98? Can anyone give an example of different behaviour when inline is used?
EDIT: Just to clarify, since this question has been marked as a duplicate: I'm not asking about named inline namespaces in general. I understand the use-case there, and I think they're great. I'm specifically asking what it means to declare an unnamed namespace as inline. Since unnamed namespaces are necessarily always local to a TU, the symbol versioning rational doesn't seem to apply, so I'm curious about what adding inline actually does.
As an aside, the standard [7.3.1.1], regarding unnamed namespaces, says:
inline appears if and only if it appears in the unnamed-namespace-definition
but this seems like a tautology to my non-language lawyer eyes -- "it appears in the definition iff it appears in the definition"! For bonus points, can anyone explain what this bit of standardese is actually saying?
EDIT: Cubbi claimed the bonus point in the comments:
the standard is saying that unnamed-namespace-definition behaves as if it were replaced by X where inline appears in X iff it appears in the unnamed-namespace-definition
I don't know whether it's the done thing to answer your own question on SO, but after some playing around my curiosity has been satisfied, so I may as well share it.
The definition of an inline namespace includes not only the hoisting of names into the enclosing namespace (which happens anyway for unnamed namespaces), but also allows templates defined within an inline namespace to be specialised outside it. It turns out that this applies to unnamed namespaces too:
inline // comment this out to change behaviour
namespace {
template <typename T> struct A {};
}
template <> struct A<int> {};
Without the inline, g++ complains about trying to specialise a template from a different namespace (though Clang does not). With inline, it compiles just fine. With both compilers, anything defined within the specialisation is still marked as having internal linkage (according to nm), as if it were within the unnamed namespace, but I guess this is to be expected. I can't really think of any reason why this would be useful, but there we go.
An arguably more useful effect comes from the change regarding argument-dependent lookup for inline namespaces, which also affects unnamed inline namespaces. Consider the following case:
namespace NS {
// Pretend this is defined in some header file
template <typename T>
void func(const T&) {}
// Some type definition private to this TU
inline namespace {
struct A {};
}
} // end namespace NS
int main()
{
NS::A a;
func(a);
}
Without inline, ADL fails and we have to explicitly write NS::func(a). Of course, if we defined the unnamed namespace at the toplevel (as one normally would), then we wouldn't get ADL whether it was inline or not, but still...
Here is one use that I have found:
namespace widgets { inline namespace {
void foo();
} } // namespaces
void widgets::foo()
{
}
In this example, foo has internal linkage and we can define the function later on by using the namespace::function syntax to ensure that the function's signature is correct. If you were to not use the widgets namespace then the void foo() definition would define a totally different function. You also don't need to re-open the namespace saving you a level of indentation.
If there is another function called foo in the widgets namespace already then this will give you an ambiguity instead rather than a nasty ODR violation.
C++11 allows inline namespaces, all members of which are also automatically in the enclosing namespace. I cannot think of any useful application of this -- can somebody please give a brief, succinct example of a situation where an inline namespace is needed and where it is the most idiomatic solution?
(Also, it is not clear to me what happens when a namespace is declared inline in one but not all declarations, which may live in different files. Isn't this begging for trouble?)
Inline namespaces are a library versioning feature akin to symbol versioning, but implemented purely at the C++11 level (ie. cross-platform) instead of being a feature of a specific binary executable format (ie. platform-specific).
It is a mechanism by which a library author can make a nested namespace look and act as if all its declarations were in the surrounding namespace (inline namespaces can be nested, so "more-nested" names percolate up all the way to the first non-inline namespace and look and act as if their declarations were in any of the namespaces in between, too).
As an example, consider the STL implementation of vector. If we had inline namespaces from the beginning of C++, then in C++98 the header <vector> might have looked like this:
namespace std {
#if __cplusplus < 1997L // pre-standard C++
inline
#endif
namespace pre_cxx_1997 {
template <class T> __vector_impl; // implementation class
template <class T> // e.g. w/o allocator argument
class vector : __vector_impl<T> { // private inheritance
// ...
};
}
#if __cplusplus >= 1997L // C++98/03 or later
// (ifdef'ed out b/c it probably uses new language
// features that a pre-C++98 compiler would choke on)
# if __cplusplus == 1997L // C++98/03
inline
# endif
namespace cxx_1997 {
// std::vector now has an allocator argument
template <class T, class Alloc=std::allocator<T> >
class vector : pre_cxx_1997::__vector_impl<T> { // the old impl is still good
// ...
};
// and vector<bool> is special:
template <class Alloc=std::allocator<bool> >
class vector<bool> {
// ...
};
};
#endif // C++98/03 or later
} // namespace std
Depending on the value of __cplusplus, either one or the other vector implementation is chosen. If your codebase was written in pre-C++98 times, and you find that the C++98 version of vector is causing trouble for you when you upgrade your compiler, "all" you have to do is to find the references to std::vector in your codebase and replace them by std::pre_cxx_1997::vector.
Come the next standard, and the STL vendor just repeats the procedure again, introducing a new namespace for std::vector with emplace_back support (which requires C++11) and inlining that one iff __cplusplus == 201103L.
OK, so why do I need a new language feature for this? I can already do the following to have the same effect, no?
namespace std {
namespace pre_cxx_1997 {
// ...
}
#if __cplusplus < 1997L // pre-standard C++
using namespace pre_cxx_1997;
#endif
#if __cplusplus >= 1997L // C++98/03 or later
// (ifdef'ed out b/c it probably uses new language
// features that a pre-C++98 compiler would choke on)
namespace cxx_1997 {
// ...
};
# if __cplusplus == 1997L // C++98/03
using namespace cxx_1997;
# endif
#endif // C++98/03 or later
} // namespace std
Depending on the value of __cplusplus, I get either one or the other of the implementations.
And you'd be almost correct.
Consider the following valid C++98 user code (it was permitted to fully specialize templates that live in namespace std in C++98 already):
// I don't trust my STL vendor to do this optimisation, so force these
// specializations myself:
namespace std {
template <>
class vector<MyType> : my_special_vector<MyType> {
// ...
};
template <>
class vector<MyOtherType> : my_special_vector<MyOtherType> {
// ...
};
// ...etc...
} // namespace std
This is perfectly valid code where the user supplies its own implementation of a vector for a set of type where she apparently knows a more efficient implementation than the one found in (her copy of) the STL.
But: When specializing a template, you need to do so in the namespace it was declared in. The Standard says that vector is declared in namespace std, so that's where the user rightfully expects to specialize the type.
This code works with a non-versioned namespace std, or with the C++11 inline namespace feature, but not with the versioning trick that used using namespace <nested>, because that exposes the implementation detail that the true namespace in which vector was defined was not std directly.
There are other holes by which you could detect the nested namespace (see comments below), but inline namespaces plug them all. And that's all there is to it. Immensely useful for the future, but AFAIK the Standard doesn't prescribe inline namespace names for its own standard library (I'd love to be proven wrong on this, though), so it can only be used for third-party libraries, not the standard itself (unless the compiler vendors agree on a naming scheme).
http://www.stroustrup.com/C++11FAQ.html#inline-namespace (a document written by and maintained by Bjarne Stroustrup, who you'd think should be aware of most motivations for most C++11 features.)
According to that, it is to allow versioning for backward-compatibility. You define multiple inner namespaces, and make the most recent one inline. Or anyway, the default one for people who don't care about versioning. I suppose the most recent one could be a future or cutting-edge version which is not yet default.
The example given is:
// file V99.h:
inline namespace V99 {
void f(int); // does something better than the V98 version
void f(double); // new feature
// ...
}
// file V98.h:
namespace V98 {
void f(int); // does something
// ...
}
// file Mine.h:
namespace Mine {
#include "V99.h"
#include "V98.h"
}
#include "Mine.h"
using namespace Mine;
// ...
V98::f(1); // old version
V99::f(1); // new version
f(1); // default version
I don't immediately see why you don't put using namespace V99; inside namespace Mine, but I don't have to entirely understand the use-case in order to take Bjarne's word for it on the committee's motivation.
In addition to all the other answers.
Inline namespace can be used to encode ABI information or Version of the functions in the symbols. It is due to this reason they are used to provide backward ABI compatibility. Inline namespaces let you inject information into the mangled name (ABI) without altering the API because they affect linker symbol name only.
Consider this example:
Suppose you write a function Foo that takes a reference to an object say bar and returns nothing.
Say in main.cpp
struct bar;
void Foo(bar& ref);
If you check your symbol name for this file after compiling it into an object.
$ nm main.o
T__ Z1fooRK6bar
The linker symbol name may vary but it will surely encode the name of function and argument types somewhere.
Now, it could be that bar is defined as:
struct bar{
int x;
#ifndef NDEBUG
int y;
#endif
};
Depending upon Build type, bar can refer to two different types/layouts with same linker symbols.
To prevent such behavior we wrap our struct bar into an inline namespace, where depending upon the Build type the linker symbol of bar will be different.
So, we could write:
#ifndef NDEBUG
inline namespace rel {
#else
inline namespace dbg {
#endif
struct bar{
int x;
#ifndef NDEBUG
int y;
#endif
};
}
Now if you look at the object file of each object you build one using release and other with debug flag. You will find that linker symbols include inline namespace name as well. In this case
$ nm rel.o
T__ ZROKfoo9relEbar
$ nm dbg.o
T__ ZROKfoo9dbgEbar
Linker Symbol names may be different.
Notice presence of rel and dbg in the symbol names.
Now, if you try to link debug with release mode or vise-versa you will get a linker error as contrary to runtime error.
So to sum up the main points, using namespace v99 and inline namespace were not the same, the former was a workaround to version libraries before a dedicated keyword (inline) was introduced in C++11 which fixed the problems of using using, whilst providing the same versioning functionality. Using using namespace used to cause problems with ADL (although ADL now appears to follow using directives), and out-of-line specialisation of a library class / function etc. by the user wouldn't work if done outside of the true namespace (whose name the user wouldn't and shouldn't know, i.e. the user would have to use B::abi_v2:: rather than just B:: for the specialisation to resolve).
//library code
namespace B { //library name the user knows
namespace A { //ABI version the user doesn't know about
template<class T> class myclass{int a;};
}
using namespace A; //pre inline-namespace versioning trick
}
// user code
namespace B { //user thinks the library uses this namespace
template<> class myclass<int> {};
}
This will show a static analysis warning first declaration of class template specialization of 'myclass' outside namespace 'A' is a C++11 extension [-Wc++11-extensions]. But if you make namespace A inline, then the compiler correctly resolves the specialisation. Although, with the C++11 extensions, the issue goes away.
Out-of-line definitions don't resolve when using using; they have to be declared in a nested/non-nested extension namespace block (which means the user needs to know the ABI version again, if for whatever reason they were permitted to provide their own implementation of a function).
#include <iostream>
namespace A {
namespace B{
int a;
int func(int a);
template<class T> class myclass{int a;};
class C;
extern int d;
}
using namespace B;
}
int A::d = 3; //No member named 'd' in namespace A
class A::C {int a;}; //no class named 'C' in namespace 'A'
template<> class A::myclass<int> {}; // works; specialisation is not an out-of-line definition of a declaration
int A::func(int a){return a;}; //out-of-line definition of 'func' does not match any declaration in namespace 'A'
namespace A { int func(int a){return a;};} //works
int main() {
A::a =1; // works; not an out-of-line definition
}
The issue goes away when making B inline.
The other functionality inline namespaces have is allowing the library writer to provide a transparent update to the library 1) without forcing the user to refactor code with the new namespace name and 2) preventing lack of verbosity and 3) providing abstraction of API-irrelevant details, whilst 4) giving the same beneficial linker diagnostics and behaviour that using a non-inline namespace would provide. Let's say you're using a library:
namespace library {
inline namespace abi_v1 {
class foo {
}
}
}
It allows the user to call library::foo without needing to know or include the ABI version in the documentation, which looks cleaner. Using library::abiverison129389123::foo would look dirty.
When an update is made to foo, i.e. adding a new member to the class, it will not affect existing programs at the API level because they wont already be using the member AND the change in the inline namespace name will not change anything at the API level because library::foo will still work.
namespace library {
inline namespace abi_v2 {
class foo {
//new member
}
}
}
However, for programs that link with it, because the inline namespace name is mangled into symbol names like a regular namespace, the change will not be transparent to the linker. Therefore, if the application is not recompiled but is linked with a new version of the library, it will present a symbol abi_v1 not being found error, rather than it actually linking and then causing a mysterious logic error at runtime due to ABI incompatibility. Adding a new member will cause ABI compatibility due to the change in type definition, even if it doesn't affect the program at compile time (API level).
In this scenario:
namespace library {
namespace abi_v1 {
class foo {
}
}
inline namespace abi_v2 {
class foo {
//new member
}
}
}
Like using 2 non-inline namespaces, it allows for a new version of the library to be linked without needing to recompile the application, because abi_v1 will be mangled in one of the global symbols and it will use the correct (old) type definition. Recompiling the application would however cause the references to resolve to library::abi_v2.
Using using namespace is less functional than using inline (in that out of line definitions don't resolve) but provides the same 4 advantages as above. But the real question is, why continue to use a workaround when there is now a dedicated keyword to do it. It's better practice, less verbose (have to change 1 line of code instead of 2) and makes the intention clear.
I actually discovered another use for inline namespaces.
With Qt, you get some extra, nice features using Q_ENUM_NS, which in turn requires that the enclosing namespace has a meta-object, which is declared with Q_NAMESPACE. However, in order for Q_ENUM_NS to work, there has to be a corresponding Q_NAMESPACE in the same file⁽¹⁾. And there can only be one, or you get duplicate definition errors. This, effectively, means that all of your enumerations have to be in the same header. Yuck.
Or... you can use inline namespaces. Hiding enumerations in an inline namespace causes the meta-objects to have different mangled names, while looking to users like the additional namespace doesn't exist⁽²⁾.
So, they're useful for splitting stuff into multiple sub-namespaces that all look like one namespace, if you need to do that for some reason. Of course, this is similar to writing using namespace inner in the outer namespace, but without the DRY violation of writing the name of the inner namespace twice.
It's actually worse than that; it has to be in the same set of braces.
Unless you try to access the meta-object without fully qualifying it, but the meta-object is hardly ever used directly.
Inline namespaces can also be used to provide fine(r)-grained access to features/names within namespaces.
This is used in std::literals. The literals namespaces in std are all inline namespaces, so that:
if you use using namespace std; somewhere, you also get access to all user defined literals in the std.
but if you just need a set of udl in you local code, you can also do using namespace std::literals::string_literals; and you will just get the udl symbols defined in that namespace.
This seems a useful technique for symbols that you would want to access unqualified (udl, operators, etc.) where you can just bundle them up in an inline namespace so that you can do a specific using on just that (sub-) namespace instead of the namespace of the whole library.