In C++ can one rename/alias a non-namespace, non-class name? - c++

I have
namespace src {
struct src_bar;
void src_baz();
template<class T> class src_qux;
}
which I'd like to reference as
namespace dst {
struct dst_bar;
void dst_baz();
template<class T> class dst_qux;
}
meaning that I'd like to "rename" or "alias" or "relabel" names from src.
For dst_bar one can of course use namespace dst { typedef src_bar dst_bar; }. Is there some (non-macro) equivalent allowing me to rename src::src_baz as dst::dst_baz and src::src_qux as dst::dst_qux?
Unless I'm mistaken, neither a using statement nor a namespace alias can accomplish the other two. Writing templated forwarding functions for dst_baz() is a possibility but requires knowledge of the arity of src_baz. Template typedefs could have dst_qux<T>::type be src_qux<T> but the indirection adds verbosity.
FWIW, my use case is taking C names like somepackage_someportion_somefuncA and providing a namespaced-friendly version somepackage::someportion::somefuncA so that other folks can employ using somepackage::someportion for brevity.

For functions you will have to manually forward the requests. For non-template types you can just typedef. For template types, y can use the new using feature of c++11 To create a typedef-style alias for a template, if your compiler supports it, or else you are basically out of luck.

No, you cannot "rename" something so that it no longer uses its original name.
However, you can get the same effect via trickery.
First, import the header for the defines you want to corral inside a namespace block:
namespace foo {
#include "src_stuff.h"
}
Now you have foo::src::src_bar and friends.
Then, use typedef liberally:
namespace bar {
typedef foo::src::src_bar dst_bar;
}
Voila, you've got a bar::dst_bar which is the same as src::src_bar would have been. If you don't care about the definitions remaining available under the old names in addition, skip the first step and just make typedefs for convenience. That's what they're for, after all.

If usage of macros is an acceptable option for you then here is one solution:
// declare this macro in some common file
#define CREATE(SCOPE) \
namespace SCOPE { \
struct SCOPE## _bar; \
void SCOPE## _baz(); \
template<class T> class SCOPE## _qux; \
}
Usage:
CREATE(src);
CREATE(dst);

Related

C++ macro value that behaves differently in function vs class/namespace scope

I am trying to define a macro that behaves differently when used in a function context vs a class body or namespace. The purpose of this is to selectively include BOOST_LOG_NAMED_SCOPE (or achieve equivalent behavior) in the LOG_CONTEXT macro referenced here.
I have tried two approaches:
Using BOOST_PP_IF along with a preprocessor function to test whether e.g. __func__ is non-empty.
The output of g++ -E on a basic source file still contains the literal text __func__ in both global and function scope so that probably rules out any approach targeting the preprocessing phase.
Using something like sizeof(__func__) to select a specialized template that implements behavior similar to BOOST_LOG_NAMED_SCOPE. I can re-use the boost::log::attributes::named_scope::sentry object from boost log, but I'm stuck trying to figure out how to instantiate it conditionally in a function context, or at least in a way that works everywhere. The following construction seems to work fine in class definitions and functions, but fails with a "multiple definition" error when linking together multiple translation units that include a header with a LOG_CONTEXT at global or namespace scope:
#include <boost/log/attributes/named_scope.hpp>
#include <type_traits>
namespace logging {
namespace attrs = boost::log::attributes;
namespace detail {
// Default no-op on construction when not in a function.
template<typename ScopeTraits, bool InFunction>
class named_scope_helper
{};
// Specialization when in a function.
template<typename ScopeTraits>
class named_scope_helper<ScopeTraits, true> :
public attrs::named_scope::sentry
{
public:
named_scope_helper() BOOST_NOEXCEPT
: sentry(
ScopeTraits::scope_name(),
ScopeTraits::filename(),
ScopeTraits::lineno() )
{}
};
template<size_t N>
class not_1 : public std::true_type
{};
template<>
class not_1<1> : public std::false_type
{};
#define LOGGING_LOG_IN_FUNCTION \
::logging::detail::not_1<sizeof(__func__)>::value
} // namespace detail
} // namespace logging
// scope_name_t/filename_t are required since attrs::named_scope::sentry
// requires string literals.
#define LOG_CONTEXT( name_ ) \
struct __logging_log_scope_traits__ \
{ \
using scope_name_t = const char (&)[sizeof(name_)]; \
static scope_name_t scope_name() \
{ return name_; } \
using filename_t = const char (&)[sizeof(__FILE__)]; \
static filename_t filename() \
{ return __FILE__; } \
static size_t lineno() \
{ return __LINE__; } \
}; \
::logging::detail::named_scope_helper< \
__logging_log_scope_traits__, LOGGING_LOG_IN_FUNCTION> \
BOOST_LOG_UNIQUE_IDENTIFIER_NAME(_log_named_scope_sentry_);
My normal workaround for this would be to use static in the declaration of the _log_named_scope_sentry_, but that defeats the purpose.
I could add another macro that is strictly for non-execution contexts, but wanted to investigate this approach first since it would be an interesting trick to have. How can I proceed on one of the two approaches I've started above, or is there another option I haven't considered?
A general solution would be preferred but I'm only really concerned with GCC and Clang.

How to wrap a C struct in a C++ class and keep the same name?

Let's say the Acme company releases a useful library with an extremely ugly C API. I'd like to wrap the structs and related functions in C++ classes. It seems like I can't use the same names for the wrapper classes, because the original library is not inside a namespace.
Something like this is not possible, right?
namespace AcmesUglyStuff {
#include <acme_stuff.h> // declares a struct Thing
}
class Thing {
public:
...
private:
AcmesUglyStuff::Thing thing;
};
Linking will be a problem.
The only way I can think to wrap the library, and not pollute my namespace with the C library names, is a hack like this, reserving space in the class:
// In mything.h
namespace wrapper {
class Thing {
public:
...
private:
char impl[SIZE_OF_THING_IN_C_LIB];
};
}
// In thing.cc
#include <acme_stuff.h>
wrapper::Thing::Thing() {
c_lib_function((::Thing*)impl); // Thing here referring to the one in the C lib
}
Is that the only way? I'd like to avoid putting prefixes on all my class names, like XYThing, etc.
Seems like you're making this harder than it needs to be.
#include "acme_stuff.h" // puts all of its names in global namespace
namespace acme {
class Thing {
public:
// whatever
private:
::Thing thing;
};
}
Now just use acme::Thing rather than Thing.
If it's really important to you to not have the C names in the global namespace, then you need a level of indirection:
namespace acme {
class Thing {
public:
Thing();
~Thing();
// whatever
private:
void *acme_thing;
};
}
In your implementation file, #include "acme_stuff.h", in your constructor create a new ::Thing object and store its address in acme_thing, in your destructor delete it, and in your member functions cast acme_thing to type ::Thing*.
It's not a good idea to try to name something the exact same thing as something else. (I mean equal fully-qualified names, including all namespaces.) If some library has already grabbed the obvious best name in the global namespace, you'll need to pick a different name.
You could put your class Thing in a namespace as Pete Becker suggests, and then use ::Thing to access Acme's Thing. That would be fine if you're prepared to always access your class through it's fully namespace-qualified name (e.g. My::Thing). It's tempting to try using My::Thing; or using namespace My;, but that won't work, because any translation unit that includes the definition of your class (e.g. via a header file you create) must necessarily pull Acme's Thing into the global namespace first (otherwise an "Undefined symbol" compilation error would occur when parsing the definition of My::Thing).
Is it really a C API? Try to extern "C" {} to whole included header to solve the linking problem.
namespace AcmesUglyStuff {
extern "C" {
#include <acme_stuff.h>
}
}

How can I find out, if type is created inside namespace?

I have created Tag system, which allows tag entity in ma engine. It works based on c++ types, so every tag is a type (I have my reasons).
Now, when I want to create new tag, I have to call struct tagname{};.
If I want to use tag in other files, I use forward declaration struct name;
To make it more user friendly I created macros
#define CREATE_TAG(name) struct name{};
#define USE_TAG(name) struct name;
But when user call one macro inside namespace and other outside namespace, I have a problem, because that are 2 different types.
I want to ensure, that user call macros outside any namespace. Since our project is in one namespace namespace root, I can use it for chcecking.
This is what I came up with so far:
namespace root
{
const unsigned TAGS_CAN_NOT_BE_DEFINED_INSIDE_ANY_NAMESPACE = 1;
};
#define CREATE_TAG(name) namespace root{ struct name{}; static_assert(root::TAGS_CAN_NOT_BE_DEFINED_INSIDE_ANY_NAMESPACE,""); }
#define USE_TAG(name) namespace root{ struct name; static_assert(root::TAGS_CAN_NOT_BE_DEFINED_INSIDE_ANY_NAMESPACE,""); }
Now when macro is used outside root namespace, everything is fine. When called inside root namespace or other nested namespaces, it gives the error TAGS_CAN_BE... not a member.
But I do not really like it. I don't like the whole macro thing, but that is not the question. Question is: Is there a better way to find out, if macro is called in namespace?
Since my macros passing in Type, I can use some typetraits, but I did not find any suitable one.
You could try something like this:
#include <type_traits>
#define DECLARE_ME(name) struct tag_##name { static_assert(!std::is_same<tag_##name, ::tag_##name>::value, "Not in global namespace"); }
You'll probably get a compiler-error that's not even the static assertion in case of a misuse, but it should catch the problem. You could probably make it more user-friendly with a more elaborate trait class.
So it looks that
namespace root
{
const unsigned TAGS_CAN_NOT_BE_DEFINED_INSIDE_ANY_NAMESPACE = 1;
};
#define CREATE_TAG(name) namespace root{ struct name{}; static_assert(root::TAGS_CAN_NOT_BE_DEFINED_INSIDE_ANY_NAMESPACE,""); }
#define USE_TAG(name) namespace root{ struct name; static_assert(root::TAGS_CAN_NOT_BE_DEFINED_INSIDE_ANY_NAMESPACE,""); }
is the answer...
I will wait a few days before I accept it. I am still open to your answers

Creating a C++ namespace in header and source (cpp)

Is there any difference between wrapping both header and cpp file contents in a namespace or wrapping just the header contents and then doing using namespace in the cpp file?
By difference I mean any sort performance penalty or slightly different semantics that can cause problems or anything I need to be aware of.
Example:
// header
namespace X
{
class Foo
{
public:
void TheFunc();
};
}
// cpp
namespace X
{
void Foo::TheFunc()
{
return;
}
}
VS
// header
namespace X
{
class Foo
{
public:
void TheFunc();
};
}
// cpp
using namespace X;
{
void Foo::TheFunc()
{
return;
}
}
If there is no difference what is the preferred form and why?
The difference in "namespace X" to "using namespace X" is in the first one any new declarations will be under the name space while in the second one it won't.
In your example there are no new declaration - so no difference hence no preferred way.
Namespace is just a way to mangle function signature so that they will not conflict. Some prefer the first way and other prefer the second version. Both versions do not have any effect on compile time performance. Note that namespaces are just a compile time entity.
The only problem that arises with using namespace is when we have same nested namespace names (i.e) X::X::Foo. Doing that creates more confusion with or without using keyword.
There's no performance penalties, since the resulting could would be the same, but putting your Foo into namespace implicitly introduces ambiguity in case you have Foos in different namespaces. You can get your code fubar, indeed. I'd recommend avoiding using using for this purpose.
And you have a stray { after using namespace ;-)
If you're attempting to use variables from one to the other, then I'd recommend externalizing them, then initializing them in the source file like so:
// [.hh]
namespace example
{
extern int a, b, c;
}
// [.cc]
// Include your header, then init the vars:
namespace example
{
int a, b, c;
}
// Then in the function below, you can init them as what you want:
void reference
{
example::a = 0;
}
If the second one compiles as well, there should be no differences. Namespaces are processed in compile-time and should not affect the runtime actions.
But for design issues, second is horrible. Even if it compiles (not sure), it makes no sense at all.
The Foo::TheFunc() is not in the correct namespacein the VS-case. Use 'void X::Foo::TheFunc() {}' to implement the function in the correct namespace (X).
In case if you do wrap only the .h content you have to write using namespace ... in cpp file otherwise you every time working on the valid namespace. Normally you wrap both .cpp and .h files otherwise you are in risk to use objects from another namespace which may generate a lot of problems.
I think right thing to do here is to use namespace for scoping.
namespace catagory
{
enum status
{
none,
active,
paused
}
};
void func()
{
catagory::status status;
status = category::active;
}
Or you can do the following:
// asdf.h
namespace X
{
class Foo
{
public:
void TheFunc();
};
}
Then
// asdf.cpp
#include "asdf.h"
void X::Foo::TheFunc()
{
return;
}

C++ compiler unable to find function (namespace related)

I'm working in Visual Studio 2008 on a C++ programming assignment. We were supplied with files that define the following namespace hierarchy (the names are just for the sake of this post, I know "namespace XYZ-NAMESPACE" is redundant):
(MAIN-NAMESPACE){
a bunch of functions/classes I need to implement...
(EXCEPTIONS-NAMESPACE){
a bunch of exceptions
}
(POINTER-COLLECTIONS-NAMESPACE){
Set and LinkedList classes, plus iterators
}
}
The MAIN-NAMESPACE contents are split between a bunch of files, and for some reason which I don't understand the operator<< for both Set and LinkedList is entirely outside of the MAIN-NAMESPACE (but within Set and LinkedList's header file).
Here's the Set version:
template<typename T>
std::ostream& operator<<(std::ostream& os,
const MAIN-NAMESPACE::POINTER-COLLECTIONS-NAMESPACE::Set<T>& set)
Now here's the problem: I have the following data structure:
Set A
Set B
Set C
double num
It's defined to be in a class within MAIN-NAMESPACE. When I create an instance of the class, and try to print one of the sets, it tells me that:
error C2679: binary '<<' : no operator found which takes a right-hand operand of type 'const MAIN-NAMESPACE::POINTER-COLLECTIONS-NAMESPACE::Set' (or there is no acceptable conversion)
However, if I just write a main() function, and create Set A, fill it up, and use the operator- it works.
Any idea what is the problem? (note: I tried any combination of using and include I could think of).
Strange - even though putting free functions associated with a type to a different namespace is a bad practice, the global namespace declarations are always visible.
The only thing I can think of is that declaration with the same name in MAIN-NAMESPACE would shadow the one in the global namespace - isn't there an operator<<, possibly for totally unrelated type, in MAIN-NAMESPACE? If so, you should fix that by using ::operator<< declaration in MAIN-NAMESPACE. Example:
namespace A
{
namespace B
{
class C{};
}
}
void f(A::B::C*);
namespace A
{
void f(int*); // try commenting
using ::f; // these two lines
void g()
{
B::C* c;
f(c);
}
}
OK I figured this out.
jpalecek's intuition about there existing another operator<< in the namespace was correct (apparently I forgot to comment it out).
The lookup rules for namespaces first start the search in the function call's namespace and search up the enclosing namespaces, right up to the global namespace (then it does the Argument dependent lookup if no match is found). However, if along the way it finds some match for operator<<, it stops the search, regardless of the fact that the types used in those functions may be incompatible, as was the case here.
The solution is either to include it into the MAIN-NAMESPACE (which I'm not allowed to), or import it from the global namespace with "using ::operator<<".
Try calling the function explicitly?
::operator<<( cout, myObj );
As SoaBox pointed out, try calling it explicitly.
For your information, if you wish to call a global function which has been hidden in the current namespace precede the function with :: to bypass the local function and call the global function.
Try calling the function explicitly?
::operator<<( cout, myObj );
Yes, that does work!
it will try to find the f function in
the current namespace (at the place of
call) or in the enclosing namespaces
of c1 and c2 types (namespace1,
namespace2::namespace3), but it will
not try other namespaces in the
search.
So let's see if I got this right: the reason invoking the operator<< from a main() function worked is because I was in the global namespace (as was operator<<).
The reason it failed when invoking from the class I implemented is because the class was in a not global namespace and there were no variables in it that pointed the compiler towards the global namespace.
OK people asked for a specific examples, so here's the relevant part of the code.
//Disclamer: in the slim case someone from my uni sees this, encounters it in the submission file, and decides I copied it or something, my student number is 311670137
This is the header file Set.h:
namespace MTM {//This is the MAIN-NAMESPACE
namespace PointerCollections {
(ITERATORS AND PREDICATE CLASSES)
template<typename T>
class Set {
public:
/////////////////////////////////
// Definitions
/////////////////////////////////
private:
/////////////////////////////////
// Definitions
/////////////////////////////////
};
///////////////////////////////////////////////////////////////////////////////
// The implementation part.
///////////////////////////////////////////////////////////////////////////////
}
}
// operator<< - the same a Set::print(std::ostream& os,
// const BinaryPredicate<T>& predicate)
// function called with the 'predicate' parameter omitted
template<typename T>
std::ostream& operator<<(std::ostream& os,
const MTM::PointerCollections::Set<T>& set){
set.print(os);
return os;
}
This is what I defined in a different file:
namespace MTM {
using std::ostream;
class Schedule {
public:
///////////////////
//Definitions, including:
///////////////////
void registerStation(string stationName);
void reportRegisteredStations(std::ostream& outputStream) const;
private: //My database
//All the classes Set recieves are defined elsewhere
Set<RegisteredStation> places;
Set<BusLine> busses;
Set<TrainLine> trains;
double tarifForBuses;
double tarifForTrains;
};
}
And here's from the main:
Schedule s();
s.registerStation("1");
s.reportRegisteredStations(cout);//This invokes the error. Definition follows:
reportRegisteredStations is defined as:
void Schedule::reportRegisteredStations(std::ostream& outputStream) const{
outputStream<<places;
}
This works for me
#include <iostream>
#include <string>
using std::string;
namespace MTM {//This is the MAIN-NAMESPACE
namespace PointerCollections {
template<typename T>
class Set {
};
}
}
template<typename T>
std::ostream& operator<<(std::ostream& os,
const MTM::PointerCollections::Set<T>& set){
return os;
}
namespace MTM {
using std::ostream;
using PointerCollections::Set;
class Schedule {
public:
///////////////////
//Definitions, including:
///////////////////
void registerStation(string stationName);
void reportRegisteredStations(std::ostream& outputStream) const;
private: //My database
//All the classes Set recieves are defined elsewhere
Set<int> places;
Set<int> busses;
Set<int> trains;
double tarifForBuses;
double tarifForTrains;
};
void Schedule::reportRegisteredStations(std::ostream& outputStream) const{
outputStream<<places;
}
}
int main()
{
MTM::Schedule s;
s.reportRegisteredStations(std::cout);
}
CORRECTION: The text below is based on experience with the g++ family of compilers. After the comment to the answer I have reread the standard (which states that ADL will be used over regular name lookup, and regular name lookup should find the operator<<). I have also tried with comeau compiler (the most standard compliant compiler I know of) and the symbol is found. It seems as a problem with g++ (tried versions 3.3, 4.1, 4.3).
Original answer:
Search for Koening lookup (technically ADL: Argument dependent lookup).
The short answer is that if you have the following class:
namespace test {
class A {};
}
the stream insertion operator should be defined as:
namespace test {
std::ostream& operator<<( std::ostream&, A const & );
}
Functions or operators should be defined in the same namespace as one of the arguments that it takes. (*)
When the compiler finds a function call such as:
namespace test2 {
void g() {
namespace1::class1 c1;
namespace2::namespace3::class2 c2;
f( c1, c2 );
}
}
it will try to find the f function in the current namespace (at the place of call) or in the enclosing namespaces of c1 and c2 types (namespace1, namespace2::namespace3), but it will not try other namespaces in the search.
(*) In this case, you are pretty much limited to the test namespace, as you are not allowed to add a function to the std namespace (only template specializations).
End of original post.
Even if as commented before this may just be a problem with the compiler, it is common usage and recommended to define all free functions that operate on a user defined type in the same namespace as the type itself.