The compiler says it can't find the reference for the function when I do this:
// link.h
template <class T>
T *Link(T *&, T *(*)())
// link.cpp
template <class T>
T c:Link(T *&ChildNodeReference, T *(*ObjectCreator)()){
}
If I implement inside the class on the header it goes smoothly.
Please, I will work on the header until someone enlightens me about this.
There are somethings in C++ that are weirdly annoying. I know, there is a reason for this and etc. Even so, can't the compilers help you out about it -_-"
Templates are essentially semi-type-safe macros, hence the limitation.
Normal (non-template) functions can be compiled to native code residing object/library files, and then referenced with only a prototype available in the header, solely because there's only a single version of such a function.
With templates, C++ compiler has to compile each instantiation of the function separately. Obviously, it cannot do it "in advance", because the set of types for which you can instantiate the function is effectively unbounded (you can always define a new type in your code before calling the function). In fact, two instantiations of the same function template can be completely different. Consider this extreme case:
struct t1 {
template <int>
struct a {};
};
struct t2 {
enum { a = 123 };
};
enum { b = 456, c = 789 };
template <class T>
void foo() {
T::a<b>c;
}
Now if we call foo<t1>(), the statement inside it is a local variable declaration, because t1::a is a class template:
T::a<b> c;
But if we call foo<t2>(), the statement inside is an expression, because t2::a is an integral constant:
(T::a < b) > c;
This is just to show that compiler cannot meaningfully "compile" a template; it really has to mostly preserve the tokens.
Now, all that said, ISO C++ does actually provide the ability to separate declaration and definition of templates, so that they can be treated as normal functions, with declarations in .h files, and definitions in .cpp files. This is called "export templates", because you have to precede both declaration and definiton with keyword export:
// link.h
export template <class T>
T *Link(T *&, T *(*)());
// link.cpp
export template <class T>
T *Link(T *&ChildNodeReference, T *(*ObjectCreator)()) {
}
This is, however, a controversial feature of the Standard because of very high burden on implementation, and most popular implementations refuse to implement it; notably, g++, MSVC, and C++Builder do not implement it. The only compiler I know of that supports it is Comeau C++.
Programming non-template code or non-inlined functions in headers is a Bad Thing™. The reason you have cpp files is to prevent redefinition of the same function code more than once, amongst other things.
The difference with templates is that the compiler practically doesn't touch them until your code instantiates a specialisation of that template which is why they need to have their source inside the header.
When the compiler finds an instantiation of a template specialisation (say List<int>), it goes back to the included template code and compiles it with that specialisation, which is why you don't have issues with redefinition of function code.
What you don't seem to understand is that this doesn't apply to non-templated code. All non-template code is compiled as normal and thus CPP files are needed to only define the code once then link it all together.
If you define functions inside a header, your linker will not link the compiled translation units because they have defined the same function more than once.
Templated implementations (not only definitions) have to be available at compile time.
So, the full template code is normally put in the header file.
Template code must be in the header. Sorry, I completely missed that! (and I thought I'd been coding C++ for years :P)
You forgot the * for the return type. So implementation is not matching definition. Add it and it should work:
T *c:Link(T *&ChildNodeReference, T *(*ObjectCreator)())
{
}
Implementation must be too in the header file under the class definition in order to be available at compile time.
Related
I try to make a generic function in C++, that creating an array of a type which depends on usage (when calling it), like array of int or char.
This is the code in my header file:
template<class T>
T** makeArray(...);
This is in the .cpp file:
template<class T>
T** ClassA::makeArray(...){
//...
}
And this is how I call it in the main file:
char** charArr = cg.makeArray<char>(...);
//...
int** intArr = cg.makeArray<int>(...);
But when compiling, I get these errors:
undefined reference to `char** ClassA::makeArray<char>( ...
undefined reference to `int** ClassA::makeArray<int>( ...
What do the errors mean and how do I fix it?
You have to include the entire body of a C++ template function in the header.
This is because of the way that C++ templates work: template parameter substitution happens only when you actually use the template, and the entire template definition has to be visible at that time. Another way of looking at it is to say that you cannot compile and link to a templateFunction<T>—you can only compile a template function once it's been instantiated as a templateFunction<char> or whatever. For this reason, you cannot put template function definitions in .cpp files, but rather have to put them in .h files.
The compiler needs to have access to the entire template definition (not just the signature) in order to generate code for each instantiation of the template, so you need to move the definitions of the functions to your header.
This is called the inclusion model. Also read this FAQ
In general, you cannot have template definitions in a source file; they should be in the header file so that the compiler can see them.
This comes up over and over again here, so I can't be bothered to give the full explanation again; I recommend you read this from the C++ FAQ: Why can't I separate the definition of my templates class from its declaration and put it inside a .cpp file?.
[Incidentally, why are you returning a pointer-to-pointer? A pointer should be enough...]
If your code is using the template on a diferent cpp file, the template should go on the header file.
I'm using Code::Blocks to build my project, which contains three files: main.cpp, TimeSeries.cpp, TimeSeries.h. TimeSeries.h provides declarations for the TimeSeries class as follows:
template<class XType, class YType> class TimeSeries {
public:
TimeSeries(void);
~TimeSeries(void);
};
Then TimeSeries.cpp contains:
#include "TimeSeries.h"
template<class XType, class YType>
TimeSeries<XType, YType>::TimeSeries(void) {
}
template<class XType, class YType>
TimeSeries<XType, YType>::~TimeSeries(void) {
}
And finally, main.cpp contains
#include "TimeSeries.h"
typedef TimeSeries<float, float> FTimeSeries;
int main(int argc, char** argv) {
FTimeSeries input_data;
return 0;
}
When building with C::B, I get the following error:
undefined reference to `TimeSeries<float, float>::TimeSeries()'
What can I do?
Thanks,
CFP.
Basically all templated code should be defined in a header, otherwise it will not be built since nothing uses it in the compiled unit.
Each cpp file is compiled as a separate unit, and thus constructor and destructor are not compiled. The compiler has no way of knowing what type of template argument you will use in main.cpp when it compiles TimeSeries.cpp.
The reason for splitting code into header- and source-files is so that the declaration and the implementation are separated. The compiler can translate the source-file (compilation unit) into an object file, and other compilation-units that want to use the classes and functions just include the header-file, and link the object file. This way, the code has to be compiled only once, and can be reused by linking it.
The problem with templates is that, as long as there are no parameters provided for them, the compiler cannot compile them. The same template instantiated with different parameters results in different types. std::vector<int> and std::vector<float> are, from the compilers perspective, not related in any way. Because of this, template-classes usually have to reside completely in a header-file, because, when the template is used, the compiler needs the complete definition in order to generate the class depending on the parameters.
As #Gabriel Schreiber pointed out in his answer, you can tell the compiler that he should compile the template with a specific set of parameters, making that available to other compilation units just by linking. However, this does not make the template available for other parameter sets.
You need to add this in your .cpp-file (below the definitions):
template class TimeSeries<float, float>;
When the compiler compiles TimeSeries.cpp it doesn't know which for which types the template is need because it is used in another source file. You need to tell the compiler explicitly.
Read about Explicit Template Instantiation in your copy of the Stroustrup or on the internet.
Template code is not compiled until the template function is used. But where does it save the compiled code, is it saved in the object file from which used the template function in the first place?
For example,
main.cpp is calling a template function from the file test.h, the compiler generates an object file main.o,
Is the template function inside the main.o file? because template code is not inlined, is it?
It's totally compiler implementation dependant. Most compilers will generate code around, inline or in cpp-like files and then compile with that. Sometimes, with optimization setup, some compilers will even reuse the same code instead of recreate it for each cpp.
So you have to see your compiler's doc for more details.
Yes, the template function code is emitted in the main.o file. Some of it may be inlined, as any other code may be inlined, but in general, code for templates is emitted in any file in which the template is instantiated. Think of it as first instantiating the template code to produce normal non-template functions, and then compiling those functions with whatever inlining and optimization the compiler applies to any other function.
When the same template instantiation (e.g. std::vector<int>) occurs in multiple compilation units (.cpp files), there is a bit of difficulty, because the code is instantiated in each of them. There are various ways of handling this, sometimes involving a cleanup step in the linking phase where duplicate template instantiations are resolved into a single one; your compiler manual can provide more information on exactly how it handles that situation.
Template code is compiled even if it is never instantiated. Otherwise, compilers couldn't be required to emit a diagnostic for this:
template< typename T >
void f()
{
blah
}
Template compilation happens in two phases. Besides the basic checks, everything that depends on template parameters can only be checked when a template is instantiated and the formal parameters are filled in with actual types. For example, here
template< typename T >
void f()
{
typename T::nested_type x;
}
T::nested_type can only be checked after the template is instantiated and an actual type is given for T. However, a basic check ("given T::nested_type is a type, is this a valid variable definition?") is performed the moment the compiler encounters the template's definition. (That's why that typename is required, BTW. Depending on T, T::nested_type might just as well be the name of member of T or a static data member - which would make T::nested_type x; a syntax error. So we have to tell the compiler to treat T::nested_type as the name of a type.)
You mean instantiated, not compiled. At compile time the compiler finds out each and every version that your code uses and instatiates (in object files) all the required versions.
It is always inlined (meaning, it is always internal linkage, having inline semantics). It may in fact be not inlined after all, just as an inline function, however, template is not code. It is a "template for making code". Therefore, it will normally reside in a header, except special cases, see below.
There was an idea to make something else, codenamed "export keyword". It was removed from standard.
Special cases: you can compile template instantiations into an object file, without having them used. This is the only way to avoid having all template code inlined.
This is how it is done:
template class std::vector<MyClass>;
This will force the compiler to instantiate a template in the current location. C++0x will have a syntax to force compiler not to do it, and have the linker search for template instantiation elsewhere:
extern template class std::vector<MyClass>; // C++0x only
From what I understand template classes and template functions (for the most part) must be declared and defined in the same header file. With that said:
Are there any other ways to achieve separate compilation of template files other than using particular compilers? If yes, what are those?
What, if any, are the drawbacks of having the declaration and definition in the same file?
What is considered best-practice when it comes to template declaration & definition?
How To Organize Template Source Code
Basically, you have the following options:
Make the template definition visible to compiler in the point of instantiation.
Instantiate the types you need explicitly in a separate compile unit, so that linker can find it.
Use keyword export (if available)
One of the drawbacks encountered by implementing templates in the .h files is that any time you make a small change to the implementation, all the code that uses the template must be recompiled. There's really no way around this aside from not using templates, or declaring & defining them in the CPP file where you use them.
You can implement templates in a seperate file, and then include that file from the .h file. Such as:
templ.h
template<class V> V foo(const V& rhs);
#include "templ.inc"
templ.inc
template<class V> V foo*const V& rhs)
{
// do something...
return val;
}
My personal preference is to implement templates right in the h file unless they become large, and then I'll break it up in to h and inc files.
Not really. The definition of the template must be available at compile time, since templates are instantiated depending on the template arguments you give them. This is why they must be placed in headers, so the compiler can have the code to write a new instantiation. You pretty much need a compiler with support for the export keyword.
People can see your code, if that's a drawback to you. It might also be less "neat" to some people, but I don't think that's an issue.
A further problem is the compile times every time you change the .h (especially if it is included in a lot of places)
Not really. There is the export keyword, but most compilers don't support this. The only mainstream one that I know of that does support this is the Comeau compiler.
If your template is part of a public API then you're exposing your code to the world. (most people don't consider this a problem, but some do. It depends on your business).
Put them both in the same header file.
Suppose I have fileA.h which declares a class classA with template function SomeFunc<T>(). This function is implemented directly in the header file (as is usual for template functions). Now I add a specialized implementation of SomeFunc() (like for SomeFunc<int>()) in fileA.C (ie. not in the header file).
If I now call SomeFunc<int>() from some other code (maybe also from another library), would it call the generic version, or the specialization?
I have this problem right now, where the class and function live in a library which is used by two applications. And one application correctly uses the specialization, while another app uses the generic form (which causes runtime problems later on). Why the difference? Could this be related to linker options etc? This is on Linux, with g++ 4.1.2.
It is an error to have a specialization for a template which is not visible at the point of call. Unfortunately, compilers are not required to diagnose this error, and can then do what they like with your code (in standardese it is "ill formed, no diagnostic required").
Technically, you need to define the specialization in the header file, but just about every compiler will handle this as you might expect: this is fixed in C++11 with the new "extern template" facility:
extern template<> SomeFunc<int>();
This explicitly declares that the particular specialization is defined elsewhere. Many compilers support this already, some with and some without the extern.
Have you added a prototype with parameters to your header file?
I mean is there somewhere in fileA.h
template<> SomeFunc<int>();
If not that's probably the reason.
I had the same problem with gcc4, here is how i solved it. It was more simple a solution than what i was lead to believe by previous comments. The previous posts ideas were correct but their syntax didn't work for me.
----------header-----------------
template < class A >
void foobar(A& object)
{
std::cout << object;
}
template <>
void foobar(int);
---------source------------------
#include "header.hpp"
template <>
void foobar(int x)
{
std::cout << "an int";
}
Per the specs, your specialized function template should never be called outside fileA.C, unless you export the template definition, which no compiler (except Comeau) currently supports (or has it planned for the forseeable future).
On the other hand, once the function template is instantiated, there is a function visible to the compiler that is no longer a template. GCC may re-use this definition across different compiler units because the standard states that each template shall only be instantiated once for a given set of type arguments [temp.spec]. Still, since the template is not exported, this should be limited to the compilation unit.
I believe that GCC may expose a bug here in sharing its list of instantiated templates across compilation units. Normally, this is a reasonable optimization but it should take function specializations into account which it doesn't seem to do correctly.
In Microsoft C++, I did an experiment with inline functions. I wanted to know what would happen if I defined incompatible versions of a function in different sources. I got different results depending on whether I was using a Debug build or a Release build. In Debug, the compiler refuses to inline anything, and the linker was linking the same version of the function no matter what was in scope in the source. In Release, the compiler inlined whichever version had been defined at the time, and you got differing versions of the function.
In neither case were there any warnings. I kind of suspected this, which is why I did the experiment.
I assume that template functions would behave the same, as would other compilers.
As Anthony Williams says, the extern template construct is the correct way to do this, but since his sample code is incomplete and has multiple syntax errors, here's a complete solution.
fileA.h:
namespace myNamespace {
class classA {
public:
template <class T> void SomeFunc() { ... }
};
// The following line declares the specialization SomeFunc<int>().
template <> void classA::SomeFunc<int>();
// The following line externalizes the instantiation of the previously
// declared specialization SomeFunc<int>(). If the preceding line is omitted,
// the following line PREVENTS the specialization of SomeFunc<int>();
// SomeFunc<int>() will not be usable unless it is manually instantiated
// separately). When the preceding line is included, all the compilers I
// tested this on, including gcc, behave exactly the same (throwing a link
// error if the specialization of SomeFunc<int>() is not instantiated
// separately), regardless of whether or not the following line is included;
// however, my understanding is that nothing in the standard requires that
// behavior if the following line is NOT included.
extern template void classA::SomeFunc<int>();
}
fileA.C:
#include "fileA.h"
template <> void myNamespace::classA::SomeFunc<int>() { ... }
Brandon: that's what I thought - the specialized function should never be called. Which is true for the second application I mentioned. The first app, however, clearly calls the specialized form even though the specialization is not declared in the header file!
I mainly seek enlightenment here :-) because the first app is a unit test, and it's unfortunate to have a bug that doesn't appear in the test but in the real app...
(PS: I have fixed this specific bug, indeed by declaring the specialization in the header; but what other similar bugs might still be hidden?)
#[anthony-williams],
are you sure you're not confusing extern template declarations with extern template instantiations? From what I see, extern template may only be used for explicit instantiation, not for specialization (which implies implicit instantiation). [temp.expl.spec] doesn't mention the extern keyword:
explicit-specialization:
template < > declaration
Unless the specialized template function is also listed in the header file, the other application will have no knowledge of the specialized version. The solution is the add SomeFunc<int>() to the header as well.