Lets say that I have this:
struct foo {
template <typename T>
void bar(const T param) { cout << param << endl; }
};
Now I want to add the specialization:
template <>
void bar<char>(const char param) { cout << static_cast<int>(param) << endl; }
Can I just add the declaration to the header that foo is declared in and defince bar<char> in the implementation file foo is implemented in?
The reason that I'm asking is that locally I'm seeing it work both ways. I think the difference is: Specialized template methods that I'm only using internally to the defining class can be defined in the implementation. Specialized template methods I'm using externally seem to need to be defined in the header. I haven't been able to find anything conclusive on this though so I thought I'd ask.
The one thing thy shan't do is to declare specialization in .cpp file. There are good reasons for it, discussed in
Explicit specialization of member function template in source file
However, if you put declaration in header file (as I understand you intend to do), the specialization can go into .cpp file, as long as it is called from there at least once - and than you can have calls to the specialization outside of the translation units.
The reason for at least one requirement is that compiler is only going to instantiate the template if it is called and the definition is available, but once template is instantiated, it can be used from any translation units.
Related
I have some header file with declared template class and template function:
print.h:
template<typename T> struct Printer {
static void print();
};
template<typename T> void print();
In the implementation file, I have some base implementation of Printer, some specializations of it, and implementation of the function print that calls Printer<T>::print.
print.cpp
template<typename T> void Printer<T>::print() {
cout << "base";
}
template<> struct Printer<int> {
static void print() {
cout << "int";
}
};
template<> struct Printer<bool> {
static void print() {
cout << "bool";
}
};
template<typename T> void print() {
return Printer<T>::print();
}
In another file, I add one more specialization of Printer and a call of function print after it.
another.cpp
template<> struct Printer<char> {
static void print() {
cout << "char";
}
};
void printChar() {
print<char>();
}
Which Printer<T>::print does printChar call? Base or specialization Printer<char>? Can I change the behavior of printChar in this way?
(print.cpp is compiled first)
I compiled this code in a single file in the same order and called printChar. The output is 'char'. But I am not sure that it will work in different files as I described. Because for me it looks like it opens the ability to extend existing code by something like a cheating injection. But maybe it is OK to practice (Swift is based on a similar kind of extensions). If it works, Is it ok to practice it?
I'm not sure which previous usages you're referring to. I see only one usage of print<char>, after the specialization of print<char>.
Templates work kind of like #define, they stamp out definitions on-demand as needed. So the right template specialization definition had better be available at the time the code needs to be stamped out, which by the way takes place independently per TU. And these definitions must be identical across TU's.
See [temp.expl.spec]/pp.6-7:
6 If a template, a member template or a member of a class template is explicitly specialized then that specialization shall be declared before the first use of that specialization that would cause an implicit instantiation to take place, in every translation unit in which such a use occurs; no diagnostic is required.
7 The placement of explicit specialization declarations ... can affect whether a program is well-formed according to the relative positioning of the explicit specialization declarations and their points of instantiation in the translation unit as specified above and below. When writing a specialization, be careful about its location; or to make it compile will be such a trial as to kindle its self-immolation.
So as long as print<char> is instantiated after the specialization of print<char> in every translation unit where it is used (including implicit instantiations), the program is fine. The printChar() case is OK. But if print<char> happens to be instantiated in any other TU, the program would become ill-formed NDR.
For this reason all specializations are usually bundled (included) in the same header file that declares the primary template. It might increase compile times but can save a lot of headache during program development and maintenance later on.
I have some template code that I would prefer to have stored in a CPP file instead of inline in the header. I know this can be done as long as you know which template types will be used. For example:
.h file
class foo
{
public:
template <typename T>
void do(const T& t);
};
.cpp file
template <typename T>
void foo::do(const T& t)
{
// Do something with t
}
template void foo::do<int>(const int&);
template void foo::do<std::string>(const std::string&);
Note the last two lines - the foo::do template function is only used with ints and std::strings, so those definitions mean the app will link.
My question is - is this a nasty hack or will this work with other compilers/linkers? I am only using this code with VS2008 at the moment but will be wanting to port to other environments.
The problem you describe can be solved by defining the template in the header, or via the approach you describe above.
I recommend reading the following points from the C++ FAQ Lite:
Why can’t I separate the definition of my templates class from its declaration and put it inside a .cpp file?
How can I avoid linker errors with my template functions?
How does the C++ keyword export help with template linker errors?
They go into a lot of detail about these (and other) template issues.
For others on this page wondering what the correct syntax is (as did I) for explicit template specialisation (or at least in VS2008), its the following...
In your .h file...
template<typename T>
class foo
{
public:
void bar(const T &t);
};
And in your .cpp file
template <class T>
void foo<T>::bar(const T &t)
{ }
// Explicit template instantiation
template class foo<int>;
Your example is correct but not very portable.
There is also a slightly cleaner syntax that can be used (as pointed out by #namespace-sid, among others).
However, suppose the templated class is part of some library that is to be shared...
Should other versions of the templated class be compiled?
Is the library maintainer supposed to anticipate all possible templated uses of the class?
An Alternate Approach
Add a third file that is the template implementation/instantiation file in your sources.
lib/foo.hpp - from library
#pragma once
template <typename T>
class foo {
public:
void bar(const T&);
};
lib/foo.cpp - compiling this file directly just wastes compilation time
// Include guard here, just in case
#pragma once
#include "foo.hpp"
template <typename T>
void foo::bar(const T& arg) {
// Do something with `arg`
}
foo.MyType.cpp - using the library, explicit template instantiation of foo<MyType>
// Consider adding "anti-guard" to make sure it's not included in other translation units
#if __INCLUDE_LEVEL__
#error "Don't include this file"
#endif
// Yes, we include the .cpp file
#include <lib/foo.cpp>
#include "MyType.hpp"
template class foo<MyType>;
Organize your implementations as desired:
All implementations in one file
Multiple implementation files, one for each type
An implementation file for each set of types
Why??
This setup should reduce compile times, especially for heavily used complicated templated code, because you're not recompiling the same header file in each
translation unit.
It also enables better detection of which code needs to be recompiled, by compilers and build scripts, reducing incremental build burden.
Usage Examples
foo.MyType.hpp - needs to know about foo<MyType>'s public interface but not .cpp sources
#pragma once
#include <lib/foo.hpp>
#include "MyType.hpp"
// Declare `temp`. Doesn't need to include `foo.cpp`
extern foo<MyType> temp;
examples.cpp - can reference local declaration but also doesn't recompile foo<MyType>
#include "foo.MyType.hpp"
MyType instance;
// Define `temp`. Doesn't need to include `foo.cpp`
foo<MyType> temp;
void example_1() {
// Use `temp`
temp.bar(instance);
}
void example_2() {
// Function local instance
foo<MyType> temp2;
// Use templated library function
temp2.bar(instance);
}
error.cpp - example that would work with pure header templates but doesn't here
#include <lib/foo.hpp>
// Causes compilation errors at link time since we never had the explicit instantiation:
// template class foo<int>;
// GCC linker gives an error: "undefined reference to `foo<int>::bar()'"
foo<int> nonExplicitlyInstantiatedTemplate;
void linkerError() {
nonExplicitlyInstantiatedTemplate.bar();
}
Note: Most compilers/linters/code helpers won't detect this as an error, since there is no error according to C++ standard.
But when you go to link this translation unit into a complete executable, the linker won't find a defined version of foo<int>.
Alternate approach from: https://stackoverflow.com/a/495056/4612476
This code is well-formed. You only have to pay attention that the definition of the template is visible at the point of instantiation. To quote the standard, § 14.7.2.4:
The definition of a non-exported function template, a non-exported member function template, or a non-exported member function or static data member of a class template shall be present in every translation unit in which it is explicitly instantiated.
This should work fine everywhere templates are supported. Explicit template instantiation is part of the C++ standard.
That is a standard way to define template functions. I think there are three methods I read for defining templates. Or probably 4. Each with pros and cons.
Define in class definition. I don't like this at all because I think class definitions are strictly for reference and should be easy to read. However it is much less tricky to define templates in class than outside. And not all template declarations are on the same level of complexity. This method also makes the template a true template.
Define the template in the same header, but outside of the class. This is my preferred way most of the times. It keeps your class definition tidy, the template remains a true template. It however requires full template naming which can be tricky. Also, your code is available to all. But if you need your code to be inline this is the only way. You can also accomplish this by creating a .INL file at the end of your class definitions.
Include the header.h and implementation.CPP into your main.CPP. I think that's how its done. You won't have to prepare any pre instantiations, it will behave like a true template. The problem I have with it is that it is not natural. We don't normally include and expect to include source files. I guess since you included the source file, the template functions can be inlined.
This last method, which was the posted way, is defining the templates in a source file, just like number 3; but instead of including the source file, we pre instantiate the templates to ones we will need. I have no problem with this method and it comes in handy sometimes. We have one big code, it cannot benefit from being inlined so just put it in a CPP file. And if we know common instantiations and we can predefine them. This saves us from writing basically the same thing 5, 10 times. This method has the benefit of keeping our code proprietary. But I don't recommend putting tiny, regularly used functions in CPP files. As this will reduce the performance of your library.
Note, I am not aware of the consequences of a bloated obj file.
Let's take one example, let's say for some reason you want to have a template class:
//test_template.h:
#pragma once
#include <cstdio>
template <class T>
class DemoT
{
public:
void test()
{
printf("ok\n");
}
};
template <>
void DemoT<int>::test()
{
printf("int test (int)\n");
}
template <>
void DemoT<bool>::test()
{
printf("int test (bool)\n");
}
If you compile this code with Visual Studio - it works out of box.
gcc will produce linker error (if same header file is used from multiple .cpp files):
error : multiple definition of `DemoT<int>::test()'; your.o: .../test_template.h:16: first defined here
It's possible to move implementation to .cpp file, but then you need to declare class like this -
//test_template.h:
#pragma once
#include <cstdio>
template <class T>
class DemoT
{
public:
void test()
{
printf("ok\n");
}
};
template <>
void DemoT<int>::test();
template <>
void DemoT<bool>::test();
// Instantiate parametrized template classes, implementation resides on .cpp side.
template class DemoT<bool>;
template class DemoT<int>;
And then .cpp will look like this:
//test_template.cpp:
#include "test_template.h"
template <>
void DemoT<int>::test()
{
printf("int test (int)\n");
}
template <>
void DemoT<bool>::test()
{
printf("int test (bool)\n");
}
Without two last lines in header file - gcc will work fine, but Visual studio will produce an error:
error LNK2019: unresolved external symbol "public: void __cdecl DemoT<int>::test(void)" (?test#?$DemoT#H##QEAAXXZ) referenced in function
template class syntax is optional in case if you want to expose function via .dll export, but this is applicable only for windows platform - so test_template.h could look like this:
//test_template.h:
#pragma once
#include <cstdio>
template <class T>
class DemoT
{
public:
void test()
{
printf("ok\n");
}
};
#ifdef _WIN32
#define DLL_EXPORT __declspec(dllexport)
#else
#define DLL_EXPORT
#endif
template <>
void DLL_EXPORT DemoT<int>::test();
template <>
void DLL_EXPORT DemoT<bool>::test();
with .cpp file from previous example.
This however gives more headache to linker, so it's recommended to use previous example if you don't export .dll function.
This is definitely not a nasty hack, but be aware of the fact that you will have to do it (the explicit template specialization) for every class/type you want to use with the given template. In case of MANY types requesting template instantiation there can be A LOT of lines in your .cpp file. To remedy this problem you can have a TemplateClassInst.cpp in every project you use so that you have greater control what types will be instantiated. Obviously this solution will not be perfect (aka silver bullet) as you might end up breaking the ODR :).
There is, in the latest standard, a keyword (export) that would help alleviate this issue, but it isn't implemented in any compiler that I'm aware of, other than Comeau.
See the FAQ-lite about this.
Yes, that's the standard way to do specializiation explicit instantiation. As you stated, you cannot instantiate this template with other types.
Edit: corrected based on comment.
None of above worked for me, so here is how y solved it, my class have only 1 method templated..
.h
class Model
{
template <class T>
void build(T* b, uint32_t number);
};
.cpp
#include "Model.h"
template <class T>
void Model::build(T* b, uint32_t number)
{
//implementation
}
void TemporaryFunction()
{
Model m;
m.build<B1>(new B1(),1);
m.build<B2>(new B2(), 1);
m.build<B3>(new B3(), 1);
}
this avoid linker errors, and no need to call TemporaryFunction at all
Time for an update! Create an inline (.inl, or probably any other) file and simply copy all your definitions in it. Be sure to add the template above each function (template <typename T, ...>). Now instead of including the header file in the inline file you do the opposite. Include the inline file after the declaration of your class (#include "file.inl").
I don't really know why no one has mentioned this. I see no immediate drawbacks.
There is nothing wrong with the example you have given. But i must say i believe it's not efficient to store function definitions in a cpp file. I only understand the need to separate the function's declaration and definition.
When used together with explicit class instantiation, the Boost Concept Check Library (BCCL) can help you generate template function code in cpp files.
I have this code:
#include <iostream>
struct A
{
template<typename T> bool isImplemented()
{
std::cout << "Not Implemented" << std::endl;
return false;
}
};
template<> inline bool A::isImplemented<int>()
{
std::cout << "int Implemented" << std::endl;
return true;
}
I can understand why the template specialization needs the inline, in order to prevent the ODR to be violated the inline will merge the translation table avoiding a conflict.
But, why I don`t need an inline on the isImplemented inside the struct A?
Maybe that question can be extended to, why if the method is declared inside the struct/class on the header it does not need the inline?
As I can understand, it would create the symbols on every object file (.o) that it is called, violating the ODR, why that does not happen?
You do not need it for two reasons:
Every function defined within the definition of the class is implicit inline.
A template doesn't need inline keyword. Remember, your original function is a template, while specialization is not.
The point to highlight here is that template<typename T> bool isImplemented() IS NOT a function.
It is a template to a function and will only exist (as code) once specialized, like you did with template<> inline bool A::isImplemented<int>().
The inline here is not mandatory. Program compiles and runs perfectly without it.
Strangely your struct A does not depend on any template parameter neither isImplemented() method so I am still trying to figure out your intention.
A simple usage of your functions might look like this:
int main( )
{
A a;
a.isImplemented< int >( );
a.isImplemented< double >( );
}
And the output:
int Implemented
Not Implemented
And basically you are able to tell wich specializations you have explicitly implemented from the generic ones. Is it what you need?
EDIT:
Given the comment on my answer, I believe the original question is nicely answered here:
Explicit specialization in non-namespace scope
C++ syntax for explicit specialization of a template function in a template class?
Context
We develop a templated settings system class, that will be part of an API we expose to users of different architectures (fewer words : we should not rely on compiler-specific behaviors)
The generalized code would look like :
In the header
namespace Firm
{
// Class definition
class A
{
template<class T>
void foo(T* aParam);
};
// Non specialized template definition
template<class T>
void A::foo(T* aParam)
{
//...
}
// Declaration of a specialization
template<>
void A::foo<int>(int* aParam);
} // namespace
In the CPP file
namespace Firm
{
// Definition of the specialized member function
template<>
void A::foo<int>(int* aParam)
{
//...
}
} // namespace
Questions
Everything runs fine with gcc 4.x. (i.e. Different compilation units use the specialized methods when appropriate.) But I feel uncomfortable since I read the following entry :
Visibility of template specialization of C++ function
Accepted answer states, if I correctly understand it, that it is an error if the definition of the specialization of a template method is not visible from call site. (Which is the case in all compilation units that are not the CPP file listed above but include the header)
I cannot understand why a declaration would not be enough at this point (declaration provided by the header) ?
If it really is an error, is there a correct way to define the specialization for it to be :
non-inline
usable in any compilation unit that includes the header (and links with the corresponding .obj) ?
Put the declaration in the header file.
The compiler either needs to instantiate the template (needs the definition) or you need to use c++0x extern templates
look here:https://stackoverflow.com/a/8131212/258418
edit
# Single definition rule:
YOu should be fine since I expirienced the following behaviour for templates:
Create them in the different o files (multiple time, time consuming due to several compilations), when the linker comes in it removes the duplicates and uses one instantiation.
also look at this answer/wiki link: https://stackoverflow.com/a/8133000/258418
edit2
for classes it works like this:
extern template class yourTemplate<int>; //tell the compiler to just use this like a funciton stub and do no instantiation. put this in the header
template class yourTemplate<int>; //instantiation, put this in a c file/object that you link with the project
for your function it should work like this:
// Declaration of a specialization in your header
extern template<>
void A::foo(int* aParam);
//in your cpp file
template<>
void A::foo<int>(int* aParam) {
code...
}
template class A::foo<int>(int* aParam);
I have some template code that I would prefer to have stored in a CPP file instead of inline in the header. I know this can be done as long as you know which template types will be used. For example:
.h file
class foo
{
public:
template <typename T>
void do(const T& t);
};
.cpp file
template <typename T>
void foo::do(const T& t)
{
// Do something with t
}
template void foo::do<int>(const int&);
template void foo::do<std::string>(const std::string&);
Note the last two lines - the foo::do template function is only used with ints and std::strings, so those definitions mean the app will link.
My question is - is this a nasty hack or will this work with other compilers/linkers? I am only using this code with VS2008 at the moment but will be wanting to port to other environments.
The problem you describe can be solved by defining the template in the header, or via the approach you describe above.
I recommend reading the following points from the C++ FAQ Lite:
Why can’t I separate the definition of my templates class from its declaration and put it inside a .cpp file?
How can I avoid linker errors with my template functions?
How does the C++ keyword export help with template linker errors?
They go into a lot of detail about these (and other) template issues.
For others on this page wondering what the correct syntax is (as did I) for explicit template specialisation (or at least in VS2008), its the following...
In your .h file...
template<typename T>
class foo
{
public:
void bar(const T &t);
};
And in your .cpp file
template <class T>
void foo<T>::bar(const T &t)
{ }
// Explicit template instantiation
template class foo<int>;
Your example is correct but not very portable.
There is also a slightly cleaner syntax that can be used (as pointed out by #namespace-sid, among others).
However, suppose the templated class is part of some library that is to be shared...
Should other versions of the templated class be compiled?
Is the library maintainer supposed to anticipate all possible templated uses of the class?
An Alternate Approach
Add a third file that is the template implementation/instantiation file in your sources.
lib/foo.hpp - from library
#pragma once
template <typename T>
class foo {
public:
void bar(const T&);
};
lib/foo.cpp - compiling this file directly just wastes compilation time
// Include guard here, just in case
#pragma once
#include "foo.hpp"
template <typename T>
void foo::bar(const T& arg) {
// Do something with `arg`
}
foo.MyType.cpp - using the library, explicit template instantiation of foo<MyType>
// Consider adding "anti-guard" to make sure it's not included in other translation units
#if __INCLUDE_LEVEL__
#error "Don't include this file"
#endif
// Yes, we include the .cpp file
#include <lib/foo.cpp>
#include "MyType.hpp"
template class foo<MyType>;
Organize your implementations as desired:
All implementations in one file
Multiple implementation files, one for each type
An implementation file for each set of types
Why??
This setup should reduce compile times, especially for heavily used complicated templated code, because you're not recompiling the same header file in each
translation unit.
It also enables better detection of which code needs to be recompiled, by compilers and build scripts, reducing incremental build burden.
Usage Examples
foo.MyType.hpp - needs to know about foo<MyType>'s public interface but not .cpp sources
#pragma once
#include <lib/foo.hpp>
#include "MyType.hpp"
// Declare `temp`. Doesn't need to include `foo.cpp`
extern foo<MyType> temp;
examples.cpp - can reference local declaration but also doesn't recompile foo<MyType>
#include "foo.MyType.hpp"
MyType instance;
// Define `temp`. Doesn't need to include `foo.cpp`
foo<MyType> temp;
void example_1() {
// Use `temp`
temp.bar(instance);
}
void example_2() {
// Function local instance
foo<MyType> temp2;
// Use templated library function
temp2.bar(instance);
}
error.cpp - example that would work with pure header templates but doesn't here
#include <lib/foo.hpp>
// Causes compilation errors at link time since we never had the explicit instantiation:
// template class foo<int>;
// GCC linker gives an error: "undefined reference to `foo<int>::bar()'"
foo<int> nonExplicitlyInstantiatedTemplate;
void linkerError() {
nonExplicitlyInstantiatedTemplate.bar();
}
Note: Most compilers/linters/code helpers won't detect this as an error, since there is no error according to C++ standard.
But when you go to link this translation unit into a complete executable, the linker won't find a defined version of foo<int>.
Alternate approach from: https://stackoverflow.com/a/495056/4612476
This code is well-formed. You only have to pay attention that the definition of the template is visible at the point of instantiation. To quote the standard, § 14.7.2.4:
The definition of a non-exported function template, a non-exported member function template, or a non-exported member function or static data member of a class template shall be present in every translation unit in which it is explicitly instantiated.
This should work fine everywhere templates are supported. Explicit template instantiation is part of the C++ standard.
That is a standard way to define template functions. I think there are three methods I read for defining templates. Or probably 4. Each with pros and cons.
Define in class definition. I don't like this at all because I think class definitions are strictly for reference and should be easy to read. However it is much less tricky to define templates in class than outside. And not all template declarations are on the same level of complexity. This method also makes the template a true template.
Define the template in the same header, but outside of the class. This is my preferred way most of the times. It keeps your class definition tidy, the template remains a true template. It however requires full template naming which can be tricky. Also, your code is available to all. But if you need your code to be inline this is the only way. You can also accomplish this by creating a .INL file at the end of your class definitions.
Include the header.h and implementation.CPP into your main.CPP. I think that's how its done. You won't have to prepare any pre instantiations, it will behave like a true template. The problem I have with it is that it is not natural. We don't normally include and expect to include source files. I guess since you included the source file, the template functions can be inlined.
This last method, which was the posted way, is defining the templates in a source file, just like number 3; but instead of including the source file, we pre instantiate the templates to ones we will need. I have no problem with this method and it comes in handy sometimes. We have one big code, it cannot benefit from being inlined so just put it in a CPP file. And if we know common instantiations and we can predefine them. This saves us from writing basically the same thing 5, 10 times. This method has the benefit of keeping our code proprietary. But I don't recommend putting tiny, regularly used functions in CPP files. As this will reduce the performance of your library.
Note, I am not aware of the consequences of a bloated obj file.
Let's take one example, let's say for some reason you want to have a template class:
//test_template.h:
#pragma once
#include <cstdio>
template <class T>
class DemoT
{
public:
void test()
{
printf("ok\n");
}
};
template <>
void DemoT<int>::test()
{
printf("int test (int)\n");
}
template <>
void DemoT<bool>::test()
{
printf("int test (bool)\n");
}
If you compile this code with Visual Studio - it works out of box.
gcc will produce linker error (if same header file is used from multiple .cpp files):
error : multiple definition of `DemoT<int>::test()'; your.o: .../test_template.h:16: first defined here
It's possible to move implementation to .cpp file, but then you need to declare class like this -
//test_template.h:
#pragma once
#include <cstdio>
template <class T>
class DemoT
{
public:
void test()
{
printf("ok\n");
}
};
template <>
void DemoT<int>::test();
template <>
void DemoT<bool>::test();
// Instantiate parametrized template classes, implementation resides on .cpp side.
template class DemoT<bool>;
template class DemoT<int>;
And then .cpp will look like this:
//test_template.cpp:
#include "test_template.h"
template <>
void DemoT<int>::test()
{
printf("int test (int)\n");
}
template <>
void DemoT<bool>::test()
{
printf("int test (bool)\n");
}
Without two last lines in header file - gcc will work fine, but Visual studio will produce an error:
error LNK2019: unresolved external symbol "public: void __cdecl DemoT<int>::test(void)" (?test#?$DemoT#H##QEAAXXZ) referenced in function
template class syntax is optional in case if you want to expose function via .dll export, but this is applicable only for windows platform - so test_template.h could look like this:
//test_template.h:
#pragma once
#include <cstdio>
template <class T>
class DemoT
{
public:
void test()
{
printf("ok\n");
}
};
#ifdef _WIN32
#define DLL_EXPORT __declspec(dllexport)
#else
#define DLL_EXPORT
#endif
template <>
void DLL_EXPORT DemoT<int>::test();
template <>
void DLL_EXPORT DemoT<bool>::test();
with .cpp file from previous example.
This however gives more headache to linker, so it's recommended to use previous example if you don't export .dll function.
This is definitely not a nasty hack, but be aware of the fact that you will have to do it (the explicit template specialization) for every class/type you want to use with the given template. In case of MANY types requesting template instantiation there can be A LOT of lines in your .cpp file. To remedy this problem you can have a TemplateClassInst.cpp in every project you use so that you have greater control what types will be instantiated. Obviously this solution will not be perfect (aka silver bullet) as you might end up breaking the ODR :).
There is, in the latest standard, a keyword (export) that would help alleviate this issue, but it isn't implemented in any compiler that I'm aware of, other than Comeau.
See the FAQ-lite about this.
Yes, that's the standard way to do specializiation explicit instantiation. As you stated, you cannot instantiate this template with other types.
Edit: corrected based on comment.
None of above worked for me, so here is how y solved it, my class have only 1 method templated..
.h
class Model
{
template <class T>
void build(T* b, uint32_t number);
};
.cpp
#include "Model.h"
template <class T>
void Model::build(T* b, uint32_t number)
{
//implementation
}
void TemporaryFunction()
{
Model m;
m.build<B1>(new B1(),1);
m.build<B2>(new B2(), 1);
m.build<B3>(new B3(), 1);
}
this avoid linker errors, and no need to call TemporaryFunction at all
Time for an update! Create an inline (.inl, or probably any other) file and simply copy all your definitions in it. Be sure to add the template above each function (template <typename T, ...>). Now instead of including the header file in the inline file you do the opposite. Include the inline file after the declaration of your class (#include "file.inl").
I don't really know why no one has mentioned this. I see no immediate drawbacks.
There is nothing wrong with the example you have given. But i must say i believe it's not efficient to store function definitions in a cpp file. I only understand the need to separate the function's declaration and definition.
When used together with explicit class instantiation, the Boost Concept Check Library (BCCL) can help you generate template function code in cpp files.