consider the following code:
//header.h
template<class T>
class A
{
static int x;
};
template<class T>
int A<T>::x = 0;
//source1.cpp
#include "header.h"
void f(){} // dummy function
//main.cpp
#include "header.h"
int main(){}
In this case code compiles perfectly without errors, but if I remove the template qualifier from class
class A
{
static int x;
};
int A::x = 0;
In this case compiler erred with multiple definition of x. Can anybody explain this behavior?
And when the template class's static variable is initialized / instantiated??
Compiler will remove duplicate template instantiations on its own. If you turn your template class into regular one, then its your duty to make sure only one definition of static variable exists (otherwise linker error will appear). Also remember that static data members are not shared between instatiations of templates for different types. With c++11 you can control instatiations on your own using extern templates: using extern template (C++11).
As for the point of instatiation for static members:
14.6.4.1 Point of instantiation [temp.point]
1 For a function template specialization, a member function template specialization, or a specialization for a
member function or static data member of a class template, if the specialization is implicitly instantiated
because it is referenced from within another template specialization and the context from which it is referenced
depends on a template parameter, the point of instantiation of the specialization is the point of
instantiation of the enclosing specialization. Otherwise, the point of instantiation for such a specialization
immediately follows the namespace scope declaration or definition that refers to the specialization.
so point of instatiation should be ie. right after main() if you use your type for the first time inside main().
Templates as the name suggest are the the fragments of code that will be used several times for different parameters. For templates the compiler is to ensure if their methods and static fiels definition are linked only ones. So if you create a static field with its default value the compiler is obliged to provide single memory cell (for the same template parameter set) even though the template class header is included several times. Unfortunetly non-template classes you need to be managed by yourself.
As for the second question, I believe the standard does not state when the static fields need to be initialized, each compiler can implement then it in its own manner.
It is necessary to instantiate/initialize static members in cpp files not in headers. Static members are property of class not property of objects, so if you include header file in more cpp files, it looks like you are initializing it more times.
Answer to this question is more complex. Template is not one class. It is instantiating on demand. It means that every different use of template is one standalone "template instance". For example if you use A<int> and A<float> than you will have 2 different classes therefore you would need to initialize A<int>::x and A<float>::x.
For more info see this answer: https://stackoverflow.com/a/607335/1280316
A class (be it a template or not) can (and should) be declared in any compilation unit that referes to it.
A static field initialization does defines a variable, and as such it should exist only in one compilation unit -> that's the reason why you get an error when class A is not a template.
But when you declare a template, nothing is really created untill you instantiate it. As you never instantiate the template, the static field is never defined and you get no error.
If you had two different instantiations in source1.cpp (say A<B>) and main.cpp (say A<C>) all will still be fine : you would get A<B>::x in source1 and A<C>::x in main => two different variables since A<B> and A<C> are different classes.
The case where you instantiate same class in different compilation units is trickier. It should generate an error, but if it did, you could hardly declare special fields in templates. So it is processed as a special case by the compiler as it is explained in this other answer to generate no errors.
Related
During initial development of a template class, before I've written full test cases, I am finding that I would like the ability to force the compiler to generate the code for every member (including non-static ones) of a template class for a specific set of template parameters, just to make sure all the code at least compiles.
Specifically, I'm using GCC 9 (I don't really need this ability for other compilers, since it's just something I want to temporarily make happen during development); and its c++14, c++17, c++2a and c++20 standards.
For example, I might write the following code:
template <typename D> struct test_me {
D value;
void mistake1 () { value[0] = 0; }
void mistake2 () { mistake1(value); }
// and a bajillion other member functions...
};
And, given that I know in advance the finite set of possible template parameters (let's say int and float here), I just want to make sure they at least compile while I'm working.
Now I can do this, which obviously falls short:
int main () {
test_me<int> i;
test_me<float> f;
}
Since mistake1 and mistake2 aren't generated, the code compiles fine despite the indexed access attempt to a non-array type, and the incorrect function call.
So what I've been doing during development is just writing code that calls all the member functions:
template <typename D> static void testCalls () {
test_me<D> t;
t.mistake1();
t.mistake2();
// and so on... so many mistakes...
}
int main () {
testCalls<int>();
testCalls<float>();
}
But this gets to be a pain, especially when the member functions start to have complex side effects or preconditions, or require nontrivial parameters, or have non-public members and not-yet-developed friends. So, I'd like a way to test compilation without having to explicitly call everything (and, ideally, I'd like to be able to test compilation without modifying any "test" code at all as I add new members).
So my question is: With at least GCC 9, is there a way to force (potentially temporarily) the compiler to generate code for a template class's entire set of members, given template parameters?
Just explicitly instantiate the class:
template struct test_me<int>;
template struct test_me<float>;
Demo
What you are trying to do is not allowed by the language, at least with implicit instantiations of your test class. When you implicitly instantiate test_me with some type, the definitions of the member functions are not allowed to be implicitly instantiated, as per temp.inst#11:
An implementation shall not implicitly instantiate a function template, a variable template, a member template, a non-virtual member function, a member class or static data member of a templated class, or a substatement of a constexpr if statement ([stmt.if]), unless such instantiation is required.
So if you want to implicitly instantiate all the mistake member functions, you have no choice but to require their instantiation somehow. As in your example with testCalls, you can make calls to those member functions, or you can ODR-use them some other way, such as taking their addresses.
(I am sorry for the messy title. I will gladly accept suggestions to improve it.)
I will try to be as straightforward as possible. I have the folowing code:
file1.hpp
template <class val_t>
struct MatOps;
file2.hpp:
#include "file1.hpp"
template <> struct MatOps<float>{
static void method1(){
// Do something
}
static void method2(){
// Do something
}
static void method3(){
// Do something
}
}
File3.hpp:
#include "file1.hpp"
template <> struct MatOps<double>{
static void method1(){
// Do something different
}
static void method2(){
// Do something different
}
static void method3(){
// Do something different
}
}
main.cpp
#include "file2.hpp"
#include "file3.hpp"
int main(){
float a,b,c,d;
MatOps<float>::method1(a,b,...);
MatOps<float>::method2(c,d,...);
return 0;
}
Questions:
I am not using the explicit specialization MatOps<double>. However, is MatOps<double> actually instantiated? Or more roughly: does the inclusion of file3.hpp occupy any storage whatsoever?
I am not using MatOps<float>::method3(), but I am using the other methods in the class. Since I am explicitely using MatOps<float>, does the compiler generate code for MatOps<float>::method3()?
Rationale: I have been asked to follow some guidelines in the MISRA C++:2003 standard. Although obsolete, I have been encouraged to use whatever is reasonable in there. In particular, there is a rule that reads:
Header files should be used to declare objects, functinos, inline functions, function templates, typedefs, macros, classes, and class templates and shall not contain or produce definitions of objects or functions (or fragments of functions or objects) that occupy storage.
A header file is considered to be any file that is included via the #include directive, regardless of name or suffix.
My code is heavily templated and hence I can include any files according to this rule. My problem comes when I do full specializations (I only do two of them: the ones listed in file2.hpp and file3.hpp). What are full template specializations? Is code generated for them even if they are not used? Ultimately, do they occupy storage?
To answer your first question, I quote the following from cppreference.com:
A class template by itself is not a type, or an object, or any other
entity. No code is generated from a source file that contains only
template definitions. In order for any code to appear, a template must
be instantiated: the template arguments must be provided so that the
compiler can generate an actual class (or function, from a function
template).
Inclusion of file3.hpp will not result in code generation by itself.
As for the second part, again from the same page,
When code refers to a template in context that requires a completely
defined type, or when the completeness of the type affects the code,
and this particular type has not been explicitly instantiated,
implicit instantiation occurs. For example, when an object of this
type is constructed, but not when a pointer to this type is
constructed.
This applies to the members of the class template: unless the member is
used in the program, it is not instantiated, and does not
require a definition.
Unless you are doing an explicit instantiation of your class template, individual member functions of your class template will not get instantiated, i.e., the compiler will not generate code for MatOps<float>::method3().
I'm wondering how template member functions work. In particular, when there is an instantiation of the template member function, is the whole class redefined? My confusion comes from the fact that (if I'm right) template classes are not classes in the proper sense. i.e., when instantiated, the compiler creates the definition for a completely new class. The same for template functions. However, classes with a template function seem to be actual classes, so I'm not sure how they could possibly work. Thus, I'm wondering, after instantiating a template member function, what happens with the class definition? Moreover, if I pass a class with a template member function to a template class, can I use the template member function? Could that cause a problem? I tried it once but got an error saying that several functions where defined more that once, although I'm not sure if that was the reason or if there could be an other reason for my error. Is there any further caveat when using static template member functions?
The class definition remains as it is; all the template function does is generate a family of member functions for that class. As an example:
class A {
public:
template<typename T> foo (T &t);
}
Is not conceptually different from you writing:
class A {
public:
foo (bool &t);
foo (int &t);
foo (double &t);
}
...just more convenient. And in the last example, you wouldn't expect a new class to be created for each function would you?
Perhaps the confusion comes from the notion that functions are somehow part of the memory layout of a class; that each function is itself contained in the class, and will be instantiated somewhere in memory whenever an object of the class is created. This notion is incorrect. Functions (templated, global, member, lambda, or otherwise) are never created on the fly or copied around in memory; they are a static and unchanging part of the executable image. The memory layout of the class is not changed by the presence of an extra set of functions, even if those happen to be generated by a template member.
The template class definition is instantiated when you instantiate a class. Each member function of it is instantiated when used. This actually allows you to have member functions that would not work if called when the class is instantiated with some types and not others. However, you must ensure that the signature of the function is either syntactically viable or fails with SFINAE. It will be looked up during the first phase of parsing. The body, if the function isn't itself a template, will be checked for name lookup...so dependent names have to be labeled as such via typename.
Recently when I was trying to optimize my include hierarchy I stumbled upon the file a.hpp:
template<class T>
class A
{
using t = typename T::a_t;
};
class B;
extern template class A<B>;
which seems to be ill-formed. In fact it seems as if the extern template statement at the end causes an instantiation of A<B> which causes the compiler to complain about an incomplete type.
My goal would have been to define A<B> in a.cpp:
#include <b.hpp>
template class A<B>;
This way I avoid having to include b.hpp from a.hpp which seems like a good idea to reduce compile time. However it does not work (a.hpp on itself doesn't compile!) Is there a better way of doing this?
Note: Of course I could just not use explicit template instantiation but this is not what I want! I would like to "precompile" A<B> to save compilation time if it were used, but if A<B> is not used I don't want to include b.hpp in every file that uses a.hpp!
The extern template declaration prevents the instantiation of member function bodies, but it forces the instantiation of the class definition, since the compiler needs that anyway, and the class body needs a full definition of the template argument since it accesses its members. I'm afraid that hiding B's body from users of A<B> is not possible.
extern template is an optimization, but it doesn't change the fundamental workings of the instantiation mechanism.
From http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2003/n1448.pdf
The extern specifier used to declare an explicit instantiation of a
class template only suppresses explicit instantiations of definitions
of member functions and static data members not previously specialized
in the translation unit containing the declaration.
Thus, if a definition of B is required in A, you cannot use an extern template without knowledge of B. You could of course try to get rid of that requirement. In the given case, you could remove the using t declaration and have a meta function to yield that type:
template<typename T>
struct get_a_t;
template<typename T>
struct get_a_t<A<T>>
{
using type = typename T::a_t;
};
Not sure it that is feasible in your case. As soon as A needs to store a B or a B::a_t, you need B. References and pointers would be OK, though.
The final extern template class A is telling the compiler that there is in some compilation unit a declaration of such template specialization. The compiler goes on and then the linker should complain about not finding the correct class. It is not ill formed; it depends on the use case. You can define in a separate cpp file the template A. This will obviously just reduce a bit the compile time if you compile it over and over. You can do different structures:
one a.hpp with just the class A template.
one b.cpp file with the class B along with its .h file. (is it a template?)
b.cpp includes a.hpp and inside you make an explicite template instantiation template class A; (not with extern).
At this point whenever you need to use that template you can just write
extern template class A;
in your file and link the compiled b.cpp file. If you include the a.hpp file since you still need the template you won't recompile it since you have the extern command.
I just read the wiki article about CRTP, and I'm a little confused about template instantiation.
According to the wiki,
member function bodies (definitions) are not instantiated until long
after their declarations.
I don't quite understand what it means.
Suppose I got a class template:
template <typename T>
class A
{
public:
void foo(T t)
{
//...
};
};
When I instantiate the class template A, does it instantiate the member function foo()?
For example:
//in .cpp file
int main()
{
A<int> a; //question 1
//class template is instantiated here, isn't it?
//What about foo(), is it instantiated too?
a.foo(10); //question 2
//according to the quotation, foo() will not be instantiated until it is used.
//if so, foo() is instantiated right here, not in question 1, right?
}
You seem to be confusing one thing:
Instantiation happens during compilation, not during runtime. Hence you can't say "on which line" a class template or a function template was instantiated.
That said, you're right about the fact that member function templates aren't instantiated together with class templates.
You could observe it in such a case: You have the following files
template.h (defines class A and function A::foo)
a.cpp (uses A)
b.cpp (uses A and A::foo)
Then during compilation of a.cpp, only A would be instantiated. However, during compilation of b.cpp, both would be instantiated.
Because of this, in case A::foo contained some semantically invalid code for a given set of template parameters, you would get compile errors in b.cpp, but not a.cpp.
I hope that clears things up!
With class templates, the rule of thumb is that only those members are instantiated which are actually used.
If you want complete instantiation, C++ offers explicit instantiation (however, usually you don't; the fact that not every bit is fully instantiated means that your template class is even more generic as it lowers the requirements on T, note that syntax checking and lookup of non-dependent types (stuff that is not dependent on T) still happens).
You will find a more complete answer here: Template instantiation details of GCC and MS compilers