I have a template class called Speaker, with a template member function called speak. These both have a requires clause. How do I define the member function outside of the class in the same header file?
// speaker.h
#include <concepts>
namespace prj
{
template <typename T>
requires std::is_integral<T>
struct Speaker
{
template <typename U>
requires std::is_integral<U>
const void speak(const Speaker<U> &speaker);
};
// what do I put here?
const void Speaker<T>::speak(const Speaker<U> &speaker)
{
// code
}
}
The rules for defining template members of class templates are the same in principle as they were since the early days of C++.
You need the same template-head(s) grammar component(s), in the same order
template <typename T> requires std::is_integral_v<T>
template <typename U> requires std::is_integral_v<U>
const void Speaker<T>::speak(const Speaker<U> &speaker)
{
// code
}
The associated constraint expression and template <...> form the template-head together.
As an aside:
const qualified return types are redundant in the best case, or a pessimization in the worst case. I wouldn't recommend it (especially over void).
I'd also recommend using concepts (std::integral) if including the library header, not type traits (std::is_integral) that may or may not be included. The concepts allow for the abbreviated syntax, which is much more readable:
template<std::integral T>
template<std::integral U>
I have a class template that looks like this:
foo.h
template<class C>
class Foo
{
public:
void memberFunc();
};
#include "foo.tpp"
foo.tpp
void Foo::memberFunc()
{
...
}
Ignore the .tpp file, it's just something I do to give the illusion of separating declaration and implementation, obviously that's (kind of) not possible with templates.
My implementation file is much longer in reality, and inside it I have some global scope helper function templates that the member functions use, functions that don't make sense as member functions and that I don't want users of the class to have anything to do with.
template<class C> int helper1() { ... }
template <class C> void helper2() { ... }
template<class C>
void Foo<C>::memberFunc()
{
...
helper1<float>();
...
helper2<C>();
...
}
I do this all the time in .cpp-implementation files and I forgot that when I do it in this fake version of a .cpp file, the declaration and implementation of these little helper functions actually end up in the class template header file. This leads to users of the class template header getting their namespaces cluttered with helper functions that are useless outside the member function implementations.
Obviously I could just put them in a namespace:
namespace foo_helpers
{
template<class C> void helper1() {...}
template<class C> int helper2() {...}
}
But it still leads to outside code being able to use these functions. They only matter to the implementation of the member functions and I want that to be reflected.
When looking for a solution, I learned about the concept of unnamed namespaces. As I understand it, they only allow the current translation unit access to its contents. That sounds like exactly what I need, so I changed the helper functions to this:
namespace
{
template<class C> void helper1() {...}
template<class C> void helper2() {...}
}
But it doesn't work, the functions are still usable from files that include the header.
Is there any way to hide these helper functions from outside code?
I wondered if there was any advantages of declaring templates function out of line vs in the class.
I'm trying to get a clear understanding of the pros and cons of the two syntax.
Here's an example:
Out of line:
template<typename T>
struct MyType {
template<typename... Args>
void test(Args...) const;
};
template<typename T>
template<typename... Args>
void MyType<T>::test(Args... args) const {
// do things
}
Vs in class:
template<typename T>
struct MyType {
template<typename... Args>
void test(Args... args) const {
// do things
}
};
Are there language features that are easier to use with the first or second version? Does the first version would get in the way when using default template arguments or enable_if? I would like to see comparisons of how those two cases are playing with different language features like sfinae, and maybe potential future features (modules?).
Taking compiler specific behavior into account can be interesting too. I think MSVC needs inline in some places with the first code snippet, but I'm not sure.
EDIT: I know there is no difference on how these features works, that this is mostly a matter of taste. I want to see how both syntaxes plays with different techniques, and the advantage of one over the other. I see mostly answers that favors one over another, but I really want to get both sides. A more objective answer would be better.
There is no difference between the two versions regarding default template arguments, SFINAE or std::enable_if as overload resolution and substitution of template arguments work the same way for both of them. I also don't see any reason why there should be a difference with modules, as they don't change the fact that the compiler needs to see the full definition of the member functions anyway.
Readability
One major advantage of the out-of-line version is readability. You can just declare and document the member functions and even move the definitions to a separate file that is included in the end. This makes it so that the reader of your class template doesn't have to skip over a potentially large number of implementation details and can just read the summary.
For your particular example you could have the definitions
template<typename T>
template<typename... Args>
void MyType<T>::test(Args... args) const {
// do things
}
in a file called MyType_impl.h and then have the file MyType.h contain just the declaration
template<typename T>
struct MyType {
template<typename... Args>
void test(Args...) const;
};
#include "MyType_impl.h"
If MyType.h contains enough documentation of the functions of MyType most of the time users of that class don't need to look into the definitions in MyType_impl.h.
Expressiveness
But it is not just increased readibility that differentiates out-of-line and in-class definitions. While every in-class definition can easily be moved to an out-of-line definition, the converse isn't true. I.e. out-of-line definitions are more expressive that in-class definitions. This happens when you have tightly coupled classes that rely on the functionality of each other so that a forward declaration doesn't suffice.
One such case is e.g. the command pattern if you want it to support chaining of commands and have it support user defined-functions and functors without them having to inherit from some base class. So such a Command is essentially an "improved" version of std::function.
This means that the Command class needs some form of type erasure which I'll omit here, but I can add it if someone really would like me to include it.
template <typename T, typename R> // T is the input type, R is the return type
class Command {
public:
template <typename U>
Command(U const&); // type erasing constructor, SFINAE omitted here
Command(Command<T, R> const&) // copy constructor that makes a deep copy of the unique_ptr
template <typename U>
Command<T, U> then(Command<R, U> next); // chaining two commands
R operator()(T const&); // function call operator to execute command
private:
class concept_t; // abstract type erasure class, omitted
template <typename U>
class model_t : public concept_t; // concrete type erasure class for type U, omitted
std::unique_ptr<concept_t> _impl;
};
So how would you implement .then? The easiest way is to have a helper class that stores the original Command and the Command to execute after that and just calls both of their call operators in sequence:
template <typename T, typename R, typename U>
class CommandThenHelper {
public:
CommandThenHelper(Command<T,R>, Command<R,U>);
U operator() (T const& val) {
return _snd(_fst(val));
}
private:
Command<T, R> _fst;
Command<R, U> _snd;
};
Note that Command cannot be an incomplete type at the point of this definition, as the compiler needs to know that Command<T,R> and Command<R, U> implement a call operator as well as their size, so a forward declaration is not sufficient here. Even if you were to store the member commands by pointer, for the definition of operator() you absolutely need the full declaration of Command.
With this helper we can implement Command<T,R>::then:
template <typename T, R>
template <typename U>
Command<T, U> Command<T,R>::then(Command<R, U> next) {
// this will implicitly invoke the type erasure constructor of Command<T, U>
return CommandNextHelper<T, R, U>(*this, next);
}
Again, note that this doesn't work if CommandNextHelper is only forward declared because the compiler needs to know the declaration of the constructor for CommandNextHelper. Since we already know that the class declaration of Command has to come before the declaration of CommandNextHelper, this means you simply cannot define the .then function in-class. The definition of it has to come after the declaration of CommandNextHelper.
I know that this is not a simple example, but I couldn't think of a simpler one because that issue mostly comes up when you absolutely have to define some operator as a class member. This applies mostly to operator() and operator[] in expession templates since these operators cannot be defined as non-members.
Conclusion
So to conclude: It is mostly a matter of taste which one you prefer, as there isn't much of a difference between the two. Only if you have circular dependencies among classes you can't use in-class defintion for all of the member functions. I personally prefer out-of-line definitions anyway, since the trick to outsource the function declarations can also help with documentation generating tools such as doxygen, which will then only create documentation for the actual class and not for additional helpers that are defined and declared in another file.
Edit
If I understand your edit to the original question correctly, you'd like to see how general SFINAE, std::enable_if and default template parameters looks like for both of the variants. The declarations look exactly the same, only for the definitions you have to drop default parameters if there are any.
Default template parameters
template <typename T = int>
class A {
template <typename U = void*>
void someFunction(U val) {
// do something
}
};
vs
template <typename T = int>
class A {
template <typename U = void*>
void someFunction(U val);
};
template <typename T>
template <typename U>
void A<T>::someFunction(U val) {
// do something
}
enable_if in default template parameter
template <typename T>
class A {
template <typename U, typename = std::enable_if_t<std::is_convertible<U, T>::value>>
bool someFunction(U const& val) {
// do some stuff here
}
};
vs
template <typename T>
class A {
template <typename U, typename = std::enable_if_t<std::is_convertible<U, T>::value>>
bool someFunction(U const& val);
};
template <typename T>
template <typename U, typename> // note the missing default here
bool A<T>::someFunction(U const& val) {
// do some stuff here
}
enable_if as non-type template parameter
template <typename T>
class A {
template <typename U, std::enable_if_t<std::is_convertible<U, T>::value, int> = 0>
bool someFunction(U const& val) {
// do some stuff here
}
};
vs
template <typename T>
class A {
template <typename U, std::enable_if_t<std::is_convertible<U, T>::value, int> = 0>
bool someFunction(U const& val);
};
template <typename T>
template <typename U, std::enable_if_t<std::is_convertible<U, T>::value, int>>
bool A<T>::someFunction(U const& val) {
// do some stuff here
}
Again, it is just missing the default parameter 0.
SFINAE in return type
template <typename T>
class A {
template <typename U>
decltype(foo(std::declval<U>())) someFunction(U val) {
// do something
}
template <typename U>
decltype(bar(std::declval<U>())) someFunction(U val) {
// do something else
}
};
vs
template <typename T>
class A {
template <typename U>
decltype(foo(std::declval<U>())) someFunction(U val);
template <typename U>
decltype(bar(std::declval<U>())) someFunction(U val);
};
template <typename T>
template <typename U>
decltype(foo(std::declval<U>())) A<T>::someFunction(U val) {
// do something
}
template <typename T>
template <typename U>
decltype(bar(std::declval<U>())) A<T>::someFunction(U val) {
// do something else
}
This time, since there are no default parameters, both declaration and definition actually look the same.
Are there language features that are easier to use with the first or second version?
Quite trivial a case, but it's worth to be mentioned: specializations.
As an example, you can do this with out-of-line definition:
template<typename T>
struct MyType {
template<typename... Args>
void test(Args...) const;
// Some other functions...
};
template<typename T>
template<typename... Args>
void MyType<T>::test(Args... args) const {
// do things
}
// Out-of-line definition for all the other functions...
template<>
template<typename... Args>
void MyType<int>::test(Args... args) const {
// do slightly different things in test
// and in test only for MyType<int>
}
If you want to do the same with in-class definitions only, you have to duplicate the code for all the other functions of MyType (supposing test is the only function you want to specialize, of course).
As an example:
template<>
struct MyType<int> {
template<typename... Args>
void test(Args...) const {
// Specialized function
}
// Copy-and-paste of all the other functions...
};
Of course, you can still mix in-class and out-of-line definitions to do that and you have the same amount of code of the full out-of-line version.
Anyway I assumed you are oriented towards full in-class and full out-of-line solutions, thus mixed ones are not viable.
Another thing that you can do with out-of-line class definitions and you cannot do with in-class definitions at all is function template specializations.
Of course, you can put the primary definition in-class, but all the specializations must be put out-of-line.
In this case, the answer to the above mentioned question is: there exist even features of the language that you cannot use with one of the version.
As an example, consider the following code:
struct S {
template<typename>
void f();
};
template<>
void S::f<int>() {}
int main() {
S s;
s.f<int>();
}
Suppose the designer of the class wants to provide an implementation for f only for a few specific types.
He simply can't do that with in-class definitions.
Finally, out-of-line definitions help to break circular dependencies.
This has been already mentioned in most of the other answers and it doesn't worth it to give another example.
Separating the declaration from the implementation allows you to do this:
// file bar.h
// headers required by declaration
#include "foo.h"
// template declaration
template<class T> void bar(foo);
// headers required by the definition
#include "baz.h"
// template definition
template<class T> void bar(foo) {
baz();
// ...
}
Now, what would make this useful? Well, the header baz.h may now include bar.h and depend on bar and other declarations, even though the implementation of bar depends on baz.h.
If the function template was defined inline, it would have to include baz.h before declaring bar, and if baz.h depends on bar, then you'd have a circular dependency.
Besides resolving circular dependencies, defining functions (whether template or not) out-of-line, leaves the declarations in a form that works effectively as a table of contents, which is easier for programmers to read than declarations sprinkled across a header full of definitions. This advantage diminishes when you use specialized programming tools that provide a structured overview of the header.
I tend to always merge them - but you can't do that if they are codependent. For regular code you usually put the code in a .cpp file, but for templates that whole concept doesn't really apply (and makes for repeated function prototypes). Example:
template <typename T>
struct A {
B<T>* b;
void f() { b->Check<T>(); }
};
template <typename T>
struct B {
A<T>* a;
void g() { a->f(); }
};
Of course this is a contrived example but replace the functions with something else. These two classes require each other to be defined before they can be used. If you use a forward declaration of the template class, you still cannot include the function implementation for one of them. That's a great reason to put them out of line, which 100% fixes this every time.
One alternative is to make one of these an inner class of the other. The inner class can reach out into the outer class beyond its own definition point for functions so the problem is kind of hidden, which is usable in most cases when you have these codependent classes.
A template class is a multitude of possible classes, so I was asking me: the preprocessor directives
#ifndef MY_CLASS
#define MY_CLASS
template<typename T>
class My_Class
{};
#endif
are necessaries?
The include guards prevent repeated inclusion of the same file. You need them to prevent a translation unit that would contain the following definitions, which are illegal:
template <typename T> class My_Class { };
template <typename T> class My_Class { };
// Error: redefinition of ‘class Foo<T>’
Repeat inclusion is very easy to occur. For example, consider a case where you include A and B, and A already includes B for some undocumented reason. You may not be entitled to omit B from your explicit includes, but neither should A be required to not include it. Only by using include guards (or some equivalent mechanism) can you make such a file structure possible.
In a nutshell: You can have repeated declarations, but only one definition within one translation unit.
Counter example: The following is legal:
template <typename T> class Foo;
template <typename T> class Foo;
template <typename T> class Foo;
template <typename T> class Foo;
template <typename T> class Foo { };
int main() {}
The macros are probably used as "include guards", preventing multiple-declarations errors if the header happens to be included multiple times. So yes, they are "as necessary" as for everything else: if you can be sure that the header won't ever be included multiple times you could leave them out, but on the other hand, they are not really adding to file sizes or compilation times.
Is there a difference between defining member functions for a template class inside the class declaration versus outside?
Defined inside:
template <typename T>
class A
{
public:
void method()
{
//...
}
};
Defined outside:
template <typename T>
class B
{
public:
void method();
};
template <typename T>
void B<T>::method()
{
//...
}
For non-template classes, this is the difference between inlined and non-inlined methods. Is this also true for template classes?
The default for most of my colleagues is to provide definitions inside the class, but I've always preferred definitions outside the class. Is my preference justified?
Edit: Please assume all the above code is provided in the header file for the class.
Yes, the exact same is true for template classes.
The reason why method definitions for template classes are usually preferred to be inline is that with templates, the entire definition must be visible when the template is instantiated.
So if you put the function definition in some separate .cpp file, you'll get a linker error.
The only general solution is to make the function inline, either by defining it inside the class or outside with the inline keyword. but in either cases, it must be visible anywhere the function is called, which means it must typically be in the same header as the class definition.
There's no difference, aside from having to type more. That includes the template bit, the inline and having to use more "elaborate" names when referring to the class. For example
template <typename T> class A {
A method(A a) {
// whatever
}
};
template <typename T> inline A<T> A<T>::method(A a) {
// whatever
}
Note that when the method is defined inside you can always omit the template parameter list <T> when referring to A<T> and just use A. When defining it outside, you have to use the "full" name in the return type and in the name of the method (but not in the parameter list).
I know this..I think it must be some what help full to u?
defining a member function outside of its template
It is not ok to provide the definition of a member function of a template class like this:
// This might be in a header file:
template <typename T>
class xyz {
void foo();
};
// ...
// This might be later on in the header file:
void xyz<T>::foo() {
// generic definition of foo()
}
This is wrong for a few reasons. So is this:
void xyz<class T>::foo() {
// generic definition of foo()
}
The proper definition needs the template keyword and the same template arguments that the class template's definition was declared with. So that gives:
template <typename T>
void xyz<T>::foo() {
// generic definition of foo()
}
Note that there are other types of template directives, such as member templates, etc., and each takes on their own form. What's important is to know which you have so you know how to write each flavor. This is so especially since the error messages of some compilers may not be clear as to what is wrong. And of course, get a good and up to date book.
If you have a nested member template within a template:
template <typename T>
struct xyz {
// ...
template <typename U>
struct abc;
// ...
};
How do you define abc outside of xyz? This does not work:
template <typename U>
struct xyz::abc<U> { // Nope
// ...
};
nor does this:
template <typename T, typename U>
struct xyz<T>::abc<U> { // Nope
// ...
};
You would have to do this:
template <typename T>
template <typename U>
struct xyz<T>::abc {
// ...
};
Note that it's ...abc not ...abc<U> because abc is a "primary" template. IOWs, this is not good:
// not allowed here:
template template struct xyz::abc { };