Is it possible to somehow forbid using templated function for types for which specialization was not explicitly written. I mean something like that
template <typename T>
void foo(){}
template <>
void foo<int>(){}
int main(int argc, char* argv[]){
foo<int>(); //ok
foo<char>(); //Wrong - no specialized version for char.
}
I cannot skip generic version of function, cause then compiler says, that foo is not a template function when i try to specialize. I could simply write something that does not compile in generic function, and add some comment explaining why, but this would be quite non-informative. What i would like to do, is to be able to directly cause compiler to go with error like "foo() is not defined".
Sure: just don't define it and you'll get a linker error if you try to use it:
template <typename T>
void foo(); // not defined
template <>
void foo<int>() { }
Alternatively, you can use some variation of a static assert to give a "nicer" compile-time error. Here is an example using the C++0x static_assert. Note that you must make the false value dependent upon the template parameter, otherwise the static_assert may be triggered when the template is parsed.
template <typename T>
struct dependent_false { enum { value = false }; };
template <typename T>
void foo()
{
static_assert(dependent_false<T>::value, "Oops, you used the primary template");
}
Note that it's usually best not to specialize function templates. Instead, it is better to delegate to a specialized class template:
template <typename T>
struct foo_impl
{
static_assert(dependent_false<T>::value, "Oops, you used the primary template");
};
template<>
struct foo_impl<int>
{
static void foo() { }
};
template <typename T>
void foo()
{
return foo_impl<T>::foo();
}
Sure, just don't supply definition for the default generic template.
Related
I'm setting up a function that initializes tuples based on a tuple type and a functor struct For that has a size_t template argument INDEX to retain the compile-time index. This functor may also depend on other template arguments T.... Because of this the functors exist within other structures (TClass in this example) that hold these template arguments.
The initialization function (called Bar here) has a template<std::size_t> class template argument to ensure that the used class actually can store the index.
While the design I've come up with works fine when I call it from a non-template function, it does not compile if the template T2 of a function does determine the template parameter of the wrapper TClass.
Here is the definition of the functor For wrapped inside TClass:
#include <cstdlib>
template <typename T> struct TClass {
template<std::size_t INDEX> struct For {
void operator()() {}
};
};
And here are the function calls i want to use:
template <template<std::size_t> class FOR> void bar() {
//...
}
template <typename T> void foo() {
bar<TClass<T>::For>(); //Does not compile
}
int main() {
bar<TClass<int>::For>(); //Works
foo<int>();
return 0;
}
The compiler output for the faulty foo-call is:
error: dependent-name ‘TClass<T>::For’ is parsed as a non-type, but instantiation yields a type
Bar<TClass<T>::For>(); //Does not compile
I know that dependent type names usually have to be preceded by a typename but this is also not necessary for the first bar-call. I assumed it was because the template argument can only be interpreted as a type. So I thought that maybe typename would result in correct compilation but if I change foo to
template <typename T> void foo() {
bar<typename TClass<T>::For>(); //Does not compile
}
I get:
error: ‘typename TClass<int>::For’ names ‘template<long unsigned int INDEX> struct TClass<int>::For’, which is not a type
Bar<typename TClass<T>::For>(); //Does not compile
I've also come up with a design where the ()-operator of TClass depends on the template INDEX which also works fine because it is not necessary to use nested types anymore. It looks like this:
#include <cstdlib>
template <typename T> struct TClass {
template<std::size_t INDEX> void operator()() {}
};
template <typename FOR> void bar() {
//...
}
template <typename T> void foo() {
bar<TClass<T>>(); //Does compile
}
Apparently it is not possible to use dependent type names in functions where the template of the type is determined by the function's template parameters, but why? And how do I implement this correctly? To make writing future type checks with type traits easier I would prefer it if I can use a functor.
The compiler cannot know that TClass<T>::For refers to a template at the first stage of template instantiation. It needs a bit of help with template keyword. Fix:
template <typename T> void foo() {
bar<TClass<T>::template For>();
}
If we have a standard class:
class Foo {
public:
int fooVar = 10;
int getFooVar();
}
The implementation for getFooVar() would be:
int Foo::getFooVar() {
return fooVar;
}
But in a templated class:
template <class T>
class Bar {
public:
int barVar = 10;
int getBarVar();
}
The implementation for getBarVar() must be:
template <class T>
int Bar<T>::getBarVar(){
return barVar();
}
Why must we have the template <class T> line before the function implementation of getBarVar and Bar<T>:: (as opposed to just Bar::), considering the fact that the function doesn't use any templated variables?
You need it because Bar is not a class, it's a template. Bar<T> is the class.
Bar itself is a template, as the other answers said.
But let's now assume that you don't need it, after all, you specified this, and I added another template argument:
template<typename T1, typename T2>
class Bar
{
void something();
};
Why:
template<typename T1, typename T2>
void Bar<T1, T2>::something(){}
And not:
void Bar::something(){}
What would happen if you wanted to specialize your implementation for one type T1, but not the other one? You would need to add that information. And that's where this template declaration comes into play and why you also need it for the general implementation (IMHO).
template<typename T>
void Bar<T, int>::something(){}
When you instantiate the class, the compiler checks if implementations are there. But at the time you write the code, the final type (i.e. the instantiated type) is not known.
Hence the compiler instantiates the definitions for you, and if the compiler should instantiate something it needs to be templated.
Any answer to this question boils down to "because the standard says so". However, instead of reciting standardese, let's examine what else is forbidden (because the errors help us understand what the language expects). The "single template" case is exhausted pretty quickly, so let's consider the following:
template<class T>
class A
{
template<class X>
void foo(X);
};
Maybe we can use a single template argument for both?
template<class U>
void A<U>::foo(U u)
{
return;
}
error: out-of-line definition of 'foo' does not match any declaration in 'A<T>'
No, we cannot. Well, maybe like this?
template<class U>
void A<U>::foo<U>(U u)
{
return;
}
error: cannot specialize a member of an unspecialized template
No. And this?
template<class U, class V>
void A<U>::foo(V u)
{
return;
}
error: too many template parameters in template redeclaration
How about using a default to emulate the matching?
template<class U>
template<class V = U>
void A<U>::foo(V u)
{
return;
}
error: cannot add a default template argument to the definition of a member of a class template
Clearly, the compiler is worried about matching the declaration. That's because the compiler doesn't match template definitions to specific calls (as one might be used to from a functional language) but to the template declaration. (Code so far here).
So on a basic level, the answer is "because the template definition must match the template declaration". This still leaves open the question "why can we not just omit the class template parameters then?" (as far as I can tell no ambiguity for the template can exist so repeating the template parameters does not help) though...
Consider a function template declaration
tempalte <typename T>
void foo();
now a definition
void foo() { std::cout << "Hello World"; }
is either a specialization of the above template or an overload. You have to pick either of the two. For example
#include <iostream>
template <typename T>
void foo();
void foo() { std::cout << "overload\n"; }
template <typename T>
void foo() { std::cout << "specialization\n"; }
int main() {
foo();
foo<int>();
}
Prints:
overload
specialization
The short answer to your question is: Thats how the rules are, though if you could ommit the template <typename T> from a definition of the template, a different way would be required to define an overload.
Let's suppose to have a templateclass Foo:
template <typename T>
class Foo {
void foo();
};
I have another template class Bar (independent from the first one):
template <int N>
class Bar {};
Let's say, I want to specialise the foo() method for whatever Bar class.
I'd wrongly write:
template <>
template <int N>
void Foo<Bar<N> >::foo() { /* ... */ }
The compiler blames me for because the type is not complete:
error: invalid use of incomplete type 'class Foo<Bar<N> >'
void Foo<Bar<N> >::foo() { }
Code
I am using C++98, but I'd like to know if there exist different solutions in C++11.
Note
I could solve the problem specialising the entire class Foo for a generic Bar, but after I should have to define all methods.
Example Code
That's not what I want, I am looking for (if exists) more elegant solution (both C++98 and C++11) which allows me to specialise and implement only a single class method.
EDIT:
The question on SO does not explain how to specialise with a template argument. Indeed, my question shows how the compiler complains about that.
For C++11 you can SFINAE enable/disable (using std::enable_if) two differents versions of foo() inside a not specialized Foo class.
In C++98 you don't have std::enable_if but you can simulate it (give me some minutes and I try to propose an example). Sorry: my idea doesn't works because this method require the use of default template arguments for methods that is a C++11 innovation.
Another way is define a template base class for Foo(), say FooBase, insert foo() (and only foo()) in FooBase and specialize FooBase.
Another way, that works also with C++98, can be tag dispatching: you can define an unique foo(), with zero parameter, that call another foo(), with a parameter that is determined by T.
The following is a full (C++98 compilable) example
#include <iostream>
struct barWay {};
struct noBarWay {};
template <int>
struct Bar
{ };
template <typename>
struct selectType
{ typedef noBarWay type; };
template <int N>
struct selectType< Bar<N> >
{ typedef barWay type; };
template <typename T>
struct Foo
{
void foo (noBarWay const &)
{ std::cout << "not Bar version" << std::endl; }
void foo (barWay const &)
{ std::cout << "Bar version" << std::endl; }
void foo ()
{ foo(typename selectType<T>::type()); }
};
int main ()
{
Foo<int> fi;
Foo< Bar<42> > fb;
fi.foo();
fb.foo();
}
if a common base is not desirable, yet another way could be giving foo() a customization point, like a trait for example:
template <typename T>
struct foo_traits;
template <typename T>
struct Foo {
void foo(){ foo_traits<T>::foo_cp(*this); }
};
template <typename T>
struct foo_traits{ static void foo_cp(T&){/*default*/} };
template <int N>
class Bar {};
template <int N>
struct foo_traits<Bar<N>>{ static void foo_cp(Foo<Bar<N>>&){/*spec*/} };
such trait could also be an implementation detail friend, if its only purpose is to internally provide a foo() specialization for Bar's.
If you cannot specialize foo, define it so that it delegates the call to an internal foo-implementation class. Then specialize that class.
Something like this should compile in C++98 and it doesn't differ much from your original code:
template <typename T>
class Foo {
template<typename>
struct FooImpl;
public:
void foo() { FooImpl<T>()(); }
};
template <int N>
class Bar {};
template <typename T>
template <int N>
struct Foo<T>::FooImpl< Bar<N> > {
void operator()() { /* ... */ }
};
int main() {
Foo< Bar<0> > fb;
fb.foo();
Foo<int> fi;
//fi.foo();
}
The last line doesn't compile as expected (at least I got it was the expected result, just define the function call operator for FooImpl otherwise).
This way you can define selectively the specializations for which you want foo to work. In all the other cases, an attempt at using foo will result in a compilation error.
I'd like to know if there exist different solutions in C++11.
This is a classic use case for tagged dispatch, of which max66 already suggested. The approach, and even syntax, are basically the same in C++98 and C++11.
Here's a bit of a cleaner implementation than max66's, I believe (running on godbolt):
template <class T>
class Foo {
template <class>
struct tag{};
template<class U>
void foo_helper(tag<U>){std::cout << "default\n";}
void foo_helper(tag<Bar<3> >){std::cout << "specialization for Bar<3>\n";}
public:
void foo(){return foo_helper(tag<T>());}
};
The principle is the same; a client function accepting no arguments calls a helper function that constructs an empty type based on the T argument. Then normal overloading takes care of the rest.
Only here I use a templated catch-all method.
In C++11 the syntax would only change slightly; We could say tag<Bar<3>> instead of tag<Bar<3> > because new parsing rules allow the chevron for nested templates.
We could also make the tag and the templated foo_helper catch-all into variadic templates to be a little more generic:
template <class T>
class Foo {
template <class...>
struct tag{};
template<class... U>
void foo_helper(tag<U...>){std::cout << "default\n";}
void foo_helper(tag<Bar<3>>){std::cout << "specialization for Bar<3>\n";}
public:
void foo(){return foo_helper(tag<T>{});}
};
Things actually start getting pretty interesting in C++17 with the introduction of constexpr if that allows us to write what looks like normal branching logic based on T (Live Demo):
template <class T>
class Foo {
public:
void foo(){
if constexpr (std::is_same_v<T, Bar<3>>){std::cout << "Specialization for Bar<3>\n";}
else std::cout << "default\n";
}
};
As you can see, all the tag stuff goes away in favor of using a simple if statement.
We take advantage of type_traits introduced in C++11 to check the type of T against our desired type. Something like this wouldn't necessarily work previously because all branches needed to be compiled. In C++17, only the branch that is selected (at compile-time) is compiled.
Note that you could emulate this behavior as early as C++98 by using typeid (godbolt demo):
void foo(){
if (typeid(T) == typeid(Bar<3>)){std::cout << "Specialization for Bar<3>\n";}
else std::cout << "default\n";
}
However, the typeid approach is a poor choice for 2 reasons:
It's a run time check (slow) for information we know at compile-time
It's brittle because all branches must compile for all template instantiations, whereas in C++17 if constexpr only compiles the branch that is selected.
I am working on Visual Studio 2015 community edition
let's say I have, a simple class like this:
(The example below "should be" a compilable because it include all the necessary stuff, unfortunately, it produces an error).
#include <stdexcept>
template <typename T>
class class_foo
{
// members, methods, constructors. not important stuff...
// helper functions:
private:
class tag_aaa {}; // to resolve few things at compile-time, not run-time.
class tag_bbb {}; // - || -
template <typename tag>
void erase();
// for some reason this is not interpreted as an error by my compiler:
template<>
void erase<tag_aaa>();
template<>
void erase<tag_bbb>();
};
// catch-all-do-nothing "version"
// well, catch-all-throw-an-exception because call to this function is an obvious error.
// that should never occur.
template <typename T>
template <typename tag> inline
void class_foo<T>::erase()
{
throw std::runtime_error("Very weird error...");
}
template <>
template <typename T> inline
void class_foo<T>::erase<class_foo<T>::tag_aaa>()
{
// do some stuff...
}
template <>
template <typename T> inline
void class_foo<T>::erase<class_foo<T>::tag_bbb>()
{
// do some stuff...
}
int main()
{
class_foo<double> bar;
return 0;
}
The error:
1>D:/develop/workspace/visual_studio/nevada_test_site/source/workspace/nevada_test_site/start.cu(36): error : partial specialization of class "class_foo<T>::erase<class_foo<T>::tag_aaa> [with T=T]" is not allowed
1>D:/develop/workspace/visual_studio/nevada_test_site/source/workspace/nevada_test_site/start.cu(43): error : partial specialization of class "class_foo<T>::erase<class_foo<T>::tag_bbb> [with T=T]" is not allowed
1>D:/develop/workspace/visual_studio/nevada_test_site/source/workspace/nevada_test_site/start.cu(51): warning : variable "bar" was declared but never referenced
I think about myself as a junior-hobbyist programmer, so certainly I am wrong, but I believe that both erase<class_foo<T>::tag_aaa>() and erase<class_foo<T>::tag_bbb>() are explicit specializations of the template <typename tag> void erase(); function. And as such, they are allowed. I believe that this error is due to some bad syntax but I can't find an error.
Question:
Is what I am trying to do, allowed?
If yes, what am I doing wrong?
If yes, what is the correct syntax for specializing this functions (erase)?
It look like full specialization of a template function but it's still partial specialization, hence the compilation error.
Why is it? Well, look at this specialization:
template <>
template <typename T>
inline void class_foo<T>::erase<class_foo<T>::tag_bbb>() {
// do some stuff...
}
You said it's a explicit specialization, but there is still a template parameter to fill! There's the parameter T yet to be known. So a specialization... that is still a template? That's a partial specialization!
Partial specialization of function is not allowed, for many reason. One of them is that it won't play nicely with overloading.
To effectively specialize the function, you must leave no template parameter to be known, something like this:
template<>
template<>
inline void class_foo<int>::erase<class_foo<int>::tag_bbb>() {
// do some stuff...
}
But it's not what you want.
Here's how I'd fix this problem. Use overloading instead of specializing:
template<typename T>
struct class_foo {
private:
struct tag_aaa {};
struct tag_bbb {};
void erase(tag_aaa) {
// Stuff when tag_aaa
}
void erase(tag_bbb) {
// Stuff when tag_bbb
}
};
Instead of invoking those like this:
erase<tag_aaa>(); // with specialization
You must invoke it like that:
erase(tag_aaa{}); // with overloading
We have complex template classes that have some methods which will not work with certain policies or types. Therefore, when we detect those types (at compile time, using type-traits) we fire a static assertion with a nice message.
Now we do a lot of manual template instantiation as well. Partly it is so that the methods are forced to compiler to syntax check the methods. It also reduces compile time for the library user. The problem is that the static assertions are always fired and consequently we cannot manually instantiate the template class in question.
Is there a workaround for this?
EDIT: To make it clearer, here is an example (the explicit instantiation in this case will fail on someFunc1():
// header
template <typename T>
class someClass
{
void someFunc() {}
void someFunc1() { static_assert(false, assertion_failed); }
};
// source
template someClass<int>; // Explicit instantiation
EDIT2: Here is another example. This time you can compile it to see what I mean. First compile right away. The code should compile. Then Uncomment [2] and the static assertion should fire. Now comment out [2] and Uncomment [1]. The static assertion will fire because you are explicitly instantiating the template. I want to avoid removing explicit instantiation because of the benefits that come with it (see above for benefits).
namespace Loki
{
template<int> struct CompileTimeError;
template<> struct CompileTimeError<true> {};
}
#define LOKI_STATIC_CHECK(expr, msg) \
{ Loki::CompileTimeError<((expr) != 0)> ERROR_##msg; (void)ERROR_##msg; }
template <typename T>
class foo
{
public:
void func() {}
void func1() { LOKI_STATIC_CHECK(sizeof(T) == 4, Assertion_error); }
};
template foo<int>;
//template foo<double>; // [1]
int main()
{
foo<int> a;
a.func1();
foo<double> b;
//b.func1(); //[2]
return 0;
}
You can't have both: you can't have a static assertion to prevent instantiation and explicitly instantiate the type! This is an obvious contradiction. What you can have, however, is conditionally included functionality even though it is somewhat a pain in the neck: If a certain member function is not supposed to be supported for certain types, you can move this function into a base class which conditionally has it. This way you wouldn't use a static assertion but just remove the member function. I realize that this introduces interesting other problems, e.g. with respect to the location of member variables, but I think in the context you are describing this is the best you can get.
Here is a quick example of how this could look like:
template <typename T, bool = std::numeric_limits<T>::is_integer> struct foo_base;
template <typename T> struct foo_base<T, false> { /* intentionally left blank */ };
template <typename T> struct foo_base<T, true> { void foo() { /*...*/ } };
template <typename T>
struct Foo: foo_base<T> { /* .... */ };
template struct Foo<int>; // will have foo()
template struct Foo<double>; // will not have foo()
Alright, so if you're forcing the instantiation of all methods using explicit instantiation, you can't get away with any compile time tricks to prevent instantiation of the offending methods, such as enable_if. It'd be easy enough to move the error to runtime, but that's undesirable.
I think the best you can do is move the error to link time, which will statically ensure that the program does not contain a code path that could potentially call the prohibited function, but the error messages won't be very helpful to anyone that doesn't know about the restriction you're imposing. Anyway, the solution is to declare a specialization of the prohibited member functions but not define them:
template<typename T>
struct Foo {
void bar() {
std::cout << "bar\n";
}
void baz() {
std:: cout << "baz\n";
}
};
template<> void Foo<int>::baz(); // use of Foo<int>::baz() will resolve to this specialization, and linking will fail
template struct Foo<int>;
template struct Foo<char>;
int main() {
Foo<int> f;
f.bar();
// f.baz(); // uncommenting this line results in an ugly link time error
Foo<char> b;
b.bar();
b.baz(); // works with Foo<char>
}
The static asserts no longer help give nice error messages when a mistake is made in client code, but you might want to leave them in because they'll fire if you forget to provide a specialization.
enable_if is a flexible mechanism for precise template methods targeting, may be what you are after. Example:
#include <string>
#include <iostream>
#include <boost/utility.hpp>
#include <boost/type_traits.hpp>
#include <boost/static_assert.hpp>
template <class T> class mywrapper
{
T _value;
template <class V>
typename boost::enable_if<boost::is_scalar<V>, void>::type printval_(V const& value)
{
BOOST_STATIC_ASSERT(boost::is_scalar<V>::value);
std::cout << "scalar: " << value << std::endl;
}
template <class V>
typename boost::enable_if<boost::is_compound<V>, void>::type printval_(V const& value)
{
BOOST_STATIC_ASSERT(boost::is_compound<V>::value);
std::cout << "compound: " << value << std::endl;
}
public:
mywrapper(T const& value):_value(value) { }
void printval() { printval_(_value); }
};
template class mywrapper<int>;
template class mywrapper<std::string>;
int main()
{
mywrapper<int> ival(333);
mywrapper<std::string> sval("test");
ival.printval();
sval.printval();
return 0;
}
I did not get an opportunity to test enable_if as suggested by bobah but I did come up with a solution that does not require boost and that satisfies my original requirement to a good extent (I say good and not full, will explain at the end)
The solution is to put a dummy template on the code that will fail if compiled under some selected types and is fine under others. So:
struct dummyStruct {};
#define DUMMY_TEMP typename dummy
#define DUMMY_PARAM dummyStruct
namespace Loki
{
template<int> struct CompileTimeError;
template<> struct CompileTimeError<true> {};
}
#define LOKI_STATIC_CHECK(expr, msg) \
{ Loki::CompileTimeError<((expr) != 0)> ERROR_##msg; (void)ERROR_##msg; }
template <typename T>
class foo
{
public:
void func() {}
template <typename T_Dummy>
void func1() { LOKI_STATIC_CHECK(sizeof(T) == 4, Assertion_error); }
};
template foo<int>;
template foo<double>; // [1]
int main()
{
foo<int> a;
a.func1<DUMMY_PARAM>();
foo<double> b;
//b.func1<DUMMY_PARAM>(); //[2] - this is a static error
return 0;
}
In all of my template code, these kind of functions (i.e. the ones that have static asserts OR work on some types and may fail on others by using type traits [in which case there is a selection of several different functions for different types]) are hidden from the client. So in my implementation, adding the extra dummy parameter is an OK compromise.
As a bonus, it lets me know that this function is designed to be used by only certain types. Furthermore, my original problem of explicit instantiation is solved by this simple technique.