There are a few questions already similar to this already on stack overflow, but nothing that seemd to directly answer the question I have. I do apologise if I am reposting.
I'd like to overload a few methods of a templated class (with 2 template parameters) with a partial template specialisation of those methods. I haven't been able to figure out the correct syntax, and am starting to think that it's not possible. I thought I'd post here to see if I can get confirmation.
Example code to follow:
template <typename T, typename U>
class Test
{
public:
void Set( T t, U u );
T m_T;
U m_U;
};
// Fully templated method that should be used most of the time
template <typename T, typename U>
inline void Test<T,U>::Set( T t, U u )
{
m_T=t;
m_U=u;
}
// Partial specialisation that should only be used when U is a float.
// This generates compile errors
template <typename T>
inline void Test<T,float>::Set( T t, float u )
{
m_T=t;
m_U=u+0.5f;
}
int _tmain(int argc, _TCHAR* argv[])
{
Test<int, int> testOne;
int a = 1;
testOne.Set( a, a );
Test<int, float> testTwo;
float f = 1.f;
testTwo.Set( a, f );
}
I know that I could write a partial specialisation of the entire class, but that kinda sucks. Is something like this possible?
(I'm using VS2008)
Edit: Here is the compile error
error C2244: 'Test::Set' : unable to match function definition to an existing declaration
Thanks :)
You cannot partially specialize a member function without defining partial specialization of the class template itself. Note that partial specialization of a template is STILL a template, hence when the compiler sees Test<T, float>, it expects a partial specialization of the class template.
--
$14.5.4.3/1 from the C++ Standard (2003) says,
The template parameter list of a
member of a class template partial
specialization shall match the
template parameter list of the class
template partial specialization. The
template argument list of a member of
a class template partial
specialization shall match the
template argument list of the class
template partial specialization. A
class template specialization is a
distinct template. The members of the
class template partial specialization
are unrelated to the members of the
primary template. Class template
partial specialization members that
are used in a way that requires a
definition shall be defined; the
definitions of members of the primary
template are never used as definitions
for members of a class template
partial specialization. An explicit
specialization of a member of a class
template partial specialization is
declared in the same way as an
explicit specialization of the primary
template.
Then the Standard itself gives this example,
// primary template
template<class T, int I> struct A {
void f();
};
template<class T, int I> void A<T,I>::f() { }
// class template partial specialization
template<class T> struct A<T,2> {
void f();
void g();
void h();
};
// member of class template partial specialization
template<class T> void A<T,2>::g() { }
I hope the quotation from the Standard along with the example answers your question well.
The particular problem you're sketching is easy:
template< class T >
inline T foo( T const& v ) { return v; }
template<>
float foo( float const& v ) { return v+0.5; }
Then call foo from your Test::Set implementation.
If you want the full generality, then similarly use a helper class with static helper member functions, and partially specialize that helper class.
Cheers & hth.,
There's also another solution to the partial specialization problem, if you don't want to introduce additional functions, methods or classes to your code.
#include <type_traits>
template <typename T1, typename T2>
class C
{
void f(T1 t1);
}
template <typename T1, typename T2>
void C<T1, T2>::f(T1 t1)
{
if (std::is_same<T2, float>::value)
// Do sth
else
// Do sth
}
Related
Note: this seems to be a repost of a problem: C++ - Overload templated class method with a partial specilization of that method
I have boiled down a problem I am having with C++ template specialization down to a simple case.
It consists of a simple 2-parameter template class Thing, where I would like to specialize Thing<A,B>::doSomething() for B=int.
#include <cstdio>
// A 3-parameter template class.
template <class A, class B>
class Thing
{
public:
Thing(A a, B b) : a_(a), b_(b) {}
B doSomething();
private:
A a_;
B b_;
};
// The generic case works as expected.
template <class A, class B>
B Thing<A,B>::doSomething()
{
return b_;
}
// This specialization does not work!
template <class A>
int Thing<A,int>::doSomething()
{
return b_+1;
}
int main() {
// Setup our thing.
Thing<double,int> thing(1.0,2);
// This doesn't compile - but works with the generic case.
printf("Expecting 3, and getting %i\n", thing.doSomething());
// Clean up.
return 0;
}
Unfortunately, g++ exits with the error:
partial_specialization.cpp:30: error: invalid use of incomplete type ‘class Thing<A, int>’
partial_specialization.cpp:8: error: declaration of ‘class Thing<A, int>’
The clang++ compiler is a bit more verbose, but has the same problem:
partial_specialization.cpp:30:19: error: nested name specifier 'Thing<A, int>::' for declaration does not
refer into a class, class template or class template partial specialization
int Thing<A,int>::doSomething()
~~~~~~~~~~~~~~^
partial_specialization.cpp:32:12: error: use of undeclared identifier 'b_'
return b_+1;
^
2 errors generated.
I have read and understood that partial template specializations on functions aren't allowed - but I thought I was partially specializing over classes of Thing in this case.
Any ideas?
What I did: A workaround, as determined from the link provided by the accepted answer:
template< class T >
inline T foo( T const & v ) { return v; }
template<>
inline int foo( int const & v ) { return v+1; }
// The generic case works as expected.
template <class A, class B>
B Thing<A,B>::doSomething()
{
return foo(b_);
}
Partial specialization of a function template, whether it is member function template or stand-alone function template, is not allowed by the Standard:
template<typename T, typename U> void f() {} //okay - primary template
template<typename T> void f<T,int>() {} //error - partial specialization
template<> void f<unsigned char,int>() {} //okay - full specialization
But you can partially specialize the class template itself. You can do something like this:
template <class A>
class Thing<A,int> //partial specialization of the class template
{
//..
int doSomething();
};
template <class A>
int Thing<A,int>::doSomething() { /* do whatever you want to do here */ }
Note that when you partially specialize a class template, then the template parameter-list of member function (in its definition outside the class), must match the template parameter list of the class template partial specialization. That means, for the above partial specialization of the class template, you cannot define this:
template <class A>
int Thing<A,double>::doSomething(); //error
Its not allowed, because the template parameter-list in function definition didn't match the template parameter-list of the class template partial specialization. §14.5.4.3/1 from the Standard (2003) says,
The template parameter list of a member of a class template partial specialization shall match the template parameter list of the class template partial specialization.[...]
For more on this, read my answer here:
C++ - Overload templated class method with a partial specilization of that method
So what is the solution? Would you partially specialize your class along with all the repetitive work?
A simple solution would be work delegation, instead of partially specializing the class template. Write a stand-alone function template and specialize this as:
template <class B>
B doTheActualSomething(B & b) { return b; }
template <>
int doTheActualSomething<int>(int & b) { return b + 1; }
And then call this function template from doSomething() member function as:
template <class A, class B>
B Thing<A,B>::doSomething() { return doTheActualSomething<B>(b_); }
Since in your particular case, doTheActualSomething needs to know the value of only one member, namely b_, the above solution is fine, as you can pass the value to the function as argument whose type is the template type argument B, and specialization for int is possible being it full-specialization.
But imagine if it needs to access multiple members, type of each depends on the template type argument-list, then defining a stand-alone function template wouldn't solve the problem, because now there will be more than one type argument to the function template, and you cannot partially specialize the function for just, say, one type (as its not allowed).
So in this case you can define a class template instead, which defines a static non-template member function doTheActualSomething. Here is how:
template<typename A, typename B>
struct Worker
{
B doTheActualSomething(Thing<A,B> *thing)
{
return thing->b_;
}
};
//partial specialization of the class template itself, for B = int
template<typename A>
struct Worker<A,int>
{
int doTheActualSomething(Thing<A,int> *thing)
{
return thing->b_ + 1;
}
};
Notice that you can use thing pointer to access any member of the class. Of course, if it needs to access private members, then you've to make struct Worker a friend of Thing class template, as:
//forward class template declaration
template<typename T, typename U> struct Worker
template <class A, class B>
class Thing
{
template<typename T, typename U> friend struct Worker; //make it friend
//...
};
Now delegate the work to the friend as:
template <class A, class B>
B Thing<A,B>::doSomething()
{
return Worker<A,B>::doTheActualSomething(this); //delegate work
}
Two points to be noted here:
In this solution, doTheActualSomething is not a member function template. Its not enclosing class which is template. Hence we can partially specialize the class template anytime, to get the desired effect of the partial member function template specialization.
Since we pass this pointer as argument to the function, we can access any member of the class Thing<A,B>, even private members, as Worker<T,U> is also a friend.
Complete online demo : http://www.ideone.com/uEQ4S
Now there is still a chance of improvement. Now all instantiations of Worker class template are friends of all instantiation of Thing class template. So we can restrict this many-to-many friendship as:
template <class A, class B>
class Thing
{
friend struct Worker<A,B>; //make it friend
//...
};
Now only one instantiation of Worker class template is a friend of one instantiation of Thing class template. That is one-to-one friendship. That is, Worker<A,B> is a friend of Thing<A,B>. Worker<A,B> is NOT a friend of Thing<A,C>.
This change requires us to write the code in somewhat different order. See the complete demo, with all the ordering of class and function definitions and all:
http://www.ideone.com/6a1Ih
This is a very often found problem, and there is a surprisingly simple solution. I will show it in an artificial example, because it's more clearer than to use your code, and you will have to understand it to adapt it to your code
template<typename A, typename B>
struct TwoTypes { };
template<typename A, typename B>
struct X {
/* forwards ... */
void f() { fImpl(TwoTypes<A, B>()); }
/* special overload for <A, int> */
template<typename A1>
void fImpl(TwoTypes<A1, int>) {
/* ... */
}
/* generic */
template<typename A1, typename B1>
void fImpl(TwoTypes<A1, B1>) {
/* ... */
}
};
Explicitly specializing functions is never (almost never?) the right way. In my work as a programmer, I've never explicitly specialized a function template. Overloading and partial ordering is superior.
I want to define function outside the template class as described below.
Already tried a lot of combinations for the second argument which is a template and takes default argument as well.
template <typename T>
class CustomAllocator
{
//My custom allocator
};
template <typename T, typename Allocator = CustomAllocator<T> >
class CustomContainer
{
void push_back();
};
/*I want to define push_back outside my class, tried everything.
Almost 4 hours spent through stackoverflow, fluentcpp and all sites*/
// What should be specified for Allocator here ?
template <typename T>
void CustomContainer<T,Allocator>::push_back(T value)
{
}
//OR
template <typename T>
void CustomContainer<T,CustomAllocator<> >::push_back(T value)
{
}
I expect it to be defined outside class
Actual getting compiler error, if it is simple type I could easily mention int,float etc. in the second argument.
Outside of your class definition, it will be unclear to a function what type Allocator is, so you have to redeclare it just like you redeclared T
template <class T, class Allocator>
void CustomContainer<T,Allocator>::push_back(T value)
{
// ...
}
(I'm assuming that DataType should be T)
Note that your declaration of push_back, in the class should match the definition:
template <typename T, typename Allocator = CustomAllocator<T> >
class CustomContainer
{
void push_back(T);
};
You may not use default template arguments for a member function of a template defined outside the template definition.
From the C++ 17 Standard (17.1 Template parameters)
... A default template-argument shall not be specified in the template-
parameter-lists of the definition of a member of a class
template that appears outside of the member’s class.
So just write
template <typename T, typename Allocator>
void CustomContainer<T, Allocator>::push_back( const T &value )
{
//...
}
Pay attention to the argument of the function. Your declaration of the function does not correspond to its definition.
So first, apologies for terminology - I'm not sure if template prototype is the correct term. By this I mean :
template <class T, class X>
class TemplatePrototype
{
// code
};
I have a situation where I have a function that creates a template object based upon template arguments to that function.
template <class T, class X>
void doSomething()
{
TemplatePrototype<T, X> aTemplateTX;
aTemplateTX.doSomethingElse();
}
However, there are about 15 different versions of TemplatePrototype, which all have the same interface but different execution (TemplatePrototype is provided by another library). As a result, I have a lot of code that looks like this:
template <class T, class X>
void doSomethingWithOne()
{
TemplatePrototypeOne<T, X> aTemplateTX;
aTemplateTX.doSomethingElse();
}
template <class T, class X>
void doSomethingWithTwo()
{
TemplatePrototypeTwo<T, X> aTemplateTX;
aTemplateTX.doSomethingElse();
}
As a consequence of the architecture, I must know which TemplatePrototype I am going to use before I know the actual types T and X. I would like to see something like this:
template <class T, class X, class Prototype>
void doSomething()
{
Prototype<T, X> aPrototype;
aPrototype.doSomething();
}
But where I have specified part of the template arguments in advance - i.e I specify Prototype before I know T and X. Obviously, this is not possible in C++.
Equally, I cannot pass the Prototype as a template argument because it will still result in huge amounts of duplicate code.
Some important facts : I know the range of all possible inputs.
So I could theoretically use a macro to define each possible template specialisation and insert them into a container, which I would then use to access the specialisation I need. However, I am looking for a more 'elegant' solution - is it possible to pass template prototypes without specialising them as an argument to a template class, and then instantiate later when a function is called? Example:
template <class Prototype>
class Holder
{
template <class T, class X>
void doSomething()
{
Prototype<T, X> aPrototype;
aPrototype.doSomethingElse();
}
};
As far as I know this is impossible, but I was wondering if the SO community had some folks who know a solution?
EDIT:
So I have implemented this as my solution, thanks to the answers below!
#include <iostream>
template <typename T>
struct Foo
{
Foo() { aPtr = 0; }
T* aPtr;
};
template <template<typename> class C>
struct Bar
{
template <class T>
void doSomething()
{
C<T> aClass;
if (aClass.aPtr)
std::cout << "Hello world" << std::endl;
}
};
int main()
{
Bar<Foo> aFoo;
aFoo.doSomething<int>();
return 0;
}
This enables me to specify which TemplatePrototype I wish to use, before I can know the template parameters.
Yes, use a template template parameter, e.g.
template <typename T>
struct Foo
{
};
template <template<typename> class C>
struct Bar
{
};
then
Bar<Foo> b;
You're looking for template template parameters.
In the template parameter list, instead of just:
class TemplatePrototype
specify your prototype as a class template which itself has two template type parameters (without giving them a name here), like:
template<class,class> class TemplatePrototype
//^^^^^^^^^^^^^^^^^^^
This will result in a function like:
template <class T, class X,
template<class,class> class TemplatePrototype>
void doSomething()
{
TemplatePrototype<T, X> aTemplateTX;
aTemplateTX.doSomethingElse();
}
Invocation example:
doSomething<T, X, TemplatePrototypeOne>();
To become independent of the number of template parameters you pass to your "prototype" (here it was 2, namely T and X), you can use variadic templates (since C++11).
For this, first move the prototype template parameter to the first position:
template <template<class,class> class TemplatePrototype,
class T, class X>
Then, replace class T, class X with class ...Ts, which is a placeholder of an arbitrary number of type parameters. Also, in the template template parameter list, replace class,class with class.... And in the instantiation within the function implementation, replace <T, X> with <Ts...> to "expand" the parameter pack.
The result then looks like this:
template <template<class...> class TemplatePrototype,
class ... Ts>
void doSomething()
{
TemplatePrototype<Ts...> aTemplateTs;
aTemplateTs.doSomethingElse();
}
Live demo
Note: this seems to be a repost of a problem: C++ - Overload templated class method with a partial specilization of that method
I have boiled down a problem I am having with C++ template specialization down to a simple case.
It consists of a simple 2-parameter template class Thing, where I would like to specialize Thing<A,B>::doSomething() for B=int.
#include <cstdio>
// A 3-parameter template class.
template <class A, class B>
class Thing
{
public:
Thing(A a, B b) : a_(a), b_(b) {}
B doSomething();
private:
A a_;
B b_;
};
// The generic case works as expected.
template <class A, class B>
B Thing<A,B>::doSomething()
{
return b_;
}
// This specialization does not work!
template <class A>
int Thing<A,int>::doSomething()
{
return b_+1;
}
int main() {
// Setup our thing.
Thing<double,int> thing(1.0,2);
// This doesn't compile - but works with the generic case.
printf("Expecting 3, and getting %i\n", thing.doSomething());
// Clean up.
return 0;
}
Unfortunately, g++ exits with the error:
partial_specialization.cpp:30: error: invalid use of incomplete type ‘class Thing<A, int>’
partial_specialization.cpp:8: error: declaration of ‘class Thing<A, int>’
The clang++ compiler is a bit more verbose, but has the same problem:
partial_specialization.cpp:30:19: error: nested name specifier 'Thing<A, int>::' for declaration does not
refer into a class, class template or class template partial specialization
int Thing<A,int>::doSomething()
~~~~~~~~~~~~~~^
partial_specialization.cpp:32:12: error: use of undeclared identifier 'b_'
return b_+1;
^
2 errors generated.
I have read and understood that partial template specializations on functions aren't allowed - but I thought I was partially specializing over classes of Thing in this case.
Any ideas?
What I did: A workaround, as determined from the link provided by the accepted answer:
template< class T >
inline T foo( T const & v ) { return v; }
template<>
inline int foo( int const & v ) { return v+1; }
// The generic case works as expected.
template <class A, class B>
B Thing<A,B>::doSomething()
{
return foo(b_);
}
Partial specialization of a function template, whether it is member function template or stand-alone function template, is not allowed by the Standard:
template<typename T, typename U> void f() {} //okay - primary template
template<typename T> void f<T,int>() {} //error - partial specialization
template<> void f<unsigned char,int>() {} //okay - full specialization
But you can partially specialize the class template itself. You can do something like this:
template <class A>
class Thing<A,int> //partial specialization of the class template
{
//..
int doSomething();
};
template <class A>
int Thing<A,int>::doSomething() { /* do whatever you want to do here */ }
Note that when you partially specialize a class template, then the template parameter-list of member function (in its definition outside the class), must match the template parameter list of the class template partial specialization. That means, for the above partial specialization of the class template, you cannot define this:
template <class A>
int Thing<A,double>::doSomething(); //error
Its not allowed, because the template parameter-list in function definition didn't match the template parameter-list of the class template partial specialization. §14.5.4.3/1 from the Standard (2003) says,
The template parameter list of a member of a class template partial specialization shall match the template parameter list of the class template partial specialization.[...]
For more on this, read my answer here:
C++ - Overload templated class method with a partial specilization of that method
So what is the solution? Would you partially specialize your class along with all the repetitive work?
A simple solution would be work delegation, instead of partially specializing the class template. Write a stand-alone function template and specialize this as:
template <class B>
B doTheActualSomething(B & b) { return b; }
template <>
int doTheActualSomething<int>(int & b) { return b + 1; }
And then call this function template from doSomething() member function as:
template <class A, class B>
B Thing<A,B>::doSomething() { return doTheActualSomething<B>(b_); }
Since in your particular case, doTheActualSomething needs to know the value of only one member, namely b_, the above solution is fine, as you can pass the value to the function as argument whose type is the template type argument B, and specialization for int is possible being it full-specialization.
But imagine if it needs to access multiple members, type of each depends on the template type argument-list, then defining a stand-alone function template wouldn't solve the problem, because now there will be more than one type argument to the function template, and you cannot partially specialize the function for just, say, one type (as its not allowed).
So in this case you can define a class template instead, which defines a static non-template member function doTheActualSomething. Here is how:
template<typename A, typename B>
struct Worker
{
B doTheActualSomething(Thing<A,B> *thing)
{
return thing->b_;
}
};
//partial specialization of the class template itself, for B = int
template<typename A>
struct Worker<A,int>
{
int doTheActualSomething(Thing<A,int> *thing)
{
return thing->b_ + 1;
}
};
Notice that you can use thing pointer to access any member of the class. Of course, if it needs to access private members, then you've to make struct Worker a friend of Thing class template, as:
//forward class template declaration
template<typename T, typename U> struct Worker
template <class A, class B>
class Thing
{
template<typename T, typename U> friend struct Worker; //make it friend
//...
};
Now delegate the work to the friend as:
template <class A, class B>
B Thing<A,B>::doSomething()
{
return Worker<A,B>::doTheActualSomething(this); //delegate work
}
Two points to be noted here:
In this solution, doTheActualSomething is not a member function template. Its not enclosing class which is template. Hence we can partially specialize the class template anytime, to get the desired effect of the partial member function template specialization.
Since we pass this pointer as argument to the function, we can access any member of the class Thing<A,B>, even private members, as Worker<T,U> is also a friend.
Complete online demo : http://www.ideone.com/uEQ4S
Now there is still a chance of improvement. Now all instantiations of Worker class template are friends of all instantiation of Thing class template. So we can restrict this many-to-many friendship as:
template <class A, class B>
class Thing
{
friend struct Worker<A,B>; //make it friend
//...
};
Now only one instantiation of Worker class template is a friend of one instantiation of Thing class template. That is one-to-one friendship. That is, Worker<A,B> is a friend of Thing<A,B>. Worker<A,B> is NOT a friend of Thing<A,C>.
This change requires us to write the code in somewhat different order. See the complete demo, with all the ordering of class and function definitions and all:
http://www.ideone.com/6a1Ih
This is a very often found problem, and there is a surprisingly simple solution. I will show it in an artificial example, because it's more clearer than to use your code, and you will have to understand it to adapt it to your code
template<typename A, typename B>
struct TwoTypes { };
template<typename A, typename B>
struct X {
/* forwards ... */
void f() { fImpl(TwoTypes<A, B>()); }
/* special overload for <A, int> */
template<typename A1>
void fImpl(TwoTypes<A1, int>) {
/* ... */
}
/* generic */
template<typename A1, typename B1>
void fImpl(TwoTypes<A1, B1>) {
/* ... */
}
};
Explicitly specializing functions is never (almost never?) the right way. In my work as a programmer, I've never explicitly specialized a function template. Overloading and partial ordering is superior.
// InternalTemplate.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
template<class T>
struct LeftSide
{
static void insert(T*& newLink, T*& parent)
{
parent->getLeft() = newLink;
newLink->parent = newLink;
}
};
template<class T>
struct Link
{
T* parent_;
T* left_;
T* right_;
T*& getParent()const
{
return parent_;
}
template<class Side>
void plugIn(Link<T>*& newLink);
};
template<class T>
template<class Side>
void Link<T>::plugIn(Link<T>*& newLink)//<<-----why can't I type
//void Link<T>::plugIn<Side>(Link<T>*& newLink)<---<Side> next to plugIn
{
Side::insert(newLink,this);
}
int _tmain(int argc, _TCHAR* argv[])
{
return 0;
}
I find it strange that I have to specify parameter for a class but cannot specify parameter for a function. Is there any reason why?
Function templates and class templates are complementary (I call them orthogonal, but you are free not to agree), and in template metaprogramming they actually serve orthogonal purposes.
Class templates allow you to pattern match on the template argument, ie. they provide partial specialization.
Function templates, to the contrary, don't allow partial specialization, but they allow template argument deduction, which means you don't have to write the template arguments explicitly (except for extra arguments, as in your example).
This, I think, explains the differences in syntax since they are different in what they can achieve. Moreover, function templates can have overloads, class templates cannot.
The way to combine both concepts is
1) to have helper class templates with static non template functions if you want partial specialization for function templates:
template <typename T>
struct doSomethingWithPointersHelper
{
static void act(T x) { ... }
};
template <typename T>
struct doSomethingWithPointersHelper<T*>
{
static void act(T* x) { ... }
};
// This acts as if we had a partial specialization
// for pointer types
template <typename T>
doSomethingWithPointers(T x)
{ return doSomethingWithPointersHelper<T>::act(x); }
There are other ways to achieve this in particular cases, but this approach always works.
2) To have helper template functions if you want to make use of argument deduction when constructing complex classes:
template <typename T, typename U>
struct MyComplexClass
{ ... };
template <typename T, typename U>
MyComplexClass<T, U> makeComplex(T t, U u)
{ return MyComplexClass<T, U>(t, u); }
in the standard library, you find make_pair, bind1st or mem_fun which make use of this technique.
$14/2 -
A template-declaration can appear only as a namespace scope or class scope declaration. In a function template declaration, the last component of the declarator-id shall be a template-name or operator-functionid (i.e., not a template-id). [ Note: in a class template declaration, if the class name is a simple-template-id, the declaration declares a class template partial specialization (14.5.5). —end note ]"
The standard forbids such a syntax explicitly. Refer this for more idea about template id / template name
You need to specialize on the Link struct in order to define it's template member function.
template<>
template<class Side>
void Link<int>::plugIn(Link<int>*& newLink)
{
Side::insert(newLink,this);
}
Gotta be honest, this makes my brain explode a little.