Suppose we have template class
template <typename T>
class MyTem{
public:
bool is_T_Pointer(){
<...>
}
};
class Cls : MyTem<Cls>{
<...>
};
int main(void){
Cls* classOnHeap = new Cls(); /* T is pointer */
Cls classOnStack; /* T is not pointer */
<...>
}
I know this is a bad example but if someone could help me find out if T is pointer from template class that would be great.
Remember we have inheritance with template of same class as base class.
Doesn't have to be complete implementation, a vague technique will be enough
You should employ partial specialization here:
template<class T>
class A
{
public:
A() {}
};
template<class T>
class A<T*>
{
public:
A(int) {}
};
Then the following will not compile, because compiler is forced to choose pointer version of template and there is no default constructor:
A<char*> a;
this does compile:
A<char> a;
If the compiler supports C++11 use std::is_pointer:
#include <iostream>
#include <type_traits>
template <typename T>
class MyTem
{
public:
static const bool IS_POINTER = std::is_pointer<T>::value;
};
int main()
{
std::cout << MyTem<char*>::IS_POINTER << "\n";
std::cout << MyTem<char>::IS_POINTER << "\n";
return 0;
}
See demo http://ideone.com/Mo394 .
Related
Let's say I have a template my_type. I want it to have general functionality, to have a few extra functions when T is not an array and to have others when T is an array.
Let's say I have the following template:
template <typename T>
class my_class<T> {
public:
int f1(); // This function is available for all T
int f2(); // This function is available when T is not an array
int f3(); // This function is available when T is an array
}
So if I try:
my_class<int> c1; my_class<int[3]> c2;
c1.f1(); c2.f1(); // both fine
c1.f2(); c2.f3(); // both fine
c1.f3(); c2.f2(); // both should give a compile error
I am aware std::unique_ptr does this internally. So how does it do it?
Another way, using enable_if. Note also the use of a base class to capture all common behaviour.
#include <type_traits>
template<class T>
struct my_base
{
int f1();
};
template<class T, typename Enable = void>
class my_class;
template<class T>
class my_class<T, std::enable_if_t<std::is_array<T>::value>>
: public my_base<T>
{
public:
int f3(); // This function is available when T is an array
};
template <typename T>
class my_class<T, std::enable_if_t<not std::is_array<T>::value>>
: public my_base<T>
{
public:
int f2(); // This function is available when T is not an array
};
int main()
{
auto a = my_class<int[]>();
a.f1();
// a.f2();
a.f3();
auto na = my_class<int>();
na.f1();
na.f2();
// na.f3();
}
I have figured it out myself. The following code will do the exact thing I have asked for.
template<typename T>
class my_class {
public:
int f1() { return 1; }
int f2() { return 2; }
};
template<typename T>
class my_class<T[]> {
public:
int f1() { return 1; }
int f3() { return 3; }
};
Note that the implementation of the common function (f1) had to be copied. Now is there a way to use a single implementation? (note that it is NOT as simple as a return 1; like in the example code and thus I can't separate functionality into a non-template function)
I'm sure there is a very easy answer, but I can't figure it out. I have written a templated class, but I want to pass that class by reference in a class function that isn't templated. Heres what I have. I get a bunch of errors. All I need to do is figure how to format the way to insert templated class into function, but I'm at a lost. Thank you and sorry if the code doesn't really help you out.
#include <iostream>
using namespace std;
template <typename T>
class Foo {
public:
Foo();
insert(const T& Item)
//And other function, just examples
};
class noFoo(){
void test(Foo <T>& foo);
int i;
int j;
int k
};
template <typename T>
void noFoo::test(Food <T>& foo)}
cout << "hi";
}
int main() {
Foo<char> wr;
test(wr);
return 0;
}
Make test a function template. I also corrected loads of syntax errors for you (class noFoo()?), removed unnecessary code, and ran clang-format for indentation.
#include <iostream>
template <typename T>
class Foo {};
class noFoo
{
public:
template <typename T>
void test(Foo<T> &);
};
template <typename T>
void noFoo::test(Foo<T> &)
{
std::cout << "hi\n";
}
int main()
{
Foo<char> wr;
noFoo{}.test(wr);
}
Since your question is tagged d, here the same code in D.
import std.stdio;
class Foo(T) {};
class noFoo
{
public:
void test(T)(Foo!(T))
{
writeln("hi");
}
};
void main()
{
auto wr = new Foo!char;
(new noFoo).test(wr);
}
Is it possible to create a class template with a member function definition only if the object created is of a specific type?
I've created a template class I will use for storing either int or doubles, but for doubles I would like to be able to set precision too (objects created with myclass < double> should have this functionality, but for myclass< int> there is no need for that to be present at all).
I know I can use a base class template, and create new classes "myInt", "myDouble" using that and implement the functionality only in the myDouble class, but I think it would be cleaner to define the functionality (both the function and a member variable) for doubles in the class template, if that's possible and preferable?
Let's add an example to show what I want to do:
#include <iostream>
#include <iomanip>
class commonBase{
public:
void setState(int state);
virtual void print() = 0;
private:
int _my_state;
};
template <typename T>
class generalObject : public commonBase {
public:
void value(T value);
void print(){ std::cout << "My value: " << _my_value << std::endl; }
private:
T _my_value;
};
template <typename T>
void generalObject<T>::value(T value){
_my_value = value;
}
// Is there any way do specialize only only whats different from the generalObject template?
// Here I thought I could specialize the case where a generalObject is created of <double>, but
// when I do, nothing is derived from generalObject (or at least not visible as far as I can tell)
template<>
class generalObject<double>{
public:
void setPrecision(int precision){ _my_precision = precision; }
// here I would like a special implementation of print(), which overrides the print() in generalObject
// and instead also prints according to the precision set when the object is of <double> type.
// Row below an example which doesn't work (compiler error, _my_value undefined)
void print(){ std::cout << "My value: " << std::setprecision(_my_precision) << _my_value << std::endl; }
private:
int _my_precision;
};
int main(int argc, char* argv[]){
generalObject<int> o1;
o1.value(1);
o1.print();
o1.setState(1); //inherited from the commonBase
generalObject<double> o2;
o2.setPrecision(2);
o2.value(2); //here value isn't available (compile error)
o2.print();
o2.setState(123); //also isn't available (compile error)
}
Sure.
template <typename T> class Poly;
void set_precision(Poly<double>* self, int a) {};
If you really want dot notation you can then add:
template <typename T> class Poly {
public: void set_precision(int a){::set_precision(this,a);}
...
However I think you should think about what you're trying to accomplish. If MyInt and MyDouble have different fields and different methods and different implementations, they should probably be different classes.
This can be solved using template specialization.
We first define a common template...
template< typename T >
struct myclass
{
// common stuff
};
... and specialize that for double:
template<>
struct myclass<double>
{
int precision = 10;
void setprecision( int p ){ precision = p; }
};
Now the setprecision() method can only be called for myclass<double>. The compiler will complain if we try to call it for anything else, like myclass<int>.
int main()
{
myclass<double> d;
d.setprecision( 42 ); // compiles
myclass<int> i;
i.setprecision( 42 ); // fails to compile, as expected
}
Demo.
The basic way to have a member function of a class template exist only for some template parameters is to create a specialization of the class template for those template parameters.
template<typename T>class X{
// general definition
};
template<>class X<double>{
// double-specific definition
};
The downside of this is that the specialization will need to duplicate anything that is common. One way to address this is to move the common things out to a base class template:
template<typename T>class Xcommon{
// common stuff
};
template<typename T>class X: public Xcommon<T>{
// general definition
};
template<>class X<double>: public Xcommon<double>{
// double-specific definition
};
Alternatively, you can do it the other way: put the common stuff in the derived class, and the extras in the base, and specialize the base:
template<typename T>class Xextras{
// empty by default
};
template<typename T>class X: public Xextras<T>{
// common definition
};
template<>class Xextras<double>{
// double-specific definition
};
Either way can work; which is better depends on the details.
Both these methods work for data members and member functions.
Alternatively, you can use enable_if to mean that member functions are not selected by overload resolution if the template parameter doesn't meet a required condition. This requires that the member function is itself a template.
template<typename T>class X{
template<typename U=T> // make it a template,
std::enable_if<std::is_same_v<U,double>> double_specific_function(){
// do stuff
}
};
I wouldn't recommend this option unless there is no other choice.
If the question is about a member function, then here is one of the ways to do it without class template specialization:
#include <iostream>
#include <type_traits>
template <typename T>
struct Type {
template <typename U = T,
typename = typename std::enable_if<std::is_same<U, double>::value>::type>
void only_for_double() {
std::cout << "a doubling" << std::endl;
}
};
int main() {
Type<int> n;
Type<double> d;
// n.only_for_double(); // does not compile.
d.only_for_double();
}
Example on ideone.com
If you require a data-member presence based on the template parameter, you will have to do some kind of specialization, in which case it is, probably, simpler to put the function into corresponding specialization.
EDIT: After OP made his question more specific
Here is one way to do it without extra class and getting rid of virtual functions. Hope it helps.
#include <iostream>
#include <iomanip>
template <typename T, typename Derived = void>
class commonBase {
public:
void setState(int state) {
_my_state = state;
}
void value(T value) {
_my_value = value;
}
template <typename U = Derived,
typename std::enable_if<std::is_same<U, void>::value,
void * >::type = nullptr>
void print() const {
std::cout << "My value: " << _my_value << std::endl;
}
template <typename U = Derived,
typename std::enable_if<!std::is_same<U, void>::value,
void * >::type = nullptr>
void print() const {
static_cast<Derived const *>(this)->_print();
}
protected:
T _my_value;
int _my_state;
};
template <typename T>
class generalObject : public commonBase<T> {
};
template<>
class generalObject<double> : public commonBase<double, generalObject<double>> {
private:
friend commonBase<double, generalObject<double>>;
void _print() const {
std::cout << "My value: " << std::setprecision(_my_precision) <<
_my_value << std::endl;
}
public:
void setPrecision(int precision){ _my_precision = precision; }
private:
int _my_precision;
};
int main(){
generalObject<int> o1;
o1.value(1);
o1.print();
o1.setState(1);
generalObject<double> o2;
o2.setPrecision(2);
o2.value(1.234);
o2.print();
o2.setState(123);
}
Same code on ideone.com
So, I have a fairly simple set of templates that I want to use together, but the compiler keeps telling me that B::a has incomplete type. Everything is forward declared, but it still doesn't work...
#include <iostream>
using namespace std;
template <typename T> class A;
template <typename T> class B;
template <typename T>
class A{
public:
void ATestFunction();
void CallBFunction();
protected:
B<T> b;
};
template <typename T>
class B{
public:
void BTestFunction();
void CallAFunction();
protected:
A<T> a;
};
template <typename T>
void A<T>::ATestFunction(){
cout << "A was used for a function call" << endl;
}
template <typename T>
void B<T>::BTestFunction(){
cout << "B was used for a function call" << endl;
}
template <typename T>
void A<T>::CallBFunction(){
b.BTestFunction();
}
template <typename T>
void B<T>::CallAFunction(){
a.ATestFunction();
}
int main()
{
A<int> dragons;
dragons.CallBFunction();
return 0;
}
I ask this because I had run into some difficulty programming some array type classes that depend on each other (implementing a two dimensional array that can be accessed like this: [][]), but this problem happened and threw a gear in the works. I made this testing program, but it still fails. I've tried both MinGW 4.7.2 and GNU g++ on Linux, and each gave me the same problem.
The core of the issue can be seen in this piece of code:
template <typename T>
class A{
B<T> b;
};
template <typename T>
class B{
A<T> a;
};
C++ is a language with value semantics, that means that B<T> b; represents an object of type B<T> (rather than a reference, like in Java or C# with reference types). That is, A<T> contains a B<T>. Now if you look at the definition of the B template you see that in turn it contains an A<T> sub object. This is basically impossible, as A<T> cannot possibly contain an object that contains A<T>. What would be the size of an A<T> object?
Without knowing the real problem to solve, I won't venture to recommend an approach, but you can consider using pointers (A<T> would contain a pointer to B<T>, not a full B<T> sub object; or similarly, B<T> could contain a pointer to A<T>; or both), or references. But it might also be the case that a deeper redesign could make more sense.
Even if you used pointers, that could not work. That would basically trigger an infinite loop of A´s and B´s being created
A creates B creates A creates B creates A...
this would work.
#include <iostream>
using namespace std;
template<typename T> class A;
template<typename T> class B;
template<typename T>
class A
{
public:
A()
{
b = new B<T>(this);
}
A(B<T>* pb)
{
b = pb;
}
void ATestFunction()
{
cout << "A was used for a function call" << endl;
}
void CallBFunction()
{
b->BTestFunction();
}
protected:
B<T>* b;
};
template<typename T>
class B
{
public:
B()
{
a = new A<T>(this);
}
B(A<T>* pa)
{
a = pa;
}
void BTestFunction()
{
cout << "B was used for a function call" << endl;
}
void CallAFunction()
{
a->ATestFunction();
}
protected:
A<T>* a;
};
int main()
{
A<int> dragons;
dragons.CallBFunction();
B<int> bdragons;
bdragons.CallAFunction();
return 0;
}
or maybe just using static functions
#include <iostream>
using namespace std;
template<typename T> class A;
template<typename T> class B;
template<typename T>
class A
{
public:
static void ATestFunction()
{
cout << "A was used for a function call" << endl;
}
void CallBFunction();
};
template<typename T>
class B
{
public:
static void BTestFunction()
{
cout << "B was used for a function call" << endl;
}
void CallAFunction();
};
template<typename T>
void A<T>::CallBFunction()
{
B<int>::BTestFunction();
}
template<typename T>
void B<T>::CallAFunction()
{
A<int>::ATestFunction();
}
int main()
{
A<int> dragons;
dragons.CallBFunction();
B<int> bdragons;
bdragons.CallAFunction();
return 0;
}
I have the following simple code:
template <typename T>
struct base
{
std::vector<T> x;
};
template <typename T>
struct derived : base<T>
{
void print()
{
using base<T>::x; // error: base<T> is not a namespace
std::cout << x << std::endl;
}
};
When I compile the code (using GCC-4.7.2) I get the error that you see in the comment above.
I read here: http://gcc.gnu.org/onlinedocs/gcc-4.7.2/gcc/Name-lookup.html#Name-lookup
that
using base<T>::x
has to be included in order to bring in the scope of the base class. Any ideas what is wrong? Thank you in advance!
Put the using declaration in the class definition, not in the function body:
template <typename T>
struct derived : base<T>
{
using base<T>::x; // !!
void print()
{
std::cout << x << std::endl;
}
};
(Of course it's your responsibility to make sure that there's actually an operator<< overload for your std::vector, for example by using the pretty printer.)
You can also make it work if you explicitly say that x is a member:
template <typename T>
struct base
{
std::vector<T> x;
base() : x(1) {}
};
template <typename T>
struct derived : base<T>
{
void print()
{
std::cout << this->x[0] << std::endl;
}
};
int main()
{
derived<int> d;
d.print();
}