Reputation: 13575
For example, I have a class:
class A
{
enum {N = 5};
double mVariable;
template<class T, int i>
void f(T& t)
{
g(mVariable); // call some function using mVariable.
f<T, i+1>(t); // go to next loop
}
template<class T>
void f<T, N>(T& t)
{} // stop loop when hit N.
};
Partial specialization is not allowed in function template. How do I work around it in my case?
I slightly changed the example of Arne Mertz, like:
template<int n>
struct A
{
enum {N = n};
...
};
and use A like:
A<5> a;
The I cannot compile on Visual Studio 2012. Is it a compiler bug or something else? It is quite strange.
EDIT: Checked. It is a Visual Studio bug. :(
I think Nim gives the most simple way to implement it.
Upvotes: 10
Views: 1047
Reputation: 7491
You can emulate partial specialization of function template with function overloading:
#include <type_traits>
class A
{
enum {N = 5};
double mVariable;
// ...
void g(double)
{
// ...
}
public:
template<class T, int i = 0>
void f(T& t, std::integral_constant<int, i> = std::integral_constant<int, i>())
{
g(mVariable);
f(t, std::integral_constant<int, i + 1>());
}
template<class T>
void f(T& t, std::integral_constant<int, N>)
{
}
};
Example of using:
A a;
int t = 0;
a.f(t);
a.f(t, std::integral_constant<int, 2>()); // if you want to start loop from 2, not from 0
It is a C++11 solution, however (not so much because of std::integral_constant
class, but because of default template parameter of function template). It can be made shorter using some additional C++11 features:
template<int i>
using integer = std::integral_constant<int, i>;
template<class T, int i = 0>
void f(T& t, integer<i> = {})
{
g(mVariable);
f(t, integer<i + 1>());
}
template<class T>
void f(T& t, integer<N>)
{
}
Upvotes: 2
Reputation: 33655
With c++11 support, you can do the following:
#include <iostream>
#include <type_traits>
using namespace std;
struct A
{
enum {N = 5};
double mVariable;
void g(int i, double v)
{ std::cout << i << " " << v << std::endl; }
template<int i, class T>
typename enable_if<i >= N>::type f(T& t)
{} // stop loop when hit N.
template<int i, class T>
typename enable_if<i < N>::type f(T& t)
{
g(i, mVariable); // call some function using mVariable.
f<i+1, T>(t); // go to next loop
}
};
int main(void)
{
A a;
int v = 0;
a.f<0>(v);
}
Main reason I like is that you don't need any of the cruft as required by the previous answers...
Upvotes: 2
Reputation: 171177
You can define a helper class:
template <int i, int M>
struct inc_up_to
{
static const int value = i + 1;
};
template <int i>
struct inc_up_to<i, i>
{
static const int value = i;
};
template<class T, int i>
void f(T& t)
{
if (i < N) {
g(mVariable); // call some function using mVariable.
f<T, inc_up_to<i, N>::value>(t);
}
}
It stops the compile-time recursion by making f<T, N>
refer to f<T, N>
, but that call is avoided by the run-time condition, breaking the loop.
A simplified and more robust version of the helper (thanks @ArneMertz) is also possible:
template <int i, int M>
struct inc_up_to
{
static const int value = (i >= M ? M : i + 1); // this caps at M
// or this:
static const int value = (i >= M ? i : i + 1); // this leaves i >= M unaffected
};
This doesn't even need the partial specialisation.
Upvotes: 4
Reputation: 24626
The most straight forward solution is to use a template class instead of a function:
class A
{
enum {N = 5};
double mVariable;
template <class T, int i>
struct fImpl {
static_assert(i<N, "i must be equal to or less than N!");
static void call(T& t, A& a) {
g(a.mVariable);
fImpl<T, i+1>::call(t, a);
}
};
template<class T>
struct fImpl<T,N> {
static void call(T&, A&) {} // stop loop when hit N.
};
public:
template<class T, int i>
void f(T& t)
{
fImpl<T, i>::call(t,*this);
}
};
Upvotes: 7