Reputation: 3083
I have three example programs, two of which compile, and one of which does not. All three were compiled using g++ 4.8.1, using the command line: g++ --std=c++11 -Wall -pedantic -o foo foo.cc
. The code in question is a throw-away example meant to exhibit the problem, and is not meant to be useful by itself.
The underlying problem is a templated class that requires a specialization to handle the fact that std::promise<void>::set_value()
takes no arguments. This problem arises in the Foo::_call()
method.
The version I want to work, call it version A, fails:
#include <future>
template <typename A, typename T>
class Foo
{
private:
std::function<T ()> _f;
std::promise<T> _p;
static void _call(std::function<T ()> &f, std::promise<T> &p) {
p.set_value(f());
}
public:
Foo(std::function<T ()> x) : _f(x) {}
void set_promise() {
_call(_f, _p);
}
};
template<typename A>
inline void Foo<A, void>::_call(
std::function<void ()> &f, std::promise<void> &p)
{
f();
p.set_value();
}
int bar()
{
return 4;
}
void baz()
{
}
int main()
{
Foo<int, int> a(bar);
a.set_promise();
Foo<int, void> b(baz);
b.set_promise();
}
The error is:
foo.cc:24:53: error: invalid use of incomplete type ‘class Foo<A, void>’
std::function<void ()> &f, std::promise<void> &p)
^
foo.cc:4:7: error: declaration of ‘class Foo<A, void>’
class Foo
^
foo.cc: In instantiation of ‘static void Foo<A, T>::_call(std::function<_Res()>&, std::promise<T>&) [with A = int; T = void]’:
foo.cc:18:21: required from ‘void Foo<A, T>::set_promise() [with A = int; T = void]’
foo.cc:44:19: required from here
foo.cc:11:9: error: invalid use of void expression
p.set_value(f());
^
The following test, B, compiles. It takes out the extra template parameter A
to Foo
. In practice, I do not want this as my non-toy class needs that parameter.
#include <future>
template <typename T>
class Foo
{
private:
std::function<T ()> _f;
std::promise<T> _p;
static void _call(std::function<T ()> &f, std::promise<T> &p) {
p.set_value(f());
}
public:
Foo(std::function<T ()> x) : _f(x) {}
void set_promise() {
_call(_f, _p);
}
};
template<>
void Foo<void>::_call(
std::function<void ()> &f, std::promise<void> &p)
{
f();
p.set_value();
}
int bar()
{
return 4;
}
void baz()
{
}
int main()
{
Foo<int> a(bar);
a.set_promise();
Foo<void> b(baz);
b.set_promise();
}
Test C also compiles. It pulls the _call()
static method out of Foo
to become a templated function. I would also rather not do this, as _call
is meant to belong to Foo
. Although this works, it is inelegant.
#include <future>
template<typename T>
inline void _call(std::function<T ()> &f, std::promise<T> &p)
{
p.set_value(f());
}
template<>
void _call(std::function<void ()> &f, std::promise<void> &p)
{
f();
p.set_value();
}
template <typename A, typename T>
class Foo
{
private:
std::function<T ()> _f;
std::promise<T> _p;
public:
Foo(std::function<T ()> x) : _f(x) {}
void set_promise() {
_call(_f, _p);
}
};
int bar()
{
return 4;
}
void baz()
{
}
int main()
{
Foo<int, int> a(bar);
a.set_promise();
Foo<int, void> b(baz);
b.set_promise();
}
So, the question is, why does A fail? Is there a way of solving this problem without resorting to B or C?
Upvotes: 2
Views: 1987
Reputation: 137770
Partial template specialization only applies to the entire class template. You can't partially specialize just one member.
One solution is to introduce a base class and specialize that instead.
#include <future>
template< typename T >
struct Foo_base {
static void _call(std::function<T ()> &f, std::promise<T> &p) {
p.set_value(f());
}
};
template<>
struct Foo_base< void > {
static void _call(std::function<void ()> &f, std::promise<void> &p)
{
f();
p.set_value();
}
};
template <typename A, typename T>
class Foo : Foo_base< T >
{
private:
using Foo::Foo_base::_call;
std::function<T ()> _f;
std::promise<T> _p;
public:
Foo(std::function<T ()> x) : _f(x) {}
void set_promise() {
_call(_f, _p);
}
};
Upvotes: 3