我一直在尝试使该_CallWithRightmostArgsInner
功能正常失败的所有事情,以便SFINAE可以正常工作,并且通过这种尝试,VS2013给了我错误:error C2039: 'type' : is not a member of 'std::enable_if<false,void>'
有任何想法吗?有没有更好的选择?这里的想法是,如果Function接受NumArgs表示的数字或参数,我想对Function进行函数调用。最后两个可变参数应转发给函数,并返回结果。
template <typename Function, int NumArgs>
class SplitParameters {
public:
typedef typename function_traits<Function>::result_type result_type;
template <typename ... RightArgs>
static result_type CallWithRightmostArgs(const Function& call, RightArgs && ... rightArgs) {
static_assert(sizeof...(RightArgs) >= NumArgs, "Unable to make function call with fewer than minimum arguments.");
return _CallWithRightmostArgs(call, std::forward<RightArgs>(rightArgs)...);
}
private:
template <typename ... RightArgs>
static result_type _CallWithRightmostArgs(const Function& call, RightArgs && ... rightArgs) {
return _CallWithRightmostArgsInner(call, std::forward<RightArgs>(rightArgs)...);
}
// note the '==' vs '!=' in these two functions. I would assume that only one could exist
template <typename LeftArg, typename ... RightArgs, typename std::enable_if<sizeof...(RightArgs) != NumArgs>::type* = 0>
static result_type _CallWithRightmostArgsInner(const Function& call, LeftArg, RightArgs && ... rightArgs) {
return _CallWithRightmostArgs(call, std::forward<RightArgs>(rightArgs)...);
}
template <typename LeftArg, typename ... RightArgs, typename std::enable_if<sizeof...(RightArgs) == NumArgs>::type* = 0>
static result_type _CallWithRightmostArgsInner(const Function& call, LeftArg, RightArgs && ... rightArgs) {
return call(std::forward<RightArgs>(rightArgs)...);
}
};
我通过将代码更改为g ++-4.8来工作
#include <iostream>
template <class T>
struct function_traits
{
typedef void result_type;
};
template <typename Function, int NumArgs>
class SplitParameters {
public:
typedef typename function_traits<Function>::result_type result_type;
template <typename ... RightArgs>
static result_type CallWithRightmostArgs(const Function& call, RightArgs && ... rightArgs) {
static_assert(sizeof...(RightArgs) >= NumArgs,
"Unable to make function call with fewer than minimum arguments.");
return _CallWithRightmostArgs(call, std::forward<RightArgs>(rightArgs)...);
}
private:
template <typename ... RightArgs>
static result_type _CallWithRightmostArgs(const Function& call, RightArgs && ... rightArgs) {
return _CallWithRightmostArgsInner(call, std::forward<RightArgs>(rightArgs)...);
}
// note the '==' vs '!=' in these two functions. I would assume that only one could exist
template <typename LeftArg, typename ... RightArgs, class = typename std::enable_if<sizeof...(RightArgs) != NumArgs -1 >::type>
static result_type _CallWithRightmostArgsInner(const Function& call, LeftArg, RightArgs && ... rightArgs) {
return _CallWithRightmostArgsInner(call, std::forward<RightArgs>(rightArgs)...);
}
template <typename ... RightArgs, class = typename std::enable_if<sizeof...(RightArgs) == NumArgs>::type>
static result_type _CallWithRightmostArgsInner(const Function& call, RightArgs && ... rightArgs) {
return call(std::forward<RightArgs>(rightArgs)...);
}
};
void f(int i, int j)
{
std::cout << i << ' ' << j << std::endl;
}
int main()
{
SplitParameters<decltype(f), 2>::CallWithRightmostArgs(f, 1, 2, 3, 4);
}
编译器不喜欢您_CallWithRightmostArgs
从调用_CallWithRightmostArgsInner
,而我认为您实际上是在尝试调用该Inner
函数。
g ++还不喜欢在模板参数列表中转换0
为void*
,因此我将其class = enable_if<...>::type
改为了。
我没有详细研究它失败的原因,希望对您来说足够了。
编辑:关于typename enable_if<...>::type* = 0
被拒绝,我记得有一个类似的问题std::array
:
template <class T, int size>
void f(const std::array<T,size>&){}
这个小片段可以自行编译,但是当您这样做时:
std::array<int,4> a;
f(a);
g++ gives:
test3.cpp: In function ‘int main()’:
test3.cpp:9:8: error: no matching function for call to ‘f(std::array<int, 4ul>&)’
f(a);
^
test3.cpp:9:8: note: candidate is:
test3.cpp:4:6: note: template<class T, int size> void f(const std::array<T, size>&)
void f(const std::array<T,size>&){}
^
test3.cpp:4:6: note: template argument deduction/substitution failed:
test3.cpp:9:8: note: mismatched types ‘int’ and ‘#‘integer_cst’ not supported by dump_type#<type error>’
f(a);
^
test3.cpp:9:8: note: ‘std::array<int, 4ul>’ is not derived from ‘const std::array<T, size>’
事实证明,问题是我将模板声明int
为size
参数的,但是编译器得到的std::size_t
与并不相同int
,即使您可以轻松地在它们之间进行转换。
在上面的示例中,我什至无法替换= 0
,= NULL
因为那只是一个0L
文字,我将不得不= (void*)0
让编译器接受它(因为默认类型enable_if<true>::type
为void
)。
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我来说两句