我想使用可变参数模板来帮助解决使用va-args的问题。基本上,我想调用一个单数函数,传入函数a"命令"以及变量参数列表,然后将参数分派给另一个函数。
我已经使用久经考验的(但不是类型安全的)va_list实现了这一点。这是我尝试使用可变参数模板进行的尝试。这个例子不在下面编译,因为你会很快找到原因......
#include <iostream>
using namespace std;
typedef enum cmd_t
{
CMD_ZERO,
CMD_ONE,
CMD_TWO,
} COMMANDS;
int cmd0(double a, double b, double c)
{
cout << "cmd0 " << a << ", " << b << ", " << c << endl;
return 0;
}
int cmd1(int a, int b, int c)
{
cout << "cmd1 " << a << ", " << b << ", " << c << endl;
return 1;
}
template<typename... Args>
int DispatchCommand(COMMANDS cmd, Args... args)
{
int stat = 0;
switch (cmd)
{
case CMD_ZERO:
cmd0(args...);
break;
case CMD_ONE:
cmd1(args...);
break;
default:
stat = -1;
break;
}
return stat;
}
int main()
{
int stat;
stat = DispatchCommand(CMD_ZERO, 1, 3.141, 4);
stat = DispatchCommand(CMD_ONE, 5, 6, 7);
stat = DispatchCommand(CMD_TWO, 5, 6, 7, 8, 9);
system("pause");
return 0;
}
有没有人知道如何修改此函数以正确使用可变参数模板?
答案 0 :(得分:0)
编写一些代码,给定一个函数指针和一组参数,用那些有效的参数的最长前缀来调用它。
template<class...>struct types{using type=types;};
template<class types0, size_t N, class types1=types<>>
struct split;
template<class t00, class...t0s, size_t N, class...t1s>
struct split<types<t00,t0s...>,N,types<t1s...>>:
split<types<t0s...>,N-1,types<t1s...,t00>>
{};
template<class...t0s, class...t1s>
struct split<types<t0s...>,0,types<t1s...>>
{
using right=types<t0s...>;
using left=types<t1s...>;
};
template<class>using void_t=void;
template<class Sig,class=void>
struct valid_call:std::false_type{};
template<class F, class...Args>
struct valid_call<F(Args...), void_t<
decltype( std::declval<F>()(std::declval<Args>()...) )
>>:std::true_type {};
template<class R, class types>
struct prefix_call;
template<class R, class...Args>
struct prefix_call<R, types<Args...>> {
template<class F, class... Extra>
std::enable_if_t< valid_call<F(Args...)>::value, R >
operator()(R default, F f, Args&&...args, Extra&&...) const
{
return std::forward<F>(f)(args...);
}
template<class F, class... Extra>
std::enable_if_t< !valid_call<F(Args...)>::value, R >
operator()(R default, F f, Args&&...args, Extra&&...) const
{
return prefix_call<R, typename split<types<Args...>, sizeof...(Args)-1>::left>{}(
std::forward<R>(default), std::forward<F>(f), std::forward<Args>(args)...
);
}
};
template<class R>
struct prefix_call<R, types<>> {
template<class F, class... Extra>
std::enable_if_t< valid_call<F()>::value, R >
operator()(R default, F f, Extra&&...) const
{
return std::forward<F>(f)();
}
template<class F, class... Extra>
std::enable_if_t< !valid_call<F()>::value, R >
operator()(R default, F f, Extra&&...) const
{
return std::forward<R>(default);
}
};
上面的代码可能包含错别字。
template<typename... Args>
int DispatchCommand(COMMANDS cmd, Args... args)
{
int stat = 0;
switch (cmd) {
case CMD_ZERO: {
stat = prefix_call<int, Args...>{}(-1, cmd0, std::forward<Args>(args)...);
} break;
case CMD_ONE: {
stat = prefix_call<int, Args...>{}(-1, cmd1, std::forward<Args>(args)...);
} break;
default: {
stat = -1;
} break;
}
return stat;
}
如果覆盖cmd0
或cmd1
,则必须使用重载设置技术。
答案 1 :(得分:0)
您可以使用以下内容:
template <COMMANDS cmd> struct command
{
template <typename ... Args>
int operator() (Args&&...) const { return -1; }
};
template <> struct command<CMD_ZERO>
{
int operator()(double a, double b, double c) const
{
std::cout << "cmd0 " << a << ", " << b << ", " << c << std::endl;
return 0;
}
};
template <> struct command<CMD_ONE>
{
int operator()(int a, int b, int c) const
{
std::cout << "cmd1 " << a << ", " << b << ", " << c << std::endl;
return 1;
}
};
template <COMMANDS cmd, typename... Args> int DispatchCommand(Args&&... args)
{
return command<cmd>()(std::forward<Args>(args)...);
}
然后使用它:
DispatchCommand<CMD_ZERO>(1., 3.141, 4.);
DispatchCommand<CMD_ONE>(5, 6, 7);
DispatchCommand<CMD_TWO>(5, 6, 7, 8, 9);
但直接使用不同的功能似乎更简单:
cmd0(1., 3.141, 4.);
cmd1(5, 6, 7);