如何提取 __VA_ARGS__?
我有一个宏来为每个参数调用静态函数.
I hava a macro to call static function for each args.
例如:
#define FOO(X) X::do();
#define FOO_1(X,Y) X::do(); Y::do();
我的问题是我需要使用带有可变数量参数的 foo,是否可以使用 __VA_ARGS__
?
My question is that I need to use foo with variable number of arguments, is it possible to use __VA_ARGS__
?
如下一行:
#define FOO(...) __VA_ARGS__::do() ?
谢谢
推荐答案
宏扩展不像使用可变参数模板的参数包扩展那样工作.您所拥有的将扩展为:
Macro expansion does not work like argument pack expansion with variadic templates. What you have will expand to:
X,Y::do();
而不是
X::do(); Y::do();
如你所愿.但在 C++11 中,您可以使用可变参数模板.例如,你可以这样做:
As you hoped. But in C++11 you could use variadic templates. For instance, you could do what you want this way:
#include <iostream>
struct X { static void foo() { std::cout << "X::foo()" << std::endl; }; };
struct Y { static void foo() { std::cout << "Y::foo()" << std::endl; }; };
struct Z { static void foo() { std::cout << "Z::foo()" << std::endl; }; };
int main()
{
do_foo<X, Y, Z>();
}
你只需要这个(相对简单的)机器:
All you need is this (relatively simple) machinery:
namespace detail
{
template<typename... Ts>
struct do_foo;
template<typename T, typename... Ts>
struct do_foo<T, Ts...>
{
static void call()
{
T::foo();
do_foo<Ts...>::call();
}
};
template<typename T>
struct do_foo<T>
{
static void call()
{
T::foo();
}
};
}
template<typename... Ts>
void do_foo()
{
detail::do_foo<Ts...>::call();
}
这是一个现场示例.
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