方法 1
#include <iostream>
// 通过递归实现
template <int Beg, int End>
struct static_for
{
template <typename Fn>
void operator()(const Fn& fn) const
{
if (Beg < End)
{
fn(Beg);
static_for<Beg+1, End>()(fn); // 最后一次,2个数相等了,会调用下面的 static_for<N, N>
}
}
};
template <int N>
struct static_for<N, N>
{
template <typename Fn>
void operator()(Fn const& fn) const
{
std::cout << "static_for<N, N>" << std::endl;
}
};
int main() {
static_for<0, 5>()([&](int i){
std::cout << i << std::endl;
});
return 0;
}
方法 2
template<bool Cond>
struct my_enable_if {
};
template<>
struct my_enable_if<true> {
typedef void type;
};
template<int Beg, int End, typename Fn>
typename my_enable_if<Beg == End>::type static_for(Fn const &fn) {
}
template<int Beg, int End, typename Fn>
typename my_enable_if<Beg != End>::type static_for(Fn const &fn) {
fn(Beg);
static_for<Beg + 1, End>(fn);
}
int main() {
static_for<0, 5>([&](int i){
std::cout << i << std::endl;
});
return 0;
}
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