类型列表的运行时类型开关作为开关而不是嵌套的 if?
这是来自 TTL:
////////////////////////////////////////////////////////////
// run-time type switch
template <typename L, int N = 0, bool Stop=(N==length<L>::value) > struct type_switch;
template <typename L, int N, bool Stop>
struct type_switch
{
template< typename F >
void operator()( size_t i, F& f )
{
if( i == N )
{
f.operator()<typename impl::get<L,N>::type>();
}
else
{
type_switch<L, N+1> next;
next(i, f);
}
}
};
它用于在 TypeList 上进行类型切换.问题是――他们通过一系列嵌套的 if 来做到这一点.有没有办法将这种类型切换作为单个选择语句来代替?
It's used for typeswitching on a TypeList. Question is -- they are doing this via a series of nested if's. Is there a way to do this type switch as a single select statement instead?
谢谢！
推荐答案
你需要预处理器来生成一个大的switch.您需要 get<> 来进行无操作越界查找.检查编译器输出以确保未使用的情况不会产生输出，如果您关心的话；根据需要进行调整；v).
You'll need the preprocessor to generate a big switch. You'll need get<> to no-op out-of-bound lookups. Check the compiler output to be sure unused cases produce no output, if you care; adjust as necessary ;v) .
如果您想精通这类事情，请查看 Boost 预处理器库……
Check out the Boost Preprocessor Library if you care to get good at this sort of thing…
template <typename L>
struct type_switch
{
template< typename F >
void operator()( size_t i, F& f )
{
switch ( i ) {
#define CASE_N( N )
case (N): return f.operator()<typename impl::get<L,N>::type>();
CASE_N(0)
CASE_N(1)
CASE_N(2)
CASE_N(3) // ad nauseam.
}
};
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