void类型在std :: tuple中 [英] Void type in std::tuple
问题描述
显然,在一个格式良好的程序中不能有类型 void
的实例,因此类似下面的声明不会编译:
Obviously, you can't have an instance of type void
in a well-formed program, so something like the following declaration won't compile:
std::tuple<void, double, int> tup;
然而,只要我们严格地处理类型而不是对象,没有任何问题。例如,我的编译器(GCC)让我说:
However, as long as we're dealing strictly with types as opposed to objects, there seems to be no issue. For example, my compiler (GCC) lets me say:
typedef std::tuple<void, double, int> tuple_type;
这很有趣,因为看起来对于C ++ 0x,我们可以使用 std :: tuple
来执行很多元编程技巧,以前需要 boost :: mpl
库。例如,我们可以使用 std :: tuple
创建一个类型的向量。
This is interesting to me, because it seems that with C++0x we can just use std::tuple
to perform a lot of the meta-programming tricks that earlier would have required the boost::mpl
library. For example, we can use std::tuple
to create a vector of types.
例如,假设我们想要创建表示函数签名的类型的向量:
For example, suppose we want to create a vector of types representing a function signature:
我们可以说:
template <class R, class... Args>
struct get_function_signature;
template <class R, class... Args>
struct get_function_signature<R(*)(Args...)>
{
typedef std::tuple<R, Args...> type;
};
这似乎工作,即使函数签名有 void
类型,只要我们从不实际实例化
get_function_signature< F> :: type
的实例。
This seems to work, even if the function signature has a void
type, as long as we never actually instantiate an instance of get_function_signature<F>::type
.
然而,C ++ 0x对我来说还是新的,当然所有的实现仍然有些实验性,所以我有点不安。我们可以使用 std :: tuple
作为元程序的类型向量吗?
However, C++0x is still new to me, and of course all implementations are still somewhat experimental, so I'm a bit uneasy about this. Can we really use std::tuple
as a vector of types for meta-programming?
推荐答案
它实际上是有意义的,你可以做
It does actually make sense that you can do
typedef std :: tuple< void,double,int& tuple_type;
只要使用它作为类型列表使用 tuple_element
。因此我可以做
as long as you only use it as a type-list to use tuple_element
on. Thus I can do
tuple_element< 0,tuple_type> :: type * param;
将声明param为 void *
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