C++ 宏中的模板? [英] Template in a Macros in C++?

问题描述

可以吗

# define abc<T1> __abc<T1, T2>

template<typename T2> void somefun() {
    ... 
    abc<int>(...);
    abc<double>(...);
    ...
}

只是为了不每次打电话给 abc 都写

Just to not write it every time i call abc

推荐答案

在 C++11 中你可以:

In C++11 you can do:

template<typename T2> void somefun() {
    template <typename T>
    using abc = __abc<T, T2>;
}

没有它，您可以使用宏，但您需要这样做:

Without that you can use a macro but you'd need to do:

#define abc(T1) __abc<T1, T2>

//usage:

abc(Type) instance;

但由于这看起来不太自然，我个人会避免使用.

but since that doesn't look very natural I'd avoid it personally.

如果您想避免 C++11 之前的宏，您可以执行以下操作:

If you want to avoid the macro pre-C++11 you can do something like:

template <typename T2>
struct type {
  template <typename T1>
  struct lookup {
    typedef __abc<T1,T2> type;
  };
};

template <typename T2> void somefun() {
  typedef type<T2> abc;
  typename abc::template lookup<int>::type();
}

但老实说，它的可读性甚至不如宏案例

But in all honesty that's less readable than even the macro case

(注:__abc 保留)

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