c++可以实现去掉一层模板么

shylockhg · 2019-11-06T04:57:49Z

就是这样 typename unwrap >::type // int 我这里粗糙的试了一下好像不行 #include #include template struct unwrap { using type = T; }; template struct unwrap > { using type = T; }; int main(int argc, char* argv[]) { typename unwrap >::type a = 0; }

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就是这样

typename unwrap<std::vector<int>>::type // int

我这里粗糙的试了一下好像不行

#include <type_traits>
#include <vector>

template <typename T>
struct unwrap {
	using type =  T;
};

template <typename wrapper, typename T>
struct unwrap<wrapper<T>> {
	using type = T;
};

int main(int argc, char* argv[]) {
	typename unwrap<std::vector<int>>::type a = 0;
}

4 replies • 2019-11-06 14:57:49 +08:00

codehz

Nov 6, 2019 via Android

template<template <typename> X, typename T>
就可以匹配模板模板了，不过对于这个例子，vector 的完整形式有两个 typename，所以还要写一个 typename 才能匹配上

codehz

Nov 6, 2019

前面少了一个 typename，下面是完整例子
https://gist.github.com/codehz/b18c62d104f2b7f1a4f3bfb44c5d249b

shylockhg

Nov 6, 2019

@codehz 翻不了墙。

codehz

Nov 6, 2019

@shylockhg #3
#include <vector>

template <typename T>
struct unwrap {
using type = T;
};

template <template <typename> typename wrapper, typename T>
struct unwrap<wrapper<T>> {
using type = T;
};

template <template <typename, typename> typename wrapper, typename T, typename R>
struct unwrap<wrapper<T, R>> {
using type = T;
};

int main() {
typename unwrap<std::vector<int>>::type a = 0;
}