所有具有特征的容器类型的模板begin（）end（）size（）(template for all container types with features begin() end() size())

为矢量编写以下模板代码。 它不会编译，因为一个调用是使用列表而不是向量。 我想重写模板，以便它适用于任何容器。 对该容器的限制是它需要包含一个类型T，有一个begin（），有一个end（）并且有一个大小（）;

这可以用模板完成吗？ 我会怎么做？

#include <iostream> #include <string> #include <vector> #include <list> #include <typeinfo> namespace std { template <typename T> string to_string(T s) { return typeid(T).name(); } string to_string(string s) { return s; } } template<typename T, typename A = std::allocator<T>> static std::string range_to_string(std::vector<T, A> const& vec) { bool more = false; std::string str = "size:" + std::to_string(vec.size()) + " ["; for (const T & item : vec) { if (more) { str += ", "; } str += std::to_string(item); more = true; } str += "]"; return str; } void demo1() { std::string dat[] = {"one", "two", "three"}; std::vector<std::string> vs(dat, dat + 3); std::cout << range_to_string(vs) << std::endl; } void demo2() { std::list<short> ls= {1, 2, 3, 4}; std::cout << range_to_string(ls) << std::endl; } int main() { demo1(); demo2(); return 0; }

The following template code is written for vectors. It does not compile because of one call is using list rather than vector. I want to rewrite the template so that it works for any container. The restrictions on that container are that it needs to contain a type T, have a begin(), have an end() and have a size();

Can this be done with templates? How would I do that?

#include <iostream> #include <string> #include <vector> #include <list> #include <typeinfo> namespace std { template <typename T> string to_string(T s) { return typeid(T).name(); } string to_string(string s) { return s; } } template<typename T, typename A = std::allocator<T>> static std::string range_to_string(std::vector<T, A> const& vec) { bool more = false; std::string str = "size:" + std::to_string(vec.size()) + " ["; for (const T & item : vec) { if (more) { str += ", "; } str += std::to_string(item); more = true; } str += "]"; return str; } void demo1() { std::string dat[] = {"one", "two", "three"}; std::vector<std::string> vs(dat, dat + 3); std::cout << range_to_string(vs) << std::endl; } void demo2() { std::list<short> ls= {1, 2, 3, 4}; std::cout << range_to_string(ls) << std::endl; } int main() { demo1(); demo2(); return 0; }

最满意答案

我在这里找到了我的答案。

看起来这是常识。

#include <iostream> #include <string> #include <vector> #include <list> #include <typeinfo> namespace std { template <typename T> string to_string(T s) { return typeid(T).name(); } string to_string(short s) { return to_string((int) s); } string to_string(string s) { return s; } } template<typename T, typename A = std::allocator<T>, template <typename, typename> class container> static std::string range_to_string(container<T, A> const& vec) { bool more = false; std::string str = "size:" + std::to_string(vec.size()) + " ["; for (const T & item : vec) { if (more) { str += ", "; } str += std::to_string(item); more = true; } str += "]"; return str; } void demo1() { std::string dat[] = {"one", "two", "three"}; std::vector<std::string> vs(dat, dat + 3); std::cout << range_to_string(vs) << std::endl; } void demo2() { std::list<short> ls= {1, 2, 3, 4}; std::cout << range_to_string(ls) << std::endl; } int main() { demo1(); demo2(); return 0; }

I found my answer here.

It looks like this is common knowledge.

#include <iostream> #include <string> #include <vector> #include <list> #include <typeinfo> namespace std { template <typename T> string to_string(T s) { return typeid(T).name(); } string to_string(short s) { return to_string((int) s); } string to_string(string s) { return s; } } template<typename T, typename A = std::allocator<T>, template <typename, typename> class container> static std::string range_to_string(container<T, A> const& vec) { bool more = false; std::string str = "size:" + std::to_string(vec.size()) + " ["; for (const T & item : vec) { if (more) { str += ", "; } str += std::to_string(item); more = true; } str += "]"; return str; } void demo1() { std::string dat[] = {"one", "two", "three"}; std::vector<std::string> vs(dat, dat + 3); std::cout << range_to_string(vs) << std::endl; } void demo2() { std::list<short> ls= {1, 2, 3, 4}; std::cout << range_to_string(ls) << std::endl; } int main() { demo1(); demo2(); return 0; }

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