将参数包转换为向量 [英] converting parameter pack into a vector

问题描述

我试图理解C ++中的可变参数模板，但在以下示例中迷失了：想象一个函数 foo（T，T，T，T），该函数采用相同参数的可变参数输入T并将其转换为向量。知道如何实现吗？

I am trying to understand variadic templates in C++ and I am lost in the following example: Imagine a function foo(T, T, T, T) which takes variadic number of arguments of the same type T and converts them into a vector. Any idea how to implement one?

它应该像这样

foo<int>(1,2,3,4) returns std::vector<int> x{1,2,3,4}
foo<double>(0.1,0.2,0.3) returns std::vector<double> x{0.1,0.2,0.3}

推荐答案

如果 T 值在编译时是已知的，您可以将它们作为模板参数传递并编写类似

If the T values as known at compile-time, you can pass they as template parameters and write something like

template<typename T, T ... Is>
void foo() {
   std::vector<T> x { { Is.. } };

   for( auto xx:x )
      std::cout << xx << std::endl;
}

称为

foo<int, 2, 3, 5, 7>();

否则，您必须将它们作为参数传递；

Otherwise you have to pass they as arguments; something like

template <typename T, typename ... ARGS>
void foo (ARGS const & ... args) {    
   std::vector<T> x { { args... } };

   for( auto xx:x ) 
      std::cout << xx << std::endl;
}

称为

foo<int>(2, 3, 5, 7);

或也（从中推导类型 T

template <typename T, typename ... ARGS>
void foo (T const & arg0, ARGS const & ... args) {    
   std::vector<T> x { { arg0, args... } };

   for( auto xx:x ) 
      std::cout << xx << std::endl;
}

称为

foo(2, 3, 5, 7);

-编辑-

OP写入

它应该像这样

It should work like this

foo<int>(1,2,3,4) returns std::vector<int> x{1,2,3,4}
foo<double>(0.1,0.2,0.3) returns std::vector<double> x{0.1,0.2,0.3}

所以我想您可以简单地写

So I suppose you can simply write

template <typename T, typename ... ARGS>
std::vector<T> foo (ARGS const & ... args)
 { return { args... }; }

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