C ++ std :: tuple type, std :: tie function, std :: get function

 

tuple: tuple

1. std::tuple

https://zh.cppreference.com/w/cpp/utility/tuple

(Since C ++ 11) 

Class template std::tupleis a heterogeneous collection of fixed size value. It is a promotion of std :: pair .

 

example:

std::tuple<int, int> foo_tuple() 
{
  return {1, -1};  // N4387 前错误
  return std::tuple<int, int>{1, -1};  // 始终有效, tuple可以使用初始化列表进行赋值。
  return std::make_tuple(1, -1); // 始终有效
}

 

example:

#include <tuple>
#include <iostream>
#include <string>
#include <stdexcept>
 
std::tuple<double, char, std::string> get_student(int id)
{
    if (id == 0) return std::make_tuple(3.8, 'A', "Lisa Simpson"); //生成一个std::tuple
    if (id == 1) return std::make_tuple(2.9, 'C', "Milhouse Van Houten");
    if (id == 2) return std::make_tuple(1.7, 'D', "Ralph Wiggum");
    throw std::invalid_argument("id");
}
 
int main()
{
    auto student0 = get_student(0);
    std::cout << "ID: 0, "
              << "GPA: " << std::get<0>(student0) << ", " //std::get用于取出tuple中的某个值
              << "grade: " << std::get<1>(student0) << ", "
              << "name: " << std::get<2>(student0) << '\n';
 
    double gpa1;
    char grade1;
    std::string name1;
    std::tie(gpa1, grade1, name1) = get_student(1);//std::tie的用法！！std::tie会将变量的引用整合成一个tuple，从而实现批量赋值
    std::cout << "ID: 1, "
              << "GPA: " << gpa1 << ", "
              << "grade: " << grade1 << ", "
              << "name: " << name1 << '\n';
 
    // C++17 结构化绑定：
    auto [ gpa2, grade2, name2 ] = get_student(2);
    std::cout << "ID: 2, "
              << "GPA: " << gpa2 << ", "
              << "grade: " << grade2 << ", "
              << "name: " << name2 << '\n';
}

 

You can also use std :: ignore to ignore certain return values ​​in certain tuples, as in the example above:

double gpa1;
std::string name1;
std::tie(gpa1, ignore, name1) = get_student(1);

can

 

2. std::get(std::tuple)

https://zh.cppreference.com/w/cpp/utility/tuple/get

Use get<常量表达式>(tuple_name)to access or modify (const type reference returned from c ++ 14, can not be modified) elements of tuple (return reference).

Example:

#include <iostream>
#include <string>
#include <tuple>
 
int main()
{
    auto t = std::make_tuple(1, "Foo", 3.14);
    // 基于下标的访问
    std::cout << "(" << std::get<0>(t) << ", " << std::get<1>(t)
              << ", " << std::get<2>(t) << ")\n";
    // 基于类型的访问（ C++14 起）
    std::cout << "(" << std::get<int>(t) << ", " << std::get<const char*>(t)
              << ", " << std::get<double>(t) << ")\n";
    // 注意： std::tie 和结构化绑定亦可用于分解 tuple
}

---

I can't.