C++ 初学者：Switch 上的快捷方式

Question

我试图为一个练习编写此代码，该练习显示并允许用户通过输入与他们选择的饮料相对应的数字 (1 - 5) 来选择饮料。然后它会打印出指定他们选择的饮料，如果他们输入了 1-5 之外的任何内容，则会显示错误消息。

下面是功能代码：

#include <vector>

std::string Vending (){
int num;
std::vector<std::string> beverage ={"Coke", "Sprite", "Water", "Pepsi", "7-Up"};
std::cin >> num;
switch (num){
  case (1) :
  std::cout << "You have chosen " << beverage[num-1];
  break;
  case (2) :
  std::cout << "You have chosen " << beverage[num-1];
  break;
  case (3) :
  std::cout << "You have chosen " << beverage[num-1];
  break;
  case (4) :
  std::cout << "You have chosen " << beverage[num-1];
  break;
  case (5) :
  std::cout << "You have chosen " << beverage[num-1];
  break;

default :
 std::cout << "Error, choice was not valid, here is your money back.\n";
 break;
}
}

该程序运行良好，但我只是想知道是否有一种方法可以使 switch() 更短，例如在 case 中放置一个范围。 case(//range from 1-5) 而不是根据案例编号单独编写每个案例。

任何提示将不胜感激！ :)

Answer 1

if 语句在这里更合适，因为switch 块并非设计用于在范围上操作。

如果您必须使用switch，并且热衷于编写可移植的 C++（因此 gcc 范围 switch 不是一个选项），那么您可以编写

switch (num){
case 1: case 2: case 3: case 4: case 5:
    std::cout << "You have chosen " << beverage[num-1];
    break;
default :
     std::cout << "Error, choice was not valid, here is your money back.\n";
}

但是写起来会远更自然

if (num >= 1 && num <= beverage.size()){
    std::cout << "You have chosen " << beverage.at(num - 1);
} else {
    std::cout << "Error, choice was not valid, here is your money back.\n";
}

还有一个额外的好处是我没有硬编码饮料的数量。使用at 还可以帮助您防止未定义的行为，因为始终进行向量边界检查（即使该语句当前由if 保护）。

Answer 2

您根本不需要switch 或任何类型的选择，只需一些基本检查输入是否在范围内，然后打印它：

if (num > 0 && num <= beverage.size())
{
    std::cout << "You have chosen " << beverage[num-1];
}
else
{
    std::cout << "Error, choice was not valid, here is your money back.\n";
}