C ++初学者：开关上的快捷方式

Question

我正在尝试为一个练习编写此代码，该练习提出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）选择饮料。然后，它将打印出来，指定他们选择了哪种饮料，如果他们输入的不是1-5，则显示错误消息。

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;
}
}

程序运行正常，但我只是想知道是否存在一种使switch()变短的方法，例如将一个范围设置为大小写。 case(//range from 1-5)而不是每个案例编号单独写入每个案例。

任何提示将不胜感激！ :)

Answer 1

您根本不需要switch或任何类型的选择，只需进行一些基本的检查即可确定输入是否在范围内，然后打印出来：

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";
}

Answer 2

if语句在这里更合适，因为switch块的设计目的不是在范围内操作。

如果您必须使用switch，并且热衷于编写可移植的C ++（因此无法选择gcc范围switch，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[[far

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";
}

它具有额外的优势，我还没有对饮料的数量进行硬编码。使用at还可以帮助您防范未定义的行为，因为始终会进行矢量边界检查。