c++ 不能出现在常量表达式中| [复制]

可能的重复：
如何在 switch 语句中选择值范围？ https://stackoverflow.com/questions/9432226/how-do-i-select-a-range-of-values-in-a-switch-statement

我遇到了一些错误，我已经搜索了一段时间，但我不知道错误的原因是什么。 （我对编程还很陌生。）

以下是我收到的错误：

error: 'Essais' cannot appear in a constant-expression| (line 200)
warning: overflow in implicit constant conversion| (line 202)

我有箱子和科特迪瓦：

  char AfficherCote (int Essais)
 {
 char Cote;
 switch (Essais)
  {
(line200)       case Essais<=20:
    {
(line 202)           Cote='Excellent';

        return (Cote);
        break;
    }
    case Essais<=40:
    {
        Cote='Très bon';
        return (Cote);
        break;
    }
    case Essais<=60:
    {
        Cote='Bon';
        return (Cote);
        break;
    }
    case Essais<=80:
    {
        Cote='Moyen';
        return (Cote);
        break;
    }
    case Essais<=100:
    {
        Cote='Muvais';
        return (Cote);
        break;
    }
    case Essais>=100:
    {
        Cote='Très mauvais';
        return (Cote);
    }
  }
}

switch-case only works with constant values^(*) (such as 3 or 'a'), not with ranges (such as <=100). You also must not include the variable name in the case statement. Correct syntax would be as follows:

switch (Essais)
{
case 1:
   /* ... */
  break;
case 2:
   /* ... */
   break;
default:
   /* ... */
}

如果您需要范围测试，请使用if代替switch-case:

if (Essais <= 80)
  return "Cote";
else if (Essais <= 100)
  return "Muvais";

另请注意，不能使用单引号'对于字符串。使用双引号"相反，并使用类型的变量std::string (not char) 来存储字符串。

^(*) To be precise, the condition given in the case statements must be a constant expression of integral type, enumeration type, or class type convertible to integer or enumeration type (see §6.4.2/2 of the C++ Standard for details).