c++ - 枚举缺失导致编译器出错

Question

遵循 here 中的代码:

enum class OS_type { Linux, Apple, Windows };

const std::string ToString(OS_type v)
{
    switch (v)
    {
        case Linux:   return "Linux";
        case Apple:   return "Apple";
        case Windows: return "Windows";
        default:      return "[Unknown OS_type]";
    }
}

我想删除 default 并强制编译器在我的枚举上的切换未完成时生成错误。

Answer 1

海湾合作委员会/ clang

您正在寻找-Wswitch-enum .

Warn whenever a switch statement has an index of enumerated type and lacks a case for one or more of the named codes of that enumeration. case labels outside the enumeration range also provoke warnings when this option is used. The only difference between -Wswitch and this option is that this option gives a warning about an omitted enumeration code even if there is a default label.

const std::string ToString(OS_type v)
{
    // warning: enumeration value 'Windows' not handled in switch [-Wswitch-enum]
    switch (v)
    {
        case OS_type::Linux:   return "Linux";
        case OS_type::Apple:   return "Apple";
        default:      return "[Unknown OS_type]";
    }
}

即使使用默认值，它也会提示缺少 Windows 枚举。只需创建适用于 Windows 的案例，然后通过默认设置来取消枚举。

Visual Studio

VS 在编译器级别 3 和 4 处理此问题。您需要启用警告 C4061/C4062。 https://msdn.microsoft.com/en-US/library/96f5t7fy.aspx