符合标准的方式来定义我的枚举
如何在不正确明确枚举范围的情况下摆脱警告？符合标准的代码将与 foo::bar::mUpload 进行比较（请参阅这里），但是显式作用域真的很长，使得该死的东西不可读。
也许还有另一种不使用 typedef 的方法？我不想修改枚举——我没有编写它并且它在其他地方使用。
警告 C4482：使用了非标准扩展：在限定名称中使用了枚举“foo::bar::baz”
namespace foo {
class bar {
enum baz {mUpload = 0, mDownload};
}
}
typedef foo::bar::baz mode_t;
mode_t mode = getMode();
if (mode == mode_t::mUpload) //C4482
{
return uploadthingy();
}
else
{
assert(mode == mode_t::mDownload); //C4482
return downloadthingy();
}
How can I get rid of the warning, without explicitly scoping the enum properly? The standards-compliant code would be to compare against foo::bar::mUpload (see here), but the explicit scopes are really long and make the darn thing unreadable.
maybe there's another way that doesn't use typedef? i don't want to modify the enum--i didn't write it and its in use elsewhere.
warning C4482: nonstandard extension used: enum 'foo::bar::baz' used in qualified name
namespace foo {
class bar {
enum baz {mUpload = 0, mDownload};
}
}
typedef foo::bar::baz mode_t;
mode_t mode = getMode();
if (mode == mode_t::mUpload) //C4482
{
return uploadthingy();
}
else
{
assert(mode == mode_t::mDownload); //C4482
return downloadthingy();
}
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如果枚举是在类中定义的，那么您能做的最好的事情就是将该类带入您自己的范围中，然后仅使用
class_name::value或定义该类的 typedef。在 C++03 中，枚举的值是封闭范围的一部分（在您的情况下是类）。在 C++0x/11 中，您将能够使用枚举名称限定值：将来，您的用法将是正确的 (C++11)：
If the enum is defined within a class, the best that you can do is bring the class into your own scope and just use
class_name::valueor define a typedef of the class. In C++03 the values of an enum are part of the enclosing scope (which in your case is the class). In C++0x/11 you will be able to qualify the values with the enum name:In the future, your usage will be correct (C++11):
您可以将枚举封装到命名空间中，然后在该命名空间上使用
using语句。这显然只适用于类范围之外的枚举。
在您的情况下，我不明白您为什么不将其称为
bar::mUpload（在using namespace foo或using foo::bar 之后）You can enclose your enum into a namespace, and then use a
usingstatement on that namespace.This obviously only works for enum's outside of class scope.
In your case I don't see why don't you refer to it as
bar::mUpload(afterusing namespace foo, orusing foo::bar)您可以将 typedef 用于 foo::bar:
或者您正在寻找不同的东西？
You can use a typedef for foo::bar:
Or are you looking for something different?