如何以不同的方式专门化 typedef 及其隐式类型？
我有这样的事情：
typedef int AnotherType;
template <typename T> Func( T Value );
// And I want to specialize these two cases separately:
template <> bool Func<int>( int Value ) {...}
template <> bool Func<AnotherType>( AnotherType Value ) {...}
我真的不需要专门针对 int，我真正需要的是为 AnotherType 执行不同的函数。我无法更改 AnotherType 或基本函数的定义。
由于 SFINAE，重载也无济于事。
I have something like this:
typedef int AnotherType;
template <typename T> Func( T Value );
// And I want to specialize these two cases separately:
template <> bool Func<int>( int Value ) {...}
template <> bool Func<AnotherType>( AnotherType Value ) {...}
I don't really need to specialize for int, what I really need is to execute a different function for AnotherType. And I cannot change the definition of AnotherType or the base function.
Overloading doesn't help either because of SFINAE.
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答案是否定的。当您 typedef 时，您为类型创建了别名，而不是其本身的实际类型。编译器会将两者视为相同。这就是为什么：
会编译。他们都是整数。
The answer is no. When you typedef you create an alias for a type, not an actual type in and of itself. The compiler will treat both the same. That's why:
Will compile. They're both ints.
您可以使用BOOST_STRONG_TYPEDEF。
You could use BOOST_STRONG_TYPEDEF.
我很确定你不能让编译器以不同的方式对待 int 和 AnotherType 。 typedef 所做的只是别名类型——它实际上并不创建新类型；而是创建新类型。根据 typedef 构造的定义，编译器在所有情况下都会同等对待 int 和 AnotherType。
如果您需要一个仅包含 int 且以不同方式处理的类型，您可能应该只创建一个单成员
struct。对包含的 int 进行的大多数操作将编译为与裸 int 相同的机器代码，但现在您的数据类型可以拥有自己的模板专业化等。I'm pretty sure you can't have the compiler treat int and AnotherType differently. All typedef does is alias types -- it doesn't actually create a new type; by definition of the typedef construct, the compiler will treat int and AnotherType equivalenty in all cases.
If you need to have a type with just an int that IS treated differently, you should probably just make a single-member
struct. Most operations on the contained int will compile to the same machine code as a bare int, but now your data type can have its own template specializations and such.而且我无法更改 AnotherType 或基本函数的定义。
那么你就完蛋了。对不起。如果您无法更改定义以使用强 typedef，那么您真正拥有的唯一选项（强 typedef）就不是一个选项。
And I cannot change the definition of AnotherType or the base function.
Then you're screwed. Sorry. The only option you really have, a strong typedef, is not an option if you can't change the definition to use a strong typedef.
编译器会将这两种特化视为完全相同，因为
AnotherType只是int的另一个名称。您说您不需要专门化int，因此只需完全删除该专门化并让它专门化AnotherType恰好适用的任何类型。The compiler will treat both of the specializations as exactly the same, since
AnotherTypeis just another name forint. You say you don't need to specialize forint, so just remove that specialization completely and let it specialize on whatever typeAnotherTypehappens to work out to.