通过专业化:

template <bool> class Foo;

template <> class Foo<true>
{
  typedef int data_t;
};

template <> class Foo<false>
{
  typedef unsigned int data_t;
};

您可以选择将两种情况之一设为主要模板，将另一种情况设为专业化，但是考虑到 bool 只能具有两个值，因此我更喜欢这种对称的版本.

You can choose to make one of the two cases the primary template and the other one the specialization, but I prefer this more symmetric version, given that bool can only have two values.

_{如果这是您第一次看到此内容，您可能还想考虑 partial 专业化:}

template <typename T> struct remove_pointer     { typedef T type; };
template <typename U> struct remove_pointer<U*> { typedef U type; };

如@Nawaz所说，最简单的方法可能是 #include< type_traits> 并说:

As @Nawaz says, the easiest way is probably to #include <type_traits> and say:

typedef typename std::conditional<S, int, unsigned int>::type data_t;