c++ - 有没有办法执行 "if (condition) typedef ..."

Question

当且仅当满足编译时条件时，我想执行 typedef。如果条件不满足，则根本不执行typedef。

这在 C++11 中可行吗？

例子:

class A {
  std::conditional_typedef<true,int,myType1>; // Performs "typedef int myType1".
  std::conditional_typedef<false,int,myType2>; // Does nothing at all.
};

我正在寻找这个虚构的 std::conditional_typedef。

Answer 1

另一种方法是从基类的特化中传递

// foo is a light struct (only a typedef or not at all) that can be
// developed in two versions

template <bool>
struct foo;

template <>
struct foo<true>
 { typedef int myType1; }; // or using myType1 = int;

template <>
struct foo<false>
 { };

template <bool B>
struct bar : public foo<B> // B can be a type_traits value derived
                           // from template arguments for bar
 {
   // potential complex struct or class, developed only one time 
 };


int main()
 {
   bar<true>::myType1 mt1 { 1 };
   // bar<false>::myType1 mt2 { 1 }; error: ‘myType1’ is not a member of ‘bar<false>’
 }