Question

我正在编写一个升级器，它应该将一个变量arity函数提升到一些类似于带有索引的SQL表的std :: vectors中。我想将f参数应用于所有向量中具有相同id的每组元素。我遇到了模板推导的问题，我在这里提炼它（减去迭代器逻辑）。我写过我认为是一个明智的递归案例，但模板演绎认为它不可行。

// Simple container type
template <typename T>
struct A {
  T x;
  T y;
  T z;
};

// Wrapper type to differentiate an extracted value from a container
// in template deduction
template <typename T>
struct B {
  T value;
};

// Base case. All A<Ts> have been extracted to B<Ts>.
template <typename F, typename... Ts>
void lift (F f, B<Ts> ...bs) {
  // Call f with extracted values
  f(bs.value...);
}

// Recursive case
template <typename F, typename T, typename... Ts, typename Us>
void lift (F f, A<T> a, A<Ts> ...as, B<Us> ...bs) {
  // Copy a value from the beheaded container A<T>.
  B<T> b = {a.x};
  // Append this B<T> to args and continue beheading A<Ts>.
  lift(f, as..., bs..., b);
}

// Test function
template <typename... Ts>
struct Test {
  void operator () (Ts...) {}
};

int main () {

  B<int> b = {1};
  // No errors for the base case
  lift(Test<>());
  lift(Test<int, int>(), b, b);

  // error: no matching function for call to 'lift'
  // The notes refer to the recursive case
  A<int> a = {1,0,0};
  lift(Test<int>(), a); // note: requires at least 3 arguments, but 2 were provided
  lift(Test<int>(), a, a); // note: requires 2 arguments, but 3 were provided
  lift(Test<int>(), a, a, b); // note: requires 2 arguments, but 4 were provided
}

这段代码出了什么问题？

为了便于阅读和写作，忽略一切都是通过值传递的。为什么不编译？

Answer 1

为什么不直接提取A和B的值？

template <typename T>
T extractValue(A<T> a)
{
    return a.x;
}
template <typename T>
T extractValue(B<T> b)
{
    return b.value;
}

template <typename F, typename... T>
void lift (F f, T... values) {
    f(extractValue(values)...);
}

此外，您的最后2个测试用例无法编译，因为参数的数量是错误的。

lift(Test<int,int>(), a, a);
lift(Test<int,int,int>(), a, a, b);

我可以在构建新模板时使用一个可变参数模板参数吗？

1 个答案: