我正在尝试编写一个函数，根据枚举的运行时值将值的枚举映射到一组类型。我意识到您不能根据枚举的运行时值返回不同的类型，因为编译器不知道要分配多少堆栈空间。但是，我正在尝试将其编写为 constexpr 函数，使用新的 if-constexpr 功能来实现它。

我收到来自 clang 的错误，提示我使用了非法指定的模板参数。有人知道如何实现吗？

编辑:这是一个更容易理解的版本，更简洁地展示了我的问题: http://coliru.stacked-crooked.com/a/2b9fef340bd167a8

旧代码:

#include <cassert>
#include <tuple>
#include <type_traits>

namespace
{

enum class shape_type : std::size_t
{
  TRIANGLE = 0u,
  RECTANGLE,
  POLYGON,
  CUBE,
  INVALID_SHAPE_TYPE
};

template<std::size_t T>
struct shape {
};

using triangle = shape<static_cast<std::size_t>(shape_type::TRIANGLE)>;
using rectangle = shape<static_cast<std::size_t>(shape_type::RECTANGLE)>;
using polygon = shape<static_cast<std::size_t>(shape_type::POLYGON)>;
using cube = shape<static_cast<std::size_t>(shape_type::CUBE)>;

template<std::size_t A, std::size_t B>
static bool constexpr same() noexcept { return A == B; }

template<std::size_t ST>
static auto constexpr make_impl(draw_mode const dm)
{
  if constexpr (same<ST, shape_type::TRIANGLE>()) {
    return triangle{};
  } else if (same<ST, shape_type::RECTANGLE>()) {
    return rectangle{};
  } else if (same<ST, shape_type::POLYGON>()) {
    return polygon{};
  } else if (same<ST, shape_type::CUBE>()) {
    return cube{};
  } else {
    assert(0 == 5);
  }
}

static auto constexpr make(shape_type const st, draw_mode const dm)
{
  switch (st) {
      case shape_type::TRIANGLE:
        return make_impl<shape_type::TRIANGLE>(dm);
      case shape_type::RECTANGLE:
        return make_impl<shape_type::RECTANGLE>(dm);
      case shape_type::POLYGON:
        return make_impl<shape_type::POLYGON>(dm);
      case shape_type::CUBE:
        return make_impl<shape_type::CUBE>(dm);
      case shape_type::INVALID_SHAPE_TYPE:
        assert(0 == 17);
  }
}

} // ns anon

////////////////////////////////////////////////////////////////////////////////////////////////////
// demo
int main()
{
}

错误:

/home/benjamin/github/BoomHS/main.cxx:42:6: warning: constexpr if is a
C++1z extension [-Wc++1z-extensions]

if constexpr (same<ST, shape_type::TRIANGLE>()) {
        ^ /home/benjamin/github/BoomHS/main.cxx:59:16: error: no matching function for call to 'make_impl'
        return make_impl<shape_type::TRIANGLE>(dm);
               ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ /home/benjamin/github/BoomHS/main.cxx:40:23: note: candidate template
ignored: invalid explicitly-specified argument for template parameter
'ST' static auto constexpr make_impl(draw_mode const dm)
                      ^ /home/benjamin/github/BoomHS/main.cxx:61:16: error: no matching function for call to 'make_impl'
        return make_impl<shape_type::RECTANGLE>(dm);
               ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ /home/benjamin/github/BoomHS/main.cxx:40:23: note: candidate template
ignored: invalid explicitly-specified argument for template parameter
'ST' static auto constexpr make_impl(draw_mode const dm)
                      ^ /home/benjamin/github/BoomHS/main.cxx:63:16: error: no matching function for call to 'make_impl'
        return make_impl<shape_type::POLYGON>(dm);
               ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ /home/benjamin/github/BoomHS/main.cxx:40:23: note: candidate template
ignored: invalid explicitly-specified argument for template parameter
'ST' static auto constexpr make_impl(draw_mode const dm)
                      ^ /home/benjamin/github/BoomHS/main.cxx:65:16: error: no matching function for call to 'make_impl'
        return make_impl<shape_type::CUBE>(dm);
               ^~~~~~~~~~~~~~~~~~~~~~~~~~~ /home/benjamin/github/BoomHS/main.cxx:40:23: note: candidate template
ignored: invalid explicitly-specified argument for template parameter
'ST' static auto constexpr make_impl(draw_mode const dm)

最佳答案

是的，您的尝试是允许的。我在 N4606 中找到的两个相关摘录是:

6.4.1/2 [stmt.if]

If the if statement is of the form if constexpr [...] If the value of the converted condition is false, the first substatement is a discarded statement, otherwise the second substatement, if present, is a discarded statement.

在 constexpr if 中指示一个未采用的分支是一个废弃的语句。此外，自动函数仅考虑非丢弃的返回语句来推导返回类型

7.1.7.4/2 [dcl.spec.auto](强调我的)

[...] If the declared return type of the function contains a placeholder type, the return type of the function is deduced from non-discarded return statements, if any, in the body of the function (6.4.1).

简化后，以下代码适用于 gcc和 clang头。

namespace {

enum class shape_type { TRIANGLE, RECTANGLE, CIRCLE};

template<shape_type>
struct shape { };

using triangle = shape<shape_type::TRIANGLE>;
using rectangle = shape<shape_type::RECTANGLE>;
using circle = shape<shape_type::CIRCLE>;

template<shape_type ST>
constexpr auto make() {
  if constexpr (ST == shape_type::TRIANGLE) {
    return triangle{};
  } else if constexpr (ST == shape_type::RECTANGLE) {
    return rectangle{};
  } else if constexpr (ST == shape_type::CIRCLE) {
    return circle{};
  } 
}

}

int main() {
  auto t = make<shape_type::TRIANGLE>();
  auto r = make<shape_type::RECTANGLE>();
  auto c = make<shape_type::CIRCLE>();
}

关于c++ - constexpr-if-else 主体能否在 constexpr auto 函数中返回不同类型？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/41273101/