c++ - 在 constexpr 函数中切换

Question

在Wiki中找到如下语句:

C++11 introduced the concept of a constexpr-declared function; a function which could be executed at compile time. Their return values could be consumed by operations that require constant expressions, such as an integer template argument. However, C++11 constexpr functions could only contain a single expression that is returned (as well as static_asserts and a small number of other declarations).

C++14 relaxes these restrictions. Constexpr-declared functions may now contain the following: The conditional

• ...
• branching statements if and switch

那么，在 c++14/c++17 的 constexpr 函数中是否真的可以有一个开关？而且，如果可能的话，它的语法是什么？例如，我想要这样的东西:

enum class Terrain : std::uintmax_t {
    ROAD,
    SOIL,
    GRASS,
    MUD,
    SNOW,
};

constexpr float
getStepPrice(Terrain const& terrain)
{
    switch constexpr (terrain)
    {
        case Terrain::ROAD: return 1.0f;
        ...
    }
}

Answer 1

不完全是。对于 if constexpr，您可以放心，生成的代码没有分支。此外，丢弃的语句不需要编译。这些是我认为您期望从真正的 switch constexpr 获得的保证。

class Dog;
class Snake;

#define USE_IF

template<typename Pet>
constexpr void foo(Pet pet) {
#ifdef USE_IF
    // This works
    if constexpr(std::is_same_v<Pet, Dog>)   pet.bark();
    else                                     pet.slither();
#else
    // This doesn't
    switch (std::is_same_v<Pet, Dog>) {
        case true:  pet.bark();    break;  // <== Error if Snake
        case false: pet.slither(); break;  // <== Error if Dog
    }
#else
}

顺便说一句，对于 Baum 接受的答案，我有点挑剔，但对于实际目的来说这很好。我怀疑您会发现好的编译器会从带有 constexpr 函数的 switch-case 语句中消除逻辑上不可能的位，甚至是非 constexpr 内联函数。