Switch in constexpr function

Not exactly. In the case of if constexpr, you can rest assured that the resulting code has no branching. Moreover, discarded statements need not compile. Those are the guarantees I think you would expect from a true switch constexpr.

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
}

BTW, I'm being a little nit-picky vis-a-vis Baum's accepted answer, which is fine, for practical puposes. I suspect you'd find good compilers will elide logically impossible bits from a switch-case statement with constexpr functions, as well as even non-constexpr inlined functions.

So, Is it actually possible to have a switch in a constexpr function in c++14/c++17?

Yes.

And, if possible, what syntax is for that?

There is absolutely nothing special about the syntax, it's just a normal switch. Like this:

constexpr int fun (int i) {
    switch(i) {
        case 0: return 7;
        default: return 5;
    }
}

int main () {
    int arr[fun(3)];
}