Why is calling a constexpr function with a member array not a constant expression?

问题

I have the following helper function:

template<typename T, std::size_t N>
constexpr std::size_t Length(const T(&)[N]) {
    return N;
}

Which returns the length of a static array. In the past this always has worked but when I do this:

struct Foo
{
    unsigned int temp1[3];
    void Bar()
    {
        constexpr std::size_t t = Length(temp1); // Error here
    }
};

I get an error when using MSVS 2017:

error C2131: expression did not evaluate to a constant

note: failure was caused by a read of a variable outside its lifetime

note: see usage of 'this'

I was hoping someone can shed light on what I'm doing wrong.

回答1:

MSVC is correct. Length(temp1) is not a constant expression. From [expr.const]p2

An expression e is a core constant expression unless the evaluation of e, following the rules of the abstract machine, would evaluate one of the following expressions:

this, except in a constexpr function or a constexpr constructor that is being evaluated as part of e;

temp1 evaluates this implicitly (because you are referring to this->temp1), and so you don't have a constant expression. gcc and clang accept it because they support VLAs as an extension (try compiling with -Werror=vla or -pedantic-errors).

Why isn't this allowed? Well, you could access the underlying elements and potentially modify them. This is completely fine if you are dealing with a constexpr array or an array that is being evaluated as a constant expression, but if you are not, then you cannot possibly have a constant expression as you will be manipulating values that are set at run time.

回答2:

Length(decltype(temp1){})

seems to work.

Unfortunately, I cannot comment, but Mehrdad 's solution is wrong. The reason: it is not technically undefined behavior but it is undefined behavior. During constexpr evaluation, the compiler must catch undefined behavior. Therefore, the code is ill-formed.

回答3:

Your question's already been answered, but in terms of how to "fix" it, a quick-and-dirty way is to replace

Length(temp1)

with

Length(*(true ? NULL : &temp1))

which I think is technically undefined behavior but practically going to work fine for MSVC.

If you need a solution that works despite the UB, you can change Length to use a pointer:

template<typename T, std::size_t N>
constexpr std::size_t Length(const T(*)[N]) {
    return N;
}

and then you can use Length(true ? NULL : &temp1).

来源：https://stackoverflow.com/questions/50917360/why-is-calling-a-constexpr-function-with-a-member-array-not-a-constant-expressio

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c++

templates

constexpr