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

Question

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.

Answer 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.