Difference between const and constexpr arrays

Question

Why is there a difference between const and constexpr when used with arrays?

int const xs[]{1, 2, 3};
constexpr int ys[]{1, 2, 3};

int as[xs[0]];  // error.
int bs[ys[0]];  // fine.

I would expect both xs[0] and ys[0] to be constant expressions but only the latter is treated as such.

Answer 1

A longer comment as community wiki.

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The expression xs[0] is defined in [expr.sub]/1 as *((xs)+(0)). (See below for the parantheses.)

One of the expressions shall have the type “pointer to T” and the other shall have unscoped enumeration or integral type.

Therefore, the array-to-pointer conversion [conv.array] is applied:

An lvalue or rvalue of type “array of N T” or “array of unknown bound of T” can be converted to a prvalue of type “pointer to T”. The result is a pointer to the first element of the array.

Note it can operate on an lvalue and the result is a prvalue, 0 as an integer literal is a prvalue as well. The addition is defined in [expr.add]/5. As both are prvalues, no lvalue-to-rvalue conversion is required.

int arr[3];
constexpr int* p = arr;  // allowed and compiles

The crucial step now seems to be the indirection * [expr.unary.op]/1

The unary * operator performs indirection: the expression to which it is applied shall be a pointer to an object type, or a pointer to a function type and the result is an lvalue referring to the object or function to which the expression points.

So, the result of xs[0] is an lvalue referring to the first element of the xs array, and is of type int const.

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N.B. [expr.prim.general]/6

A parenthesized expression is a primary expression whose type and value are identical to those of the enclosed expression. The presence of parentheses does not affect whether the expression is an lvalue.

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If we now look at the bullets in [expr.const]/2 which disallow certain expressions and conversions to appear in constant expressions, the only bullet that could apply (AFAIK) is the lvalue-to-rvalue conversion:

• an lvalue-to-rvalue conversion (4.1) unless it is applied to

  • a non-volatile glvalue of integral or enumeration type that refers to a non-volatile const object with a preceding initialization, initialized with a constant expression [Note: a string literal (2.14.5) corresponds to an array of such objects. —end note ], or

  • a non-volatile glvalue of literal type that refers to a non-volatile object defined with constexpr, or that refers to a sub-object of such an object, or

[...]

But the only true lvalue-to-rvalue conversion as per (4.1) (not 4.2, which is array-to-pointer) that appears in the evaluation of xs[0] is the conversion from the resulting lvalue referring to the first element.

For the example in the OP:

int const xs[]{1, 2, 3};
int as[xs[0]];  // error.

This element xs[0] has non-volatile const integral type, its initialization precedes the constant expression where it occurs, and it has been initialized with a constant expression.

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By the way, the added "Note" in the quoted passage of [expr.const]/2 has been added to clarify that this is legal:

constexpr char c = "hello"[0];

Note that a string literal is an lvalue as well.

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It would be great if someone (could change this to) explain why xs[0] is not allowed to appear in a constant expression.