CODEWITHSUNDEEP
c++language-lawyer
Loreto
Loreto

Reputation: 694

Variable of reference type in constant expressions

Here, at http://eel.is/c++draft/expr.const#2.7 it can be read

3 - A variable is usable in constant expressions after its initializing declaration is encountered if it is a constexpr variable, or it is of reference type or of const-qualified integral or enumeration type, and its initializer is a constant initializer.

Well, the case is that I cannot figure myself an example for the case where the variable is of reference type.

In

int main()
{
    static const double& ds = 0.5;
    constexpr double cds = ds;
}

Visual Studio 16.1.2 complains:

expression must have a constant value.

the value of variable (declared at line 11) cannot be used as a constant

But, why?

Upvotes: 1

Views: 1150

Answers (1)

M.M
M.M

Reputation: 141618

Your code is not legal in C++17 because it contains a constexpr variable whose initialization requests lvalue-to-rvalue conversion and it is not one of the listed exceptions: (C+17 [expr.const]/2.7)

an lvalue-to-rvalue conversion (7.1) unless it is applied to

  • a non-volatile glvalue of integral or enumeration type that refers to a complete non-volatile const object with a preceding initialization, initialized with a constant expression, or
  • a non-volatile glvalue that refers to a subobject of a string literal, or
  • a non-volatile glvalue that refers to a non-volatile object defined with constexpr , or that refers to a non-mutable subobject of such an object, or
  • a non-volatile glvalue of literal type that refers to a non-volatile object whose lifetime began within the evaluation of e;

You should find the code compiles if you change to an integer type instead of double.

The text you quote is from a draft and as such, may or may not eventually form part of some standard.

Upvotes: 2

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