Qfyilyi

In C++14 given the following code:

void foo() {

  double d = 5.0;

  auto p1 = new int[d];

}

clang compiles this without diagnostic while gcc on the other hand produces the following diagnostic (see it live):

error: expression in new-declarator must have integral or enumeration type

    7 |     auto p1 = new int[d];

       |            

I specifically labeled this C++14 because in C++11 mode clang treats this as ill-formed and produces the following diagnostic (see it live):

error: array size expression must have integral or unscoped enumeration type, not 'double'

    auto p1 = new int[d];

                     ^           ~

Is clang correct? If so what changed in C++14 to allow this?

clang is correct, the key wording in [expr.new]p6 changes from this in the C++11 draft:

Every constant-expression in a noptr-new-declarator shall be an integral constant expression ([expr.const]) and evaluate to a strictly positive value. The expression in a noptr-new-declarator shall be of integral type, unscoped enumeration type, or a class type for which a single non-explicit conversion function to integral or unscoped enumeration type exists ([class.conv]). If the expression is of class type, the expression is converted by calling that conversion function, and the result of the conversion is used in place of the original expression. …

to this in the C++14 draft:

Every constant-expression in a noptr-new-declarator shall be a converted constant expression ([expr.const]) of type std::size_t and shall evaluate to a strictly positive value. The expression in a noptr-new-declaratoris implicitly converted to std::size_t. …

So in C++14 the requirement for the expression noptr-new-declarator was weakened to not require an integral, unscoped enumeration or a class with a
single non-explicit conversion function to one of those types but just allow implicit conversions to size_t.

This change came from the proposal A Proposal to Tweak Certain C++ Contextual Conversions, v3.