c++ - 在 C++14 中，在新表达式的维度中使用 double 值是否有效？

Question

在 C++14 中给出以下代码:

void foo() {
  double d = 5.0;
  auto p1 = new int[d];
}

clang 在没有诊断的情况下编译它，而 gcc 产生以下诊断 (see it live in godbolt):

error: expression in new-declarator must have integral or enumeration type
    7 |     auto p1 = new int[d];
      |                       ^

我特别标记了这个 C++14，因为在 C++11 模式下，clang 将其视为格式错误并产生以下诊断 (see it live in godbolt):

error: array size expression must have integral or unscoped enumeration type, not 'double'
    auto p1 = new int[d];
              ^       ~

clang 正确吗？如果是这样，C++14 中发生了什么变化以允许这样做？

Answer 1

Clang 是正确的，[expr.new]p6 中的关键词与 C++11 草案中的以下内容不同:

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. …

到 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-declarator is implicitly converted to std::size_t. …

在 C++14 中，对 noptr-new-declarator 中的表达式的要求被削弱为不需要完整的、无作用域的枚举或具有单个非显式转换函数到其中一种类型，但只允许隐式转换到 size_t。

措辞的变化来自提案A Proposal to Tweak Certain C++ Contextual Conversions, v3 .