我的印象是下面的脚注 32 适用于整个段落 §3.4.1/8 而不仅仅是它的第三个要点

Question

[basic.lookup.unqual]/8

For the members of a class X, a name used in a member function body, in a default argument, in an exceptions-pecification, in the brace-or-equal-initializer of a non-static data member (9.2), or in the definition of a class member outside of the definition of X, following the member’s declarator-id ³¹, shall be declared in one of the following ways:

before its use in the block in which it is used or in an enclosing block (6.3), or

shall be a member of class X or be a member of a base class of X (10.2), or

if X is a nested class of class Y (9.7), shall be a member of Y, or shall be a member of a base class of Y (this lookup applies in turn to Y’s enclosing classes, starting with the innermost enclosing class),³² or

...

脚注：

_{32) This lookup applies whether the member function is defined within
the definition of class X or whether the member function is defined
in a namespace scope enclosing X’s definition.}

我说得对吗？

Answer 1

我看不出有任何理由将脚注与要点 3 联系起来。脚注唯一说的是 in-class 中定义的成员函数可以 "see" 整个 class（如果它以非 class 方式定义，就会出现这种情况）。它同样适用于所有要点。

例如脚注32再次确认此代码有效

struct S
{
  void foo() { i = 42; }
  int i;
};

即S::foo() 的定义可以 "see" S::i 的声明，即使 S::i 声明在下方 S::foo() 的定义之下].您可以立即看到，上面的示例没有嵌套 classes，因此与要点 3.

无关

Answer 2

脚注不是"this statement is only true for bullet point 3"的意思。

表示"by the way, this statement is [also] true for bullet point 3, a slightly more complex scenario with which you may have had your doubts as to the fact so stated".

因此，虽然您在客观上并没有错，但这也不是任何类型的编辑问题。

我的印象是下面的脚注 32 适用于整个段落 §3.4.1/8 而不仅仅是它的第三个要点

I have the impression that footnote 32 below applies to the entire paragraph §3.4.1/8 and not only to its third bullet point

c++

language-lawyer

c++14