重载运算符“ =”匿名访问类型？ [英] Overloading operator "=" for anonymous access types?

问题描述

我正在研究巴恩斯的出色Ada书。这是一个用于对第11.7节中的链表进行深度比较的代码示例：

I am working my way through Barnes' excellent Ada book. This is a code sample for deep comparison of linked lists from section 11.7:

type Cell is
  record
    Next: access Cell;
    Value: Integer;
  end record;

function "=" (L, R: access Cell) return Boolean is
begin
  if L = null or R = null then    -- universal =
    return L = R;                 -- universal = (Line A)
  elsif L.Value = R.Value then
    return L.Next = R.Next;       -- recurses OK (Line B)
  else
    return False;
  end if;
end "=";

我似乎无法绕开A行运算符 =的原因。 （由于使用了优先规则），被称为Universal_access类型的 ，但是，调用了用户定义的运算符 = （首先使得递归成为可能） ），这一次不希望使用Universal_access的运算符 =。

I can't seem to wrap my head around why in Line A operator "=" of the universal_access type is called (because of the preference rule), on Line B, however, the user-defined operator "=" is called (which makes recursion possible in the first place), this time with no preference for operator "=" of universal_access.

L和R以及L.Next和R.Next都具有相同的匿名性输入访问单元。为什么派遣有所不同？它与L和R是访问参数有关吗？如果是这样，那里的规则是什么？

Both L and R, as well as L.Next and R.Next are of the same anonymous type "access Cell". Why the difference in "dispatching"? Does it have to do with L and R being access parameters? If so, what is the rule there?

我竭尽全力在AARM中找到任何内容，尤其是第4.5.2节，但没有任何意义。

I did my best to find anything in the AARM, especially section 4.5.2, but could not make any sense of it.

干杯。

推荐答案

我将总结我的发现（与到目前为止，还是Simon Wright和G_Zeus的帮助）。如果我错了，请纠正我：

I will summarize my findings (with the help of Simon Wright and G_Zeus) as of now. Please correct me if I'm wrong:

根据标准， L = null ， R = null ， L = R 以及 L.Next = R.Next 每个都应明确调用用户定义的运算符=。 universal_access 运算符=绝对不能插入。

According to the standard, L = null, R = null, L = R as well as L.Next = R.Next should each unambiguously call the user-defined operator =. universal_access operator = must not kick in at all here.

原因：

操作数 L ， R ， L.Next 和 R.Next 违反了 ARM 4.5.2（9.1-9.4）用于解释其中的 = 表示 universal_access 类型的运算符=的表达式：

The operands L, R, L.Next and R.Next violate the precondition in ARM 4.5.2(9.1-9.4) for interpreting = in these expressions as to mean operator = of the universal_access type:

前提条件是两个操作数都不是对象类型（ access Cell ）的访问类型，其指定类型为 Cell （ check ）， Cell 具有用户定义的原始等式运算符（ check ），使得

The precondition says that neither of the operands are of an access-to-object type (access Cell) whose designated type is Cell (check), Cell has a user-defined primitive equality operator (check) such that

• 其结果类型为 Boolean （ check ）;


• 立即在与 Cell （ check ）相同的声明列表中声明；并且


• 至少一个操作数是具有指定类型 Cell 的访问参数（两个操作数均为 check ）。

• its result type is Boolean (check);
• it is declared immediately within the same declaration list as Cell (check); and
• at least one of its operands is an access parameter with designated type Cell (both operands are, check).

operator =的 universal_access 类型的优先规则在 ARM 8.6（29.1）不适用于此处，因为它需要两个可接受的解释。但是由于4.5.2， universal_access 类型的operator =不能被接受。

The preference rule for operator = of the universal_access type in ARM 8.6(29.1) does not apply here, since it requires "two acceptable interpretations". But because of 4.5.2, operator = of the universal_access type is not an acceptable interpretation.

别无选择：在所有情况下（甚至 L = null ），它都必须是用户定义的运算符=。

So there is no choice: in all cases (even L = null) it has to be the user-defined operator =.

@Simon Wright：所以无界递归实际上是正确的编译器行为。

@Simon Wright: So "unbounded recursion" is actually the correct compiler behavior.

@G_Zeus：为 l = r <发出歧义错误/ code>是错误的编译器行为，编译器应选择 Access_Equal。 = 。

@G_Zeus: Issuing an ambiguity error for l = r is incorrect compiler behavior, the compiler should have picked Access_Equal."=".

该示例应正确读取：

...

  if Standard."="(L, null) or Standard."="(R, null) then    -- universal =
    return Standard."="(L, R);                              -- universal =
  elsif L.Value = R.Value then
    return L.Next = R.Next;                                 -- recurses OK

...

干杯。

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