Question

我似乎在这里有一个盒子里的问题在给定键和散列表的情况下，以下代码应返回与键对应的值，如果映射中不存在该键，则返回错误：

(defun get-graph-node (key graph)
  (let ((result (gethash key graph)))
    (if (nth-value 1 result)
      (nth-value 0 result)
      (error "no node matches the key"))))

在大多数情况下它都有，但我有这种奇怪的情况：跑步：

(gethash 0 *g*)

返回

   #S(GRAPH-NODE$
      :DATA "("$
      :EDGES (#S(GRAPH-NODE :DATA "b" :EDGES NIL)$
              #S(GRAPH-NODE :DATA "a" :EDGES NIL)))
   T

但是

(get-graph-node 0 *g*)

表示get-graph-node

中定义的错误

检查* g *给了我这个：

Count: 5
Size: 16
Test: EQL
Rehash size: 1.5
Rehash threshold: 1.0
[clear hashtable]
Contents: 
0 = #S(GRAPH-NODE :DATA "(" :EDGES (#S(GRAPH-NODE :DATA "b" :EDGES NIL) #S(GRAPH-NODE :DATA "a" :EDGES NIL))) 
[remove entry]
1 = #S(GRAPH-NODE :DATA "a" :EDGES NIL) [remove entry]
2 = #S(GRAPH-NODE :DATA "|" :EDGES (#S(GRAPH-NODE :DATA ")" :EDGES (NIL)))) [remove entry]
3 = #S(GRAPH-NODE :DATA "b" :EDGES NIL) [remove entry]
4 = #S(GRAPH-NODE :DATA ")" :EDGES (NIL)) [remove entry]

所以键0应该在地图中？对我可以告诉我自己失踪的绅士来说，这是我的一顶帽子。

Answer 1

将gethash的结果赋给变量result时，只存储函数返回的多个值中的第一个。要存储它们，你应该这样做：

(defun get-graph-node (key graph)
  (multiple-value-bind (result exists)
      (gethash key graph)
    (if exists
      result
      (error "no node matches the key"))))

常见的lisp（多个）值是真还是假？

1 个答案: