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PO
primarykey
Id
20062858
data
AcceptedAnswerId
0
AnswerCount
0
ClosedDate
CommentCount
2
CommunityOwnedDate
CreationDate
2013-11-19T03:54:23.367
FavoriteCount
0
LastActivityDate
2013-11-19T04:02:13.540
LastEditDate
2013-11-19T04:02:13.540
LastEditorUserId
1322972
OwnerUserId
1322972
ParentId
20062652
PostTypeId
2
Score
1
ViewCount
0
LastEditorDisplayName
text
Body
Your code has several problems. First, you're not checking the last node. A loop condition like this: <pre><code>while (travel2->next != NULL && travel1->next != NULL) </code></pre> will break as soon as one of the enumerators reaches the last node, but never checks it. Further, it will also mean two sets of a single-node-each will always compare true. Next, You have a hard return after only the first iteration, so there is no conceivable way this ever returned false on two sets that started with the same node value. <pre><code>travel2 = travel2->next; travel1 = travel1->next; return true; // this doesn't belong here. </code></pre> Next, you pass your parameter by-value, which means the copy constructor is being invoked. I don't know whether you implemented it or not (and if you didn't, you have an entire different problem), but there is no reason to duplicate the list just to see if it is equal to <code>*this*</code>. The function should take a const-reference as the parameter. <pre><code>bool set::equalset(const set& second) </code></pre> Finally, your exit condition is correct, but you cannot assume the lists were both exhausted. you have to verify it. You can do this by returning false if either traveler is non-null (and one of them will be if the lists are uneven. Putting it all together: <pre><code>bool set::equalset(const set& second) { const Data *travel1 = top; const Data *travel2 = second.top; while (travel1 && travel2) { if (!in(travel2->value)) //in checks if the value is in the set return false; if (!second.in(travel1->value)) //same idea here return false; travel2 = travel2->next; travel1 = travel1->next; } return !(travel1 || travel2); } </code></pre> <hr> Optimized for Sorted Lists If you keep the lists sorted during input and removal methods, you can significantly make this easier, seen below: <pre><code>bool set::equalset(const set& second) { const Data *travel1 = top; const Data *travel2 = second.top; while (travel1 && travel2 && travel1->value == travel2->value) { travel1 = travel1->next; travel2 = travel2->next; } return !(travel1 || travel2); } </code></pre>
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1. This table or related slice is empty.
PostParentId
1. POTesting if a linked list is equal to each other
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PostTypePostTypeId
1. PTAnswer
UserLastEditorUserId
1. USWhozCraig
UserOwnerUserId
1. USWhozCraig
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1. POTesting if a linked list is equal to each other
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1. This table or related slice is empty.
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1. VO
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CommentsPostId
1. COThank you for explaining it very clearly, it helped a lot.
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 PostPostId
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 USuser2793442
2. CO@user2793442 no worries. glad it gave you some ideas and helped.
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 USWhozCraig

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