SortedList - KeyValuePair - InvalidCastException
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Then what should I do?
Equality... Equality... Equality... A is equal to B if A = B and B = A
"Run for your life from any man who tells you that money is evil. That sentence is the leper's bell of an approaching looter." --Ayn Rand
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X| Firstly, take more than a minute and consider the problem. Give it like ten.
"Run for your life from any man who tells you that money is evil. That sentence is the leper's bell of an approaching looter." --Ayn Rand
Which I have. This is the Compare code. I suspect it has something to do with the bold text. BUt I can´t put my finger on it.
static bool Compare(SortedList SL, SortedList SL2) { // Early check for difference if (SL.Count != SL2.Count) { return false; } // Compare each value in list 1 with each value in list 2 foreach (DictionaryEntry item in SL) { if (!SL.ContainsKey(item.Key)) { // Return the moment we find a difference return false; } } // Must be the same return true; } -
Then what should I do?
But it's a list. That'll bugger things up. A.items == B.items & B.items = A.items Duplicates could mess you up. pseudocode: foreach thingy in A{ make sure thingy is in B.Thingies } foreach thingy in B { make sure thingy is in A.Thingies } Help any?
"Run for your life from any man who tells you that money is evil. That sentence is the leper's bell of an approaching looter." --Ayn Rand
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Which I have. This is the Compare code. I suspect it has something to do with the bold text. BUt I can´t put my finger on it.
static bool Compare(SortedList SL, SortedList SL2) { // Early check for difference if (SL.Count != SL2.Count) { return false; } // Compare each value in list 1 with each value in list 2 foreach (DictionaryEntry item in SL) { if (!SL.ContainsKey(item.Key)) { // Return the moment we find a difference return false; } } // Must be the same return true; }Is this homework?
"Run for your life from any man who tells you that money is evil. That sentence is the leper's bell of an approaching looter." --Ayn Rand
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But it's a list. That'll bugger things up. A.items == B.items & B.items = A.items Duplicates could mess you up. pseudocode: foreach thingy in A{ make sure thingy is in B.Thingies } foreach thingy in B { make sure thingy is in A.Thingies } Help any?
"Run for your life from any man who tells you that money is evil. That sentence is the leper's bell of an approaching looter." --Ayn Rand
not really, if you could incorporate it into the Compare code I posted it may shed some light.
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Is this homework?
"Run for your life from any man who tells you that money is evil. That sentence is the leper's bell of an approaching looter." --Ayn Rand
Not really, I am taking a course in C# but this is more of a side project I´d like to finish.
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Which I have. This is the Compare code. I suspect it has something to do with the bold text. BUt I can´t put my finger on it.
static bool Compare(SortedList SL, SortedList SL2) { // Early check for difference if (SL.Count != SL2.Count) { return false; } // Compare each value in list 1 with each value in list 2 foreach (DictionaryEntry item in SL) { if (!SL.ContainsKey(item.Key)) { // Return the moment we find a difference return false; } } // Must be the same return true; }Here's a hint - you're reading from the SL sorted list, and then checking to see if the same sorted list contains the key. Change it to point to the right list:
foreach (DictionaryEntry item in SL)
{
if (!SL2.ContainsKey(item.Key))
{Deja View - the feeling that you've seen this post before.
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not really, if you could incorporate it into the Compare code I posted it may shed some light.
I'm not going to write code for you, call it tough love. Again, I'll write pseudo code for you, you're supposed to be writing it to learn. using SortedList, in this case you can go through by INDEX and check to see if the elements that life in the INDEXES match. for(0 to elements){ if a[index] != b[index], return false. } return true. Hope this helps.
"Run for your life from any man who tells you that money is evil. That sentence is the leper's bell of an approaching looter." --Ayn Rand
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I'm not going to write code for you, call it tough love. Again, I'll write pseudo code for you, you're supposed to be writing it to learn. using SortedList, in this case you can go through by INDEX and check to see if the elements that life in the INDEXES match. for(0 to elements){ if a[index] != b[index], return false. } return true. Hope this helps.
"Run for your life from any man who tells you that money is evil. That sentence is the leper's bell of an approaching looter." --Ayn Rand
what do you mean by 0 to elements?
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Here's a hint - you're reading from the SL sorted list, and then checking to see if the same sorted list contains the key. Change it to point to the right list:
foreach (DictionaryEntry item in SL)
{
if (!SL2.ContainsKey(item.Key))
{Deja View - the feeling that you've seen this post before.
No, that doesn't work.
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what do you mean by 0 to elements?
Alright.. you have two lists of KeyValuePairs. Say they're identical. List1 = [1,A],[2,B],[3,C],[4,D] List2 = [1,A],[2,B],[3,C],[4,D] When you have 2 sorted lists, you can use the index to do the check and not the foreach loop. if(List1.Count != List2.Count) return false; for(int i = 0 ; i < List1.Count ; i++){ if(List1[i] != List2[i]) return false; } return true; And you can see when you write the problem out exactly how it happens. Example 2 Inequal list lengths. List1 = [1,A],[2,B],[3,C],[4,D] List2 = [1,A],[2,B],[3,C],[4,D],[5,E] the comparison of lengths returns false. OK. Example 3 Inequal lists List1 = [1,A],[2,B],[3,C],[4,D] List2 = [1,A],[2,B],[3,C],[5,E] remember that these are sorted, so we're COUNTING on the order. List lengths, ok, continue. List1[0] == List2[0] TRUE List1[1] == List2[1] TRUE List1[2] == List2[2] TRUE List1[3] == List2[3] FALSE Lists inequal, ok. Remember that the contents of the list are important, so you have to check both the key and the value in your KeyValuePair. B.Key == A.Key && B.Value == A.Value Hope this helps some more.
"Run for your life from any man who tells you that money is evil. That sentence is the leper's bell of an approaching looter." --Ayn Rand
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Alright.. you have two lists of KeyValuePairs. Say they're identical. List1 = [1,A],[2,B],[3,C],[4,D] List2 = [1,A],[2,B],[3,C],[4,D] When you have 2 sorted lists, you can use the index to do the check and not the foreach loop. if(List1.Count != List2.Count) return false; for(int i = 0 ; i < List1.Count ; i++){ if(List1[i] != List2[i]) return false; } return true; And you can see when you write the problem out exactly how it happens. Example 2 Inequal list lengths. List1 = [1,A],[2,B],[3,C],[4,D] List2 = [1,A],[2,B],[3,C],[4,D],[5,E] the comparison of lengths returns false. OK. Example 3 Inequal lists List1 = [1,A],[2,B],[3,C],[4,D] List2 = [1,A],[2,B],[3,C],[5,E] remember that these are sorted, so we're COUNTING on the order. List lengths, ok, continue. List1[0] == List2[0] TRUE List1[1] == List2[1] TRUE List1[2] == List2[2] TRUE List1[3] == List2[3] FALSE Lists inequal, ok. Remember that the contents of the list are important, so you have to check both the key and the value in your KeyValuePair. B.Key == A.Key && B.Value == A.Value Hope this helps some more.
"Run for your life from any man who tells you that money is evil. That sentence is the leper's bell of an approaching looter." --Ayn Rand
static bool Compare(SortedList SL, SortedList SL2) { if(SL.Count != SL2.Count) return false; for(int i = 0 ; i < SL.Count ; i++) { if(SL[i] != SL2[i]) return false; } return true;Is that right?