How to read from dictionary?

Great job Vulpes, that's so nice of you. I must say that you didn't a brilliant job. I know you might have worked very hard for this. Because of weekend i couldn't try this out even though you replied on the same day. Today i worked out. It works with the small modifications i made on that in order to suit my application. Thank's a lot Vulpes. Really very much thanks. I mean it.

It has become too typical i guess. No if the string doesn't match with any other string it's called left over irrespective of index in any list. If the string matches with any other string in any list that is called as common string irrespective of index and list.
From the example you stated
1: Common strings are:
dog, cat, duck

2: left over strings are:
lion, tiger swan

Now it will clear for i hope.

No vulpes i don't wanna confuse you much but i still need to encounter each and every common string as i need to generate hexadecimal value for each and every occurrence of a common string also but in a different manner for common strings when compared to other two cases. Sorry for troubling you much.

Just a clarification on point 3.

So, if common strings can occur in the same list, am I right in thinking that a common string doesn't have to appear in every list?

To take an extreme example, if a string only appeared twice in the same list but in no other list, then it would still be regarded as common?

The answers are as follows:

1. I will write a for each loop to generate hexadecimal value after getting decimal value of the same. The for each loop automatically returns 0 as it doesn't contain any characters. Hence i will 0x00 to external file by appending 0x to the string.

2. Ya all the occurrences of empty strings needs to be written to the file

3. The common strings need not be in same index in all the lists to treat that as common strings. In can be any index and in any list even in the same list.

4. Only the hexadecimal value is to be written to the external file.

And the format in which i want to write in external file is as follows
1. If it's an empty string i will write
0x00     for all empty strings
2. If it's a common string i will write as
"no.of bytes occupied by current string, each character hex value separated by comma"
3. and if it's neither empty string nor it didn't match with any other string
"no.of bytes occupied by current string, each character hex value separated by comma"

I guess i answered everything you asked.

Hi Vulpes,

This has become a pretty long discussion which i haven't expected. I thought that if i explain everything you might get confused, hence i explained some part of my project. So issues are coming one after the other. Now what i intended to do completely in my project is as follows:

I have 17 languages(Keys) and each 1575 list of strings(Values) associated for each language. What i need to do exactly with this is:

1. first i need to look for empty Values in all the lists and then if there are any empty strings i should generate hexadecimal value for the same.
2. Second i need to look for common strings i.e common Values in all the lists and generate hexadecimal value for the same.
3. For the left over strings in all the lists i should generate hexadecimal value in the same way.

Note: in the 2nd point which i stated, for all the common strings i should generate hexadecimal value only once.
And i am using stream writer to write this hexadecimal values to output files.
Last post of your's wasn't looking for common strings in the same language list rather it was looking only in other lists.

OK I will check out and let you know the update.

OK, this code should do it:

  if(language_string_table->Values->Count > 1)
  { 
    for each(String ^key in language_string_table->Keys) 
    {
       for(int v = 0; v < 1500; v++)
       {
          bool areEqual = true;
          String ^next = language_string_table[key][v]; // next value to compare with other lists 

          for(int i = 0; i < 1500; i++)
          {
            for each(String ^k in language_string_table->Keys) 
            {
               if (key == k) continue; // only compare with other lists
               String ^value = language_string_table[k][i];
               if (value != next)
               {
                  areEqual = false;
                  break;
               }
            }
            if(!areEqual) break; // EDIT
          }
       
          if(areEqual)
          {
            // do something
          }
          else
          {
            // do something else
          }          
       }
    }
  }

i want to do this for each and every value of each key.?

Ya

"Are you perhaps saying that you want to test the first element of the first list for equality with all the elements of all the other lists, the second element of the first list for equality with the same set of elements and so on?"

This is want i want to do....

Hi Vulpes,

There is a small issue in the post which i accepted as answer.
The issue is that, the control doesn't come to this place more than once
first = language_string_table[key][0];
hence string doesn't get updated. So all the time the only 1st list 1st element will be compared to all other elements in all other lists. What i was expecting is i should compare 1st list 1st element with all other elements and if it matches then do some task then i should take the 2nd element in the 1st list and compare with all other elements and if it matches then do the same task which i did previously similarly for all the elements in all  the lists. Can you help me out.?

Hi Suthish Nair,

Ya i liked the way Vulpes answers all of my questions. Hence so many likes.
Appreciation makes people work more efficiently, what i feel.

If you wanted to compare all the values in every list in the dictionary for equality, you could do it like this:

  bool areEqual = true;

  if(language_string_table->Values->Count > 1)
  {

    // get the first value from the dictionary
    String ^ first;
    for each(String ^key in language_string_table->Keys) 
    {
       first = language_string_table[key][0];
       break;
    }

    for(int i = 0; i < 1500; i++)
    {
       for each(String ^key in language_string_table->Keys) 
       {
           String ^value = language_string_table[key][i];
           if (value != first)
           {
              areEqual = false;
              break;
           }
       }
    }  
  }

  if(areEqual)
  {
     // do something
  }
  else
  {
     // do something else
  }

It depends on what you mean by 'compare' but you could read the strings at the same index in each list into a temporary array for further processing as follows: