loops - Iterate over C# dictionary's keys with index? -

how iterate on dictionary's keys while maintaining index of key. i've done merge foreach-loop local variable i gets incremented 1 every round of loop.

here's code works:

public iterateovermydict() {     int i=-1;     foreach (string key in mydict.keys)     {         i++;         console.write(i.tostring() + " : " + key);     } } 

however, seems low tech use local variable i. wondering if there's way don't have use "extra" variable? not saying bad way, there better one?

there's no such concept "the index of key". should treat dictionary<tkey, tvalue> having unpredictable order - order happen when iterating on may change. (so in theory, add 1 new entry, , entries in different order next time iterated on them. in theory happen without changing data, that's less in normal implementations.)

if really want numeric index happened observe time, use:

foreach (var x in dictionary.select((entry, index) => new { entry, index })) {     console.writeline("{0}: {1} = {2}", x.index, x.entry.key, x.entry.value); } 

... aware that's misleading display, suggests inherent ordering.

from documentation:

for purposes of enumeration, each item in dictionary treated keyvaluepair<tkey, tvalue> structure representing value , key. order in items returned undefined.

edit: if don't select call here, could create own extension method:

public struct indexedvalue<t> {     private readonly t value;     private readonly int index;      public t value { { return value; } }     public int index { { return index; } }      public indexedvalue(t value, int index)     {         this.value = value;         this.index = index;     } }  public static class extensions {     public static ienumerable<indexedvalue<t>> withindex<t>         (this ienumerable<t> source)     {         return source.select((value, index) => new indexedvalue<t>(value, index));     } } 

then loop be:

foreach (var x in dictionary.withindex()) {     console.writeline("{0}: {1} = {2}", x.index, x.value.key, x.value.value); }