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evens.Add( i ); |
evens.Add( i ); |
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foreach( int i in evens ) |
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System.Console.WriteLine( i.ToString() ); |
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// .NET 2.0 solution |
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List<int> array = new List<int>( new int[] { 1, 2, 3, 4, 5 } ); |
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List<int> evens = array.FindAll( delegate( int i ) { return (i%2)==0; } ); |
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foreach( int i in evens ) |
foreach( int i in evens ) |
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System.Console.WriteLine( i.ToString() ); |
System.Console.WriteLine( i.ToString() ); |
Revision as of 13:52, 23 January 2007
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You are encouraged to solve this task according to the task description, using any language you may know.
You are encouraged to solve this task according to the task description, using any language you may know.
Select certain elements from an Array into a new Array in a generic way. E.g., select all even numbers from an Array
Ruby
ary = [1,2,3,4,5,6] even_ary = ary.select{|el|el%2==0} # => [2, 4, 6]
Perl
my @list = (1, 2, 3, 4, 5, 6); my @even = grep { 0 == $_%2; } @list;
C
int arr[5] = {1,2,3,4,5}; int *result; int memoryReqd = 0,length = sizeof(arr)/sizeof(arr[0]), i,j; for (i=0; i<length; i++) { if(0 == arr[i]%2) { memoryReqd++; } } result = (int*)malloc(memoryReqd*sizeof(int)); for(i=0,j=0; i<length; i++) { if(0 == arr[i]%2) { result[j++]=arr[i]; } }
// Test our resultant array for(i=0; i<memoryReqd; i++) { printf("%d\n",result[i]); }
C#
// .NET 1.x solution ArrayList array = new ArrayList( new int[] { 1, 2, 3, 4, 5 } ); ArrayList evens = new ArrayList(); foreach( int i in array ) { if( (i%2) == 0 ) evens.Add( i ); } foreach( int i in evens ) System.Console.WriteLine( i.ToString() );
// .NET 2.0 solution List<int> array = new List<int>( new int[] { 1, 2, 3, 4, 5 } ); List<int> evens = array.FindAll( delegate( int i ) { return (i%2)==0; } ); foreach( int i in evens ) System.Console.WriteLine( i.ToString() );