String Comparison And Alphabetic Order of Individual Characters

Question

I have a question related to string comparison vs. character comparison.

Characters > and 0 (zero) have following decimal values 62

Answer 1

it returns -1 because it is comparing str2 to str1, not the other way around. Eg, "is 48 equal to 62". No, it's less than 62 so it returns -1. It's semantically a little confusing when you read the parameter order

Answer 2

The sort order of strings depends on the culture you use.

StringComparer.CurrentCulture sorts the following 1-character strings as follows on my machine:

' -   ! " # $ % & (  ) * , . / : ; ? @ [
\ ] ^ _ ` { | } ~ +  < = > 0 1 2 3 4 5 6
7 8 9 a A b B c C d  D e E f F g G h H i
I j J k K l L m M n  N o O p P q Q r R s
S t T u U v V w W x  X y Y z Z

StringComparer.Ordinal sorts the same strings as follows:

  ! " # $ % & ' ( )  * + , - . / 0 1 2 3
4 5 6 7 8 9 : ; < =  > ? @ A B C D E F G
H I J K L M N O P Q  R S T U V W X Y Z [
\ ] ^ _ ` a b c d e  f g h i j k l m n o
p q r s t u v w x y  z { | } ~

Answer 3

When you compare the characters '>' and '0', you are comparing their ordinal values.

To get the same behaviour from a string comparison, supply the ordinal string comparison type:

  Console.WriteLine(string.Compare(">", "0", StringComparison.Ordinal));
  Console.WriteLine(string.Compare(">", "0", StringComparison.InvariantCulture));
  Console.WriteLine(string.Compare(">", "0", StringComparison.CurrentCulture));

The current culture is used by default, which has a sorting order intended to sort strings 'alphabetically' rather in strictly lexical order, for some definition of alphabetically.

Answer 4

It sounds like what you want is the comparison to not use culture-specific rules. Have you tried StringComparison.Ordinal:

Console.WriteLine( string.Compare( ">", "0", StringComparison.Ordinal ) ); // returns a positive number