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You can see how to use downto in this example: // a Shuson Kato hiaku array (backwards) let shusonKato = [| "watching"; "been"; "have"; "children"; "three"; "my"; "realize"; "and"; "ant"; "an"; "kill"; "I"; |] // loop over the array backwards printing each word for index = Arraylength shusonKato - 1 downto 0 do printf "%s " shusonKato[index] When you compile and execute this code, you get the following results: I kill an ant and realize my three children have been watching The while loop is another familiar imperative language construct It is an expression that creates a loop over a section of code until a Boolean expression changes to false To create a while loop in F#, you use the keyword while followed by a Boolean expression that determines whether the loop should continue.

This method examines a data type and returns True if the type converter can make the conversion from this data type to the custom data type. This method performs the conversion from the supplied data type to the custom data type. This method examines a data type and returns True if the type converter can make the conversion from the custom object to this data type. This method performs the conversion from the custom data type to the requested data type.

The bolded code represents the customer that exists in both lists. But what if the lists were like the following ones List<Customer> customers1 = new List<Customer>() { new Customer { Identifier = "Person 1", Points = 0 },

ConvertFrom() CanConvertTo() ConvertTo()

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As with for loops, you place the body of the loop between the keywords do and done, and the body should have the type unit; otherwise, the compiler will issue a warning This code illustrates how to create a while loop: // a Matsuo Basho hiaku in a list reference let matsuoBasho = ref [ "An"; "old"; "pond!"; "A"; "frog"; "jumps"; "in-"; "The"; "sound"; "of"; "water" ].

Remember, the key task of a type converter is to convert from your custom data type and a string representation. This example uses a string representation that includes all three values from the Gradient object separated by commas and spaces: ColorA, ColorB, LinearGradientMode Here s an example: LightBlue, Purple, ForwardDiagonal Before attempting a conversion from a GradientFill object to a string, the Properties window will first query the CanConvertTo() method. If it receives a True value, it will call the actual ConvertTo() method. All the CanConvertTo() method needs to do is check that the target type is a string. Public Overrides Function CanConvertTo(ByVal context As ITypeDescriptorContext, _ ByVal destinationType As Type) As Boolean If destinationType Is GetType(String) Then Return True Else return MyBase.CanConvertTo(context, destinationType) End If End Function Notice that if the target type isn t recognized, the type converter calls the base class implementation, which will convert it, pass it to another class higher up the inheritance chain, or throw an error. The ConvertTo() method is called if CanCovertTo() returns True. ConvertTo() simply checks that it can indeed convert to the desired type. If not, it calls the base class implementation of the ConvertTo() method (because presumably it was the base class that returned True from the CanCovertTo() method). If ConvertTo() is asked to convert a GradientFill into a string, it goes ahead by calling the implementation in the ConvertToString() method.

new Customer { Identifier = "Person 2", Points = 10 } }; List<Customer> customers2 = new List<Customer>() { new Customer { Identifier = "Person 3", Points = 20 }, new Customer { Identifier = "Person 2", Points = 20 } }; Now even though Person 2 exists in two places, the values for Points are not identical, and it begs the question, Should Equals() and GetHashCode() return the state that the objects are not identical The answer lies in your application. If your code says that two customers with the same Identifier and different Points represent the same person, then Equals() and GetHashCode() must not take into account the value of Points. You need to consider this, because if you are going to call the method Distinct(), and you take into account the value of the Points property, there will be two instances of the same person. To get a list of all unique customers, you can use Union(), as follows: var uniqueCustomers = customers1.Union(customers2); Contained within the list represented by the variable uniqueCustomers will be the three customers of the two lists.

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