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How to get friendly format for GetType().Name

Pornim de la urmatorul exemplu:
List<string> list = new List<string>();
Console.WriteLine(list.GetType().Name);
Asa cum ne-am astepta ne este returnat "List`1". Dar uneori nu ne este de ajuns atata, vrem sa vedem si tipul generci. Ceva de genul acesta "List". Pentru acest lucru avem nevoie sa apelam metoda GetGenericArguments pentru a obtine tipuriile generice.
Mai jos gasiti un extension method la Type care returneaza numele la type asa cum l-am prezentat mai sus.
public static class TypeExtensions
{
private const string TypeSeparator=",";
private const char UnusedSymbol = '`';
private const string TypeFormat = "{0}<{1}>";

public static string ToGenericTypeString(this Type type)
{
if (!type.IsGenericType)
{
return type.Name;
}

return string.Format(TypeFormat,
GetGenericTypeName(type),
GetGenericArgumentsName(type));
}

private static string GetGenericArgumentsName(Type type)
{
return string.Join(
TypeSeparator,
type.GetGenericArguments()
.Select(ToGenericTypeString).ToArray());
}

private static string GetGenericTypeName(Type type)
{
string typeName =type.GetGenericTypeDefinition().Name;
return typeName.Substring(0, typeName.IndexOf(UnusedSymbol));
}
}
Enjoy!

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