Get user-friendly name for generic type in C#

Question

Is there an easy way without writing a recursive method which will give a \'user friendly\' name for a generic type from the Type class?

E.g. For the fo

Answer 1

Building a more complete answer off of Kirk's would look like this. Modifications:

• Support for all C# keywords
• Supports custom-translations
• Arrays
• Nullables being ValueType? instead of Nullable

Here's full code:

public static class TypeTranslator
{
    private static Dictionary _defaultDictionary = new Dictionary
    {
        {typeof(int), "int"},
        {typeof(uint), "uint"},
        {typeof(long), "long"},
        {typeof(ulong), "ulong"},
        {typeof(short), "short"},
        {typeof(ushort), "ushort"},
        {typeof(byte), "byte"},
        {typeof(sbyte), "sbyte"},
        {typeof(bool), "bool"},
        {typeof(float), "float"},
        {typeof(double), "double"},
        {typeof(decimal), "decimal"},
        {typeof(char), "char"},
        {typeof(string), "string"},
        {typeof(object), "object"},
        {typeof(void), "void"}
    };

    public static string GetFriendlyName(this Type type, Dictionary translations)
    {
        if(translations.ContainsKey(type))
            return translations[type];
        else if (type.IsArray)
        {
            var rank = type.GetArrayRank();
            var commas = rank > 1 
                ? new string(',', rank - 1)
                : "";
            return GetFriendlyName(type.GetElementType(), translations) + $"[{commas}]";
        }
        else if(type.IsGenericType && type.GetGenericTypeDefinition() == typeof(Nullable<>))
            return type.GetGenericArguments()[0].GetFriendlyName() + "?";
        else if (type.IsGenericType)
            return type.Name.Split('`')[0] + "<" + string.Join(", ", type.GetGenericArguments().Select(x => GetFriendlyName(x)).ToArray()) + ">";
        else
            return type.Name;
    }

    public static string GetFriendlyName(this Type type)
    {
        return type.GetFriendlyName(_defaultDictionary);
    }
}