I have the following code:
return "[Inserted new " + typeof(T).ToString() + "]";
But
typeof(T).ToString()
returns the full name including namespace
Is there anyway to just get the class name (without any namespace qualifiers?)
Try this to get type parameters for generic types:
public static string CSharpName(this Type type)
{
var sb = new StringBuilder();
var name = type.Name;
if (!type.IsGenericType) return name;
sb.Append(name.Substring(0, name.IndexOf('`')));
sb.Append("<");
sb.Append(string.Join(", ", type.GetGenericArguments()
.Select(t => t.CSharpName())));
sb.Append(">");
return sb.ToString();
}
Maybe not the best solution (due to the recursion), but it works. Outputs look like:
Dictionary<String, Object>
make use of (Type Properties)
Name Gets the name of the current member. (Inherited from MemberInfo.)
Example : typeof(T).Name;
After the C# 6.0 (including) you can use nameof expression:
using Stuff = Some.Cool.Functionality
class C {
static int Method1 (string x, int y) {}
static int Method1 (string x, string y) {}
int Method2 (int z) {}
string f<T>() => nameof(T);
}
var c = new C()
nameof(C) -> "C"
nameof(C.Method1) -> "Method1"
nameof(C.Method2) -> "Method2"
nameof(c.Method1) -> "Method1"
nameof(c.Method2) -> "Method2"
nameof(z) -> "z" // inside of Method2 ok, inside Method1 is a compiler error
nameof(Stuff) = "Stuff"
nameof(T) -> "T" // works inside of method but not in attributes on the method
nameof(f) -> “f”
nameof(f<T>) -> syntax error
nameof(f<>) -> syntax error
nameof(Method2()) -> error “This expression does not have a name”
Note! nameof
not get the underlying object's runtime Type, it is just the compile-time argument. If a method accepts an IEnumerable then nameof simply returns "IEnumerable", whereas the actual object could be "List".
string1 + anything.ToString() + string2
is redundant. The compiler inserts the call toToString
automatically if you dostring1 + anything + string2
. – Tim RobinsonType
instance (as returned bytypeof(..)
) I'm pretty sure you'd figure out this yourself... – Peter LillevoldName
property is missing from the documentation - at least, it's not where I was looking for it. – Michael Kay