假设，我有这个界面，
interface IContact
{
IAddress address { get; set; }
}
interface IAddress
{
string city { get; set; }
}
class Person : IPerson
{
public IContact contact { get; set; }
}
class test
{
private test()
{
var person = new Person();
if (person.contact.address.city != null)
{
//this will never work if contact is itself null?
}
}
}
Person.Contact.Address.City != null(这用于检查 City 是否为 null。)
但是，如果 Address 或 Contact 或 Person 本身为 null，则此检查失败。
目前，我能想到的一种解决方案是:
if (Person != null && Person.Contact!=null && Person.Contact.Address!= null && Person.Contact.Address.City != null)
{
// Do some stuff here..
}
有没有更简洁的方法来做到这一点？
我真的不喜欢 null 检查作为 (something == null) 完成。相反，是否有另一种很好的方法来执行类似 something.IsNull() 方法的操作？
在一般情况下，您可以使用表达式树并使用扩展方法进行检查:
if (!person.IsNull(p => p.contact.address.city))
{
//Nothing is null
}
完整代码:
public class IsNullVisitor : ExpressionVisitor
{
public bool IsNull { get; private set; }
public object CurrentObject { get; set; }
protected override Expression VisitMember(MemberExpression node)
{
base.VisitMember(node);
if (CheckNull())
{
return node;
}
var member = (PropertyInfo)node.Member;
CurrentObject = member.GetValue(CurrentObject,null);
CheckNull();
return node;
}
private bool CheckNull()
{
if (CurrentObject == null)
{
IsNull = true;
}
return IsNull;
}
}
public static class Helper
{
public static bool IsNull<T>(this T root,Expression<Func<T, object>> getter)
{
var visitor = new IsNullVisitor();
visitor.CurrentObject = root;
visitor.Visit(getter);
return visitor.IsNull;
}
}
class Program
{
static void Main(string[] args)
{
Person nullPerson = null;
var isNull_0 = nullPerson.IsNull(p => p.contact.address.city);
var isNull_1 = new Person().IsNull(p => p.contact.address.city);
var isNull_2 = new Person { contact = new Contact() }.IsNull(p => p.contact.address.city);
var isNull_3 = new Person { contact = new Contact { address = new Address() } }.IsNull(p => p.contact.address.city);
var notnull = new Person { contact = new Contact { address = new Address { city = "LONDON" } } }.IsNull(p => p.contact.address.city);
}
}