Talk about the difference between & and && in the judgment expression in the c#ienumerable<t> extension method?

Background

When working with frames in the process, I write the Code "T=>t.x.equals" (model) when I encounter the criteria for writing LINQ. ID) &t.y.equals (model. Name) "is similar to this form. But the results of the program error. Then I wrote the following code.

Code 1

1     class Program2     {3         Static voidMain (string[] args)4         {5             varPersons =getpersonlist ();6             varresult = persons. Where (U = u.age = =2&u.name=="B");7             foreach(varIteminchresult)8             {9 Console.WriteLine (item. ToString ());Ten             } One Console.readkey (); A         } -  -         Private StaticList<person>getpersonlist () the         { -             return NewList<person> -             { -                 Newperson {Name ="A", age =1}, +                 Newperson {Name ="B", age =2}, -                 Newperson {Name ="C", age =2} +             }; A         } at     } -  -      Public class Person -     { -          Public stringName {Get;Set; } -  in          Public intAge {Get;Set; } -  to          Public Override stringToString () +         { -             return string. Format ("Name:{0},age:{1}", Name, age); the         } *}

Run results

Code 2

1         Static voidMain (string[] args)2         {3             varPersons =getpersonlist ();4             varresult = persons. Where (U = u.age = =2&&u.name=="B");5             foreach(varIteminchresult)6             {7 Console.WriteLine (item. ToString ());8             }9 Console.readkey ();Ten}

In the original & Add a &, the same program can also run, the results are also

Questions

The ienumerable<t> extension method inside C # is the same as the bitwise operator (&) and the operator (&&)?

Talk about the difference between & and && in the judgment expression in the c#ienumerable<t> extension method?